But it's a very valid question! Most of maths has had these seemingly arbitrary choices in the past (0! and 0^0 are some others) and we choose the option that gives us the most useful tools to work with. In this case the primes are more useful to us without 1 and that also gives us this theorem which is useful in itself.
@@piderman871 Yeah, but that's fine. Math is not something that is hardcoded into reality and we have to obey it by any means. It's a language and we can agree on the words which go into it. The universe has no concept of prime numbers; it doesn't care whether we consider 1 a prime number or not. It has, however, a concept of unique representations of positive integers as products of members of a specific set of numbers. It's up to us if we want to call this set P or P\{1}. In this sense the theorem is more fundamental than the notion.
@@iwatchyoutube544 Math is our made up language for describing nature and the universe, we need everyone on earth to speak the same language without variations in order to seemlessly explain physical phenomina to every human on earth. What he means is, just like with any other language it isn't set in stone, we add and remove words from our dictionaries all the time. In math we do the same except we need some proof or some new discovery before we add or remove parts of that language
But Pluto got accepted into a new, more fitting family where he's not alone but has a bunch of new friends! 1 on the other hand is really alone. Poor 1.
So I ordered this one product online and when the order arrived, there was nothing in it. So I called them to complain for having sent me an empty product when I had expressly ordered one. They replied that empty product equals one. Damn mathematicians!
@@harmenbreedeveld8026 I thought you "love the argument" so I thought to start the argument with the joke. Is working? Ok, I see, no like the joke? Ok, I tell another. What is the difference between vacation spots Dubai and Abu Dahbi? You don't know, because you infidel, so I tell you. The people of Dubai do not like the Flintstones. But the people of Abu Dahbi do.
That's just an idiomatic coincidence, in Spanish "Prime" is "Primo" which also means "to have come on top of a conflict or situation or "cousin", so by extension our cousins come first? well, it does apply to people in Alabama lol.
@@eduardoxenofonte4004 Actually all of those are derivations of "prae" the Latin preposition meaning "before", from that we get the superlative form "primus" meaning "first"and their vulgarizations which eventually got to Latin based languages and Portuguese that is not a real language. Now "prime" from "prime numbers" is used in the sense of "first", so they are the "first numbers" as Euclid called them since they are not divisible by others meaning that all other positive integers are measured by primes, but primes are measured only by units making primes the "first numbers". My comment was just a joke but I see there is interest in the topic so it's time for beauty.
@@ster2600 You mean 1 is multiplicative identity like 0 is additive identity. So if you add up nothing you get 0 and if you multiply nothing you get 1.
@@ayantayyab7259 i.. that doesn't look quite right. I'm relatively certain it's only when dividing that you deduct the exponents from one another, not when substracting
The other question: Why isn't 1 considered as a composite number? If you think about the prime factorisation from any number as a infinite product of all existing primes, but with different exponents (like 15 is 2^0 * 3^1 * 5^1 * 7^0 * 11^0...), 1 also has a unique factorisation: every exponent has to be 0. PS: I know I'm 9 years too late, hopefully someone sees this comment by chance.
I'll take a stab at this. A composite number is a product of prime numbers. Since 1 is not a prime, and 1's only unique factor is only 1, it is not a composite number since it is not a product of prime numbers. Also, multiplying 1 by itself infinitely many times is pretty pointless and does nothing fruitful. Because if you were to allow 1 to be prime, every number would have infinitely many prime factors. That's why in the video, Dr. Grime said it has its own category.
Not really, One is on a category by itself while Pluto is in a category that has hundreds if not thousands of other elements. In any case, if you feel for Pluto imagine Triton, demoted from Dwarf Planet to Moon xD
+Heinzriko I must admit that I have not watched it, but if they use Euclides' theorem, 1 doesn't destroy the proof. the definition is that a prime number can only be divided by itself and 1, so you don't have to divide P1*P2*P3....*Pn+1 by 1 to proof this
Excluding 1 as a prime doesn't change the absolute, undeniable fact that, Optimus is a Prime. That's because he is unique. He is the product of 1 Matrix and Orion Pax, no other combination will result in Optimus Prime.
@@samueldeandrade8535 True. The convention could also easily be changed or reformulated. This video makes it seem like it’s as set in stone, as the Fundamental Theorem of Arithmetic.
not really there are many dwarf planets in our solar system so Pluto isn't alone, i think there might be more dwarf planets than regular but don't quote me.
+ncooty All of the prime number have to be raised to the power zero: only 1*1*1*1*... = 1. If any of the primes is raised to a power greater than 0, the result is no longer 1. The fact that this is a unique way to write it follows from the fact that there's only one way to write an empty product: namely, no prime numbers :)
dslxksanta not quite yet. its still a dwarf planet and still doesn't fall within the official definition of a planet. • Is in orbit around the sun. • Is round or nearly round. • Has "cleared the neighborhood" around its orbit, meaning it is not surrounded by objects of similar size and characteristics. That last part is the problem.
Hahaha yes it is. But again, no harm done. That definition is there for the same reason 1 is not a prime. Its just there to make the wording of everything more convenient.
You've probably already seen something like this before, he didn't explicit it that well though. 16 = 2*2*2*2 = 2^4 8 = 2^3 4 = 2^2 2 = 2^1 1 = 2^0 Indeed, every number power 0 is 1. Also the real definition of prime number is : N be an integer. N is prime if, and only if, it can be divided by two *distinct* integers, that are itself and 1. This definition already excludes 1.
@@notkamui9749 Well hang on a minute, since any number raised to the 0 power is 1, then surely 1 would still have to be excluded because 2^0 and 3^0 both result in 1. The fundamental theorem of arithmetic says that there has to be a unique product.
For those asking about prime numbers, and how they fit, this is a theorem of ARITHMETIC, not ALL of mathematics. The theorem is about the prime factorization of numbers. For example, it is very explicit to say that the prime factorization of 20 = 2 x 2 x 5. But the prime factorization of eleven, you would explicitly say 11= 11. Arithmetic, however seems to be so fundamental to mathematics, that it is difficult to see a theorem that applies to just the manipulation of numbers rather than on the existence of those numbers.
Going along with what Dr Grimes said at 3:50, imagine if 1 were counted as a prime; we were constantly having to say "for all primes (except 1)" in our proofs and theorems. Mathematicians would, as they always do, come up with a new name for the set of primes not including 1, since it would have been a set that was used frequently. What would they call it? Imagine they called it the Polkadot Set, or whatever. At that point, the only thing that changed at all is the name of the set for the primes as we defined them now to the "Polkadot Set." And, like he said, prime is just a word. Who cares what they're called? So, whether or not we call 1 prime, the set not including 1 would be used more frequently, and have some name.
Fabian Herzog: Bro, doesn't that introduce a new problem? Because 1 can be written as empty product of any prime. 1 = 2^0 X 3^0 is equally valid as 1 = 13^0 or for that matter 1 = 17^0 X 5^0 X 97^0. So isn't 1 evading fundamental theorem of algebra again?
1: "Yeah, but unlike you, I won't be included in a larger category. I'm just stuck, alone." Still don't get why some people don't know about dwarf planets...
ComandanteJ It feels like that, but if you look at a bunch of other stuff in maths you'll find that defining the empty product as "1" fits perfectly in every case you can possibly use it in. It's really interesting to study various formulas and models and you discover that defining the empty product as 1 just makes all of them work flawlessly. It really seems to be a universal truth for some reason, even if it might be difficult to understand why.
MasterOfTheChainsaw , I think the last one "feels" like cheating also partly because you are taking the special case of considering the reduction of powers of two; rather than being more general and taking other cases as well (such as those composites which are multiples of more than one prime such as 6 and 12) to give a more fufilling description of why it is an empty product... which interestingly would require multiplying by all of the P_i primes to the power N_j
ComandanteJ The fact is that 1 has a special place as the neutral element for the product ; it is only natural that the empty product is 1, exactly the same way that an empty addition ("let's add nothing") is 0, aka the neutral element for the addition.
A example illustrating that the empty product = 1 (using 2, 3, 5 and 7 as sample multipliers): product(2, 3, 5, 7) = product(2, 3, 5) x 7 = product(2, 3) x 5 x 7 = product(2) x 3 x 5 x 7 = product() x 2 x 3 x 5 x 7 ... so the remaining empty product on the left must = 1. The empty sum can be shown to be zero by similar means.
Well then your number can still be factorized into product() x product() x product() x product().... x product() x 2x3x5x7, which still breaks the theorem.
2 = 2 is a product? I see no multiplication going on, which is in the definition of "product". This seems to indicate a major problem in the argument suggested.
Well, no problem if 1 is prime.. 2 = 2(x1) But apparently we should not do that because then we would not got an imaginary empty product. Its so funny because dividing by 2 in that list is like saying; "we now multiple the rest with 1" But all of a sudden we should not follow this pattern lol
I'm just sharing with the RUclips community, much to my amazement, that I'm really enjoying this series of videos. Thank you for sharing them - you may have a convert...
+ColdsideRamrod 1 can be the product of multiple factors (1x1=1), just not any prime factors. It is an empty product because it is not a product of prime numbers.
+ColdsideRamrod I find it helpful to write out what we're saying when writing a number as a product of primes. A generic formula might be: n = 2^(something) * 3^(something) * 5^(something) * 7^(something) *... The exponent of each prime number tells how many times that number is in the total product. The 'unique' part of the theorem is shown in the fact that there can only be one value for each specific exponent. Now when all these exponents are 0, we get the formula: n = 2^0 * 3^0 *5^0 *... n = 1 * 1 * 1 * ... n = 1 So there is no prime number in 1 in that sense, so it is an empty product, yet still equal to 1.
Sir, I’m always fascinated by math and numbers just as I am by music and language. I think my children have learned more from watching repetitive clips that are short from you then they have in a school setting or classroom with number of students. Please continue, maybe simplify a little more but they enjoy as do I you’re Classes.
Ok, but there is still a problem because the decomposition of 1 in primes number is not unique: 1 = 2^0 = 2^0*3^0 = 2^0*3^0*5^0 = etc... So 1 does NOT have a unique decomposition in prime numbers... THerefore the theorem is false, isn't it ?
Sure, you've just defined a "bag" or "multiset" with that definition. Also, you may want to say "$0$ for all but finitely many $n \in \mathbb{N}$". The term "almost everywhere" could mean something different (e.g., on a set with zero density).
I would not say that it is well established, in most papers they would say "for all but finitely many..." to avoid ambiguity. And yes, precisely. So you defined a multiset over the primes, as I said.
Hmmm.. So if every positive whole number can be written as a unique product of primes, and 1 is not a prime how is 2 a whole number? What am I missing?
"A prime number is a number that can only de bivided by..." had to rewind to make sure my ears weren't dyslexic. (And yes, I know ears can't be dyslexic.)
4:43 i personally think you are wrong here, because the definition specifies a PRODUCT of prime numbers, and in 2=2 there is no product if 1 is not included as a prime number
ElTito Alberto Mathematicians use the symbol ∑ for a sum of a (finite or infinite) sequence of numbers, and the symbol ∏ for a product. For example, ∑(1,2,3)=1+2+3=6, and ∏(1,2,3)=1*2*3=6. Suppose that I have one number. What is the sum of all numbers I have? The number itself: ∑(5)=5. Now suppose that I have no numbers. What is the sum now? Zero: ∑()=0; see that ∑()+5=∑(5). Now similarly, a product of only one number is the number itself, ∏(5)=5, and product of no numbers is defined to be 1 so that ∏()*5=∏(5).
MikeRosoftJH Well if you only have one number, how do you define if it is a sum, or a product? :P I agree it´s not very much better than just saying: "All prime no., except one...", but I guess it gives the mathematicians less to argue about, and more time to do actual work ;) hehe
MikeRosoftJH a product of no numbers isn't just defined to be 1, it is mathematically shown to be 1. It isn't arbitrary, it is a consequence of the mathematical concepts of multiplication and division.
I have a question. If every positive whole number is a unique PRODUCT of primes, and 1 is discounted as a prime, how do prime numbers fit into this theory as you cannot use 1 as a prime number, what would produce, say, 11? --> 1 x 11? but 1 is not prime... so how does this fit?
I always heard "prime" defined as having exactly two factors (1, and itself), meaning that 1 could not be prime. Whether that is an accurate definition, I don't know, but because of that I have never considered 1 prime.
That's what many people (particularly elementary school textbooks) have done. But this video is still useful, because it explains why someone would change the definition to that in the first place. Because you can't get around the fact that the definition of prime has _changed_ from its original definition over time. I will say this, though. When you look for prime elements in different settings (mathematical structures which act similarly to the natural numbers), the number of divisors concept does not generalize well for a definition of prime. So that's something to (potentially) keep in mind.
Ha-ha. But using this definition the 1 should be excluded from prime numbers as well. Because it devided by 1 and itself, which is 1. But this is same and only number :) I prefer traditional definition. Whole number can be devided by 1 and it self.
+Любомудр Назарьев That was my whole point, a definition that _excludes one_ but not the others. That was literally my goal since we don't need one in it, but all the others. One is not a prime.
But by the original definition of the word 'prime' 1 is the prime to all numbers due to the fact that it is the unit of reference......to change the definition of what a unit is, you would have to use that theorem mentioned above to conclude that a unit is not a positive whole number, which it is axiomatically.....in fact, a unit is required to even understand what a 'whole number' is....
In Vietnam, we have a definition of prime that a prime is a number which has only 2 divisors, 1 and itself. So we can exclude 1 because it has just 1 divisor
The explanation for how to make 1 out of prime numbers is just as ridiculous as the theorem. This needed to be presented totally differently, and as a poorly defined theorem.
For the fundamental theorem of arithmetic, I prefer the definition "that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique". Without the "prime itself", you end up pretending that the primes only have one factor, but then it's only a product if you multiply by 1, which is against the definition you've used.
ccax74 He's using a more general definition of "product." Strictly speaking, multiplication is a binary operation, meaning it takes on exactly 2 inputs. No more and no less. But clearly this is not the meaning of "product" used in the Fundamental Theorem of Arithmetic (otherwise 8 and 12 and infinitely many more numbers would not be the product of primes!). Seeing that the multiplication of whole numbers is both associative an commutative, you can generalize the definition of product to take on any finite number of inputs. Usually, when we think of this, we think of allowing more than 2 inputs, but there is a consistent way to allow 0 or 1 inputs as well. When we have 1 input into our multiplication function, the multiplication function acts as the identity function, giving back the input. Why is this the reason? Well, if we say that our multiplication function is "Prod" then Prod(a,b) = ab. Now, let's divide both sides by b. We then should get Prod(a) on the left hand side, and we certainly get a on the right hand side. So it makes sense to define Prod(a) = a. By the same argument, if you divide both sides by a, you get Prod() = 1. And so, we generalize the definition of "product" to be 1 if there are zero inputs, the identity function if there is 1 input, the standard technical definition of a product for 2 inputs, and iterated standard technical products for more than 2 (but still finitely many) inputs. This is the sense in which he is using the word "product" in the theorem.
MuffinsAPlenty Well first of all I could point out that your Prod(a,b) logic to arrive at Prod(a) or Prod() doesn't hold true if either of them is zero (and since it's perfectly fine to multiply by zero, that's not obvious), but that's not my main point. If you just google for the fundamental theorem of arithmetic, the vast majority of definitions you'll find will include something like "is either prime or the product of primes" or point out that there could only be one prime factor. He should have just used one of those definitions for the video.
ccax74 Yes, he probably should have used that version of the Fundamental Theorem of Arithmetic for this video, as it seems to be people's biggest contention with this video.
Trabber Shir I like the main idea of your argument because certainly all positive integers can be written uniquely as a linear combination of primes (where vector addition is integer multiplication and scalar multiplication is exponentiation). And thinking about it this way, it certainly makes sense why 1 should be excluded from the primes (it ruins linear independence)... However, the positive integers do not form a vector space (or a module for that matter). In particular, no vector (besides 1) has a negative vector. For example, if we wanted 2 * x = 1, x = 1/2, which is not an integer, let alone a positive integer. (Though the positive real numbers _do_ form a vector space) Also, the only way you could ever hope to have a unique basis for a vector space is if your scalars form the field of cardinality 2 (Z/2Z). i.e., the only scalars are 0 and 1. This is because given any nonzero scalar c, if {x1, x2, ..., x(n-1), xn, x(n+1), ...} is a basis, then {x1, x2, ..., x(n-1), cxn, x(n+1), ...} is also a basis. (It certainly spans, since if you would have the coefficient of xn be a to get a vector v, then you can make the coefficient of cxn be a/c, and the set is certainly linearly independent as well since a(cxn) = (ac)xn, and then we can apply the linear independence of the original basis.) So it looks like you are maybe thinking of the wrong thing. If you do happen to remember, I would love to hear it (since this is certainly a wonderful explanation if it can be salvaged!).
Trabber Shir This is probably not what you were thinking of, but I figured out something similar... The positive rational numbers (under multiplication) can be represented as a Z-module (a module is like a vector space, but your nonzero scalars don't have to have multiplicative inverses: in this case, the scalars are just the integers). Again, scalar multiplication is exponentiation. The prime numbers form a Z-module basis for the positive rational numbers. And of course, 1 cannot be a part of this basis, otherwise it would be linearly dependent. (A point about uniqueness of bases: my previous argument that bases are not unique in a vector space relied on the fact that nonzero scalars had multiplicative inverses, so the same argument does not work here. However, this is still not a unique basis since -1 does have a multiplicative inverse, itself, so you can have reciprocals of primes as basis elements as well, so this basis is still not unique.) Again, it seems like it isn't exactly what you were thinking of, but is in the same vein.
But the empty product isn't a prime ;) Edit: Oh wait, that doesn't work either, since then 1 wouldn't be a product of primes. In the end we can just define that this is the same thing as 5*3. We could also just say that 1*3*5 is also the same to get uniqueness but with 1 as a prime, but in the end this way looks nicer
Michael Vogel No, that would not work. The empty product is not a prime number, and the empty product IS a product of primes: an empty product of primes. There is nothing controversial about this.
If you want the theorem to work for any number including 1, you have to consider the number as a list of power of every prime number. For example, you will say that : 15 = 2^0 * 3^1 * 5^1 * 7^0 * 11^0... The notation is still unique (but infinite). So, for 1, you have a unique notation too, which is : 1 = 2^0 * 3^0 * 5^0 * 7^0...
For anyone trying to make the empty product make sense you can think of it as equivalent multiplication. Multiplying by 6 is the same as multiplying by 2 and then 3. Multiplying by 1 is the same as multiplying by _nothing_.
I taught myself (a bit) differently when, in programming, I wanted to find as many prime numbers as possible. A prime number is not divisible by any other product of prime numbers. In other words, if you had 11, it would not be divisible by lower prime numbers. If 1 were a prime number, then 11/1=11 and thus is not a prime number.
robotica actually, a prime number is divisible by one and itself only. And that is the defenition. Yours doesn't make much sence, and I speak as a programmer
Walwalkn Wewnrkl As i said, the code does work, but since the formal definition of a prime number is: "A number only divisible by itself and 1", logically speaking you can't say that one isn't a prime because you use another definition. Basically, 1 isn't a prime because of is definition exists, but rather his definition works because 1 isn't a prime. As they said in the video, the only reason 1 isn't a prime is because it makes other stuff in maths easier, such as his program. Hope that helps :)
m1g4s It didn't mean to state a definition, I wanted to state an explanation. You can't really say that "a prime number is a number which is not divisible by any lesser prime number", but using examples you can get another perspective to why 1 really isn't a prime number, because it follows another definition.
robotica I know what you meant, but as I explained, your code works because one is not a prime. 1 being a prime is for a different reason, hence why the video proved your code rather than your code proving the video right
m1g4s "hence why the video proved your code rather than your code proving the video right" It's actually both. The video inspired to make a program which would require the first prime number, and when 2 was the first prime number which would make the code work would help people also give an "argument" for 1 not being prime.
Why is 1 the result of an empty product? Here's why: 2^3 = 2x2x2 = 8 2^2 = 2x2 = 4 2^1 = 2 = 2 2^0 = __ = 1 3^0 = __ = 1 n^0 = __ = 1 (any number multiplied by itself 0 times is always 1 -- that was just decided some time, and I'm sure it's since been thoroughly proven, however counter intuitive it may be). I wish they included the P^0 part in the video, because that would have helped clear up why the empty product is 1. Even 0 can be excluded by this definition, because while in theory it can be divided by itself (0/0 = 0) and 1 (0/1 = 0), there is no product of primes that results in 0, even the empty product stops at 1.
Gustav Mårdby It depends on what model you use - 0/0 limits on 1 for x/x, but on 0 for 0/x. Like most situations that involve dividing by zero, it's nonfunctional - the result will change depending on the context.
Very strange theorem: Now please apply this theorem , and give me a solution for 1. Write 1 as a product or prime number. 2. Write any prime number as product a product of other primes.
1. It's in the video, it's the empty product. A product of zero primes. 2. it's just the number itself. Of course you can't wrtite it as a product of other primes, that would conradict the theorem. And yes, "2" by itself can be considered a product. If you accept that products might consist of one number, or no numbers at all, everything will make sense. Well, of course you can argue these definitions and you can disagree with them, so you can say that a product is at least two numbers, and thus prime numbers themselves can't be written as a product of primes. But that doesn't contradict the theorem - you had to rephrase it, sure, but the actual mathematical meaning would stay the same.
Rajesh VP Easily. 1 = 2^0*3^0*5^0*••• There, you have a unique factorization. Now, p(n) = 2^0*3^0*•••*p(n)^1*••• There, you have a unique factorization. Happy?
Math Wiz Yes, because by definition a prime has only 2 factors: itself and 1. If you can get there by multiplying other numbers, it’s not a prime! But (qua the broader theme of this thread) it’s still silly to say that a prime number is the product of only itself, since ‘product’ implies a generative process and by that definition primes just... *are* !
feuilletoniste *it's still silly to say that a prime number is the product of only itself, since 'product' implies a generative process...* No, it does not. In fact, mathematically speaking, the idea of a generative process you are trying to appeal to is vague and meaningless. On the other hand, 'product' has a precise mathematical definition, and that definition accounts for unary products just fine. So no, it is not silly, it is simply the way math works. Maybe you should try studying some group theory before you make blatantly false statements such as this one.
Maybe this will clear it up: For every prime p, p^2 (its square) is semiprime; 2(prime)^2 = 4(semiprime); 11(prime)^2=121(semiprime), etc. Assume 1 is prime. 1^2 would be semiprime. 1^2 = 1. So if 1 is prime, 1 is semiprime. See how that doesn't work? So either accept that 1 is not prime, or go back to saying "except 1" every time you talk about primes. I know which I prefer. Since it has FEWER factors than a prime, I'm sticking with my suggested desgination for 1 as "SUBprime", mortgage scandals be damned.
But if so, the Theorem is wrong... 2 is a whole number and can not be described as a product. Also 1 is a whole number and has the same situation, except that it does if you include one. So... they added: "greater than one and can be represented by itself number". Smh, seems like they took that on their own favor. On what you remark, you could take same card and said "For every prime p greater than 1, p^2 (its square) is semiprime". I see one as a real prime.
1's special category is a unit: anything with a multiplicative inverse in its own set (closed). 1^-1=1 an integer. 2^-1, 3^-1, etc are not integers. So for positive integers 1 is the only unit. You used negatives. Once negatives are involved, 1 and -1 are both unuts.
+Deven McKee No, As an official chemistry nerd that is simply not possible, a proton by itself is extremely unstable and cannot occur naturally. Hydrogen can only exist as a per of atoms in nature. (H2, Not H)
As a chemistry nerd you only consider elements in the presence of and interacting with other elements chemically, where, yeah, it's gonna push HARD for equilibrium. As a physics nerd you'd know how, where, and why you would be chucking around naked protons.
Same problem here... for me 0 x 0 would mean you have "zero times nothing".... but wouldn't that mean it can be everything? Or infinity? Every thought goes a different way, but none of them ends up at 1
1 is a product of no primes, which means it can be a product of 1, because it is not a prime. 1 x 1 = 1, but 1 x 1 has no primes in it, so the result, 1, is a product of no primes.
You're over-thinking it. We altered the definition of a prime number slightly just to make it more convenient for us. It's just a shortcut that doesn't affect the physical laws of the universe. It's like deciding whether to use the 12-hour or 24-hour clock. Time is not affected by our choice, only the way in which we describe it.
And here I was thinking there were no politics in math. Get out of here one, you're ruining our agenda even though you are perfectly valid in every other way.
For the same reason, 2 should be excluded as well, because there are a lot of theorems involving "odd primes". Maybe 2 is related to the "big primes" or it is such a nice guy so that the other primes don't want to exclude it...
Joseph Douglas If you think of primes as integers with exactly two factors, then that makes it work. Or another way of thinking about it is that in purposes like the fundamental theorem, multiplying by one does nothing, so it can basically be removed from the factorization.
@@KnakuanaRka but they teach a prime as a number which can only be divided by 1 and itself, and then the smartass kid says "so is 1 a prime" and the teacher says "no because it's 1 and itself" "yah and you said 1 and itself and 1 is itself" "okay, primes are numbers divisible by 1 and 1 other non-1 number." "but what happened to your other definition" "shut up". why can't they just let 1 be prime and be done with it? And there's nothing really that cool about primes except that they don't have a lot of factors.
Peter Senior You really don't know anything about applications of number theory if you think prime numbers are not important. And there are many other reasons 1 is not a prime.
Thank you so much, I wuz so confused and puzzled thinking why it isn't a prime number. 0 and 1 are two great numbers and there's a real secret behind them!
With the Fundamental Theorem of Arithmetic being the way that it is, doesn't that mean that ultimately prime numbers are the only actual numbers? And all other numbers are creations of primes?
If 1 and 0 were prime numbers, they'd be the only primes that are perfect squares. All other perfect squares are not prime because in addition to one and themselves, they have a factor that, multiplied by itself, equals the perfect square (e.g. 3 is prime, but 3x3=9, and 9 isn't prime because it's a multiple of 3).
I'm still not convinced 1 is consistent with the theorem. You're saying that 1 is the product of zero prime numbers and is thus consistent with the theorem. However, is this empty product really unique? The example shown was essentially 2^0 = 1. The product of 0 2s is 1. But that can be applied to any prime number. p^0 = 1 where p is prime.
2 is not a Gaussuan prime. 2 = (1+i)(1-i). A number is prime or not with respect to a set of numbers. In the integers, 2 is prime. Gaussian integers are a+bi where a & b are integers. In the rational numbers (or any field) nothing is prime. Rvery value would be zero or a unit (values with a multiplicative inverse in the set of numbers). By definition neither are prime. By real definition, not grade school definition.
The theorem is correct for all primes, but not 1 and has thousands of logical consequences. (It also happens to be the only reason your bank account is safe from cyberattacks.) so excluding 1 doesn’t change anything about the theorem, just that the wording can be cleaner.
right ! 1 is Prime its literally the first prime number, and it does fit the definition of what a prime is. Prima materia first matter prima primero first , 2 is not first lol its second secunda segunda matteria second matter , so why in the world would prime numbers start at 2. prime numbers are extremely important. i know theres a why it was changed, the real reason behind it but are we ever going to find out the why ? i can prove 1 is prime already visually without any fancy formulas. geometry is all you need to observe the fractions and what they become
@@JOELRODRIGUEZ-lk9gu *1 is prime, its literally the first prime number,...* No, it is not. 2 is the smallest positive prime number (with -2 being the greatest negative prime number). *...and it does fit the definition of what a prime number is.* No, it does not, and I would bet real money that you do not know what the definition of a prime number is. You probably think the definition of a prime number is "a positive integer whose only positive divisors are 1 and itself." If you do believe this, then you are in the wrong, because this is not the definition of a prime number. It has not been the definition of a prime number for at least a century now. A prime number is an integer whose only positive proper divisor is 1. The number 1 has no proper divisors at all, so it is not a prime number, and for the same reason, neither is -1. *Prima materia = first matter = prima = primero = first* I am glad you know the etymology of the name "prime number," but ultimately, the etymology is irrelevant to the mathematical definitions. My name is Ángel, and the etymology of my name is the Greek word άγγελος, which means "messenger." I am, however, not a mailman, nor a courier. The etymology of my name has nothing to do with the characteristics that define me as a person. Similarly, the etymology of a word used as an English name for a mathematical concept has nothing to do with the definition of the mathematical concept. *2 is not first lol, its secunda materia, second matter* Numbers are not made of matter. Numbers are abstract concepts which exist only in the mind of sentient beings which are capable of doing mathematics. That being said, the reason the prime numbers were named after 'prima materia' as a metaphor is that they took the atomic theory of matter as an inspiration: atoms, also called first matter, were considered to be fundamental, elementary particles comprising all matter. Any particle of matter could eventually be reduced to atoms, and atoms built all other particles. The prime numbers play a similar role with the integers: the prime numbers are analogous to atoms, in that they cannot be multiplicatively composed of other numbers, and in turn, the integers are multiplicatively generated by them. -1 and 1 are not prime numbers, because they do not fit that role: they actually do not generate any integers at all, since 1•(-1) = (-1)•1 = -1, 1•1 = (-1)•(-1) = 1. -1 and 1 are trapped in their own bubble, incapable of generating the other integers. The same happens with 0. The prime numbers 2, -2, 3, -3, 5, -5, etc., are the ones we need to generate integers. 1 is the empty product of primes, and then we have 2, 3, 2•2, 5, 2•3, 7, 2•2•2, 3•3, 2•5, 11, 2•2•3, 13, 2•7, etc. In the other direction, we have -1, with (-1)•(-1), -2, -3, -2•2, -5, etc. *Prime numbers are extremely important.* Yes. *I know there is a why it was changed, the real reason behind it, but will we ever find out the why?* I strongly recommend you watch Another Roof's video on the topic, because the truth is, for most of history, 1 was not considered to be a prime number. Prior to 1600s, it was not considered to be a prime number. It began to be considered a prime number due to a misunderstanding of the concept, but in the 1800s, they stopped considering it a prime number again. Furthermore, we know exactly why both things happened, and we have known ever since they happened. There is absolutely no mystery at all. I am sure this has actually been explained to you before, but you just were not paying attention. *I can prove 1 is a prime already visually, without any fancy formulas.* False. I know you cannot do that, because the definition of a prime number is given by a formula, not by any geometric concepts. You _think_ you can prove it, but this is how like children _think_ Santa Claus is real.
@@shaurryabaheti All prime numbers have exactly 4 divisors. Both -2 and 2 have exactly 4 divisors (they have the same divisors). Therefore, if 2 is a prime number, then so is -2, and vice versa.
1:34 Every positive whole number can be written as a unique product of primes... wouldn't that exclude 2, 3, 5, 7, or any prime number for that matter? I'll continue the video now.
Knicksboncho Nothing. A multiplication can be done with a single factor. In fact, it can be done with 0 factors. That is exactly what x^0 stands for. And we know x^0 = 1. So there you have it.
Wait, if "every positive whole number can be written as a unique product of primes", then 1 must not be a positive whole number, regardless of whether it is considered prime or not. I don't think that theorem is written down correctly.
Corpsie Corpsie Not at all. The notion of the empty product is completely sensical and consistent with the rest of mathematics. The idea even pre-exists the fundamental theorem of arithmetic. The idea dates all the way back to the invention of algebra itself and the discovery of division.
always just assumed it was because prime numbers had to have two separate factors of one and itself, whereas 1 just has one factor. Really interesting video!
Semantically the base definition implies that "the number" ISN'T one. Such ambiguity could have been avoided; however, I submit the prime word of the definition as the "real" issue: EVERY does include 1. Thus it should also be divisible by two distinct WHOLE numbers, which is false-if your interpretation of the definition… is "rigorous". It's my opinion the word "and" precludes one from the possible values of "ITSELF".
We can reformulate the fundamental theorem as "all whole number have a unique factorisation to prime numbers up to multiplication by unit " to avoid such randomly things ...
Yes, you can. But the theory is arguably much cleaner when allowing empty products/unary products. Reformulating the definition of prime to allow for arbitrary finite products (rather than solely for binary products), units _legitimately do not satisfy_ the definition of prime. So we don't have to "randomly" exclude them from the definition of prime.
I know a more excactly definition of primes: A prime is a number, which can be divided by ONLY 2 different numbers (1 and itself). Then the problem would be solved (divided by 1 = 1/1 divided by itself = 1/1 -> not 2 different numbers)
100 % agree ! if 1 is a prime number, 1-1=0 make Euler product : multiple of (p-1)/p equal to 0, make R.O.S.E(realization of sieve of Eratosthenes) impossible, for example : pi(4)=4*(2-1)/2+0/2+1-1=2, pi(9)=9*((2-1)/2)*((3-1)/3)+1/2-3/6+0/3+2-1=4, pi(25)=25*((2-1)/2)*((3-1)/3)*((5-1)/5)+1/2-1/6-5/10+25/30+1/3-10/15+0/5+3-1=9, pi(x) is exact prime counting function, formula is : x time reverse of Euler product (p-1)/p plus total of mod(x,Po)/Po+ total of p-1.(p is prime number less and equal to x^(1/2), Po is all possible combination of p)
I like the comment that calls it "the origin number", especially since it really is that. All whole numbers come from 1! (They're literally just groups of that!) :D
0^0 is top-down defined as 1 in the sense of multiplicative identity. Follows from the definition that n^-n = 1/n^n A non-empty symbol for empty no-thing is of course logically very problematic. Peano axioms start from defining 0, and whole lot of problems follow from that. 0 is nice tool for making some arithmetical operations easier, but as logical foundation of number theory... less than ideal.
I don't like this formulation of the theorem, it means that "2" is a "product" of one number. I think the theorem is better formulated that every number is either PRIME ITSELF or the product of primes.
But then you'd have to include 1, which they have decided is just in the way of most calculations using primes as a set. So when you want 1 removed from the set in the majority of the cases, that's a strong case for it not belonging there...
athox No, you're missing my point. 1 fits with the definition here too. Whether 1 should or should not be included in the definition is a separate question. I'm saying the definition James gives here forgets to mention that prime numbers should not be included. (That is the definition I have seen, I'm saying, I do not doubt James has some justification for writing it as he did.)
greg55666 Why shouldn't prime numbers be included? The prime numbers have a prime product too. You can agree that n^2 is the same as n x n, right? As in, n multiplied by itself twice. Equally, n multiplied by itself once, must also be a valid product, namely n. n = n^1 = n See? You cannot get past the n^1 as being a valid multiplication, even if it only has one factor. ------- Digression: Also, n multiplied by itself never, is also a valid product. Which is why n^0 = 1 has no factors, prime or otherwise... If you include 1 in primes then you can do 1^1 = 1, but then we're back to intentionally excluding it for practical purposes.
I think it kinda goes without saying seeing as the *definition* of primes is that they are explicitly not made up of a product of any two other numbers except themselves and 1. Seeing as the theorem includes the use of the idea of prime numbers then it implicitly also excludes primes from needing to comply with its stated definition.
athox n^1 "must" also be a valid product is just circular--I'm questioning whether it should be included. I think your digression proves my point. I'm saying "n^1" is not a product. Let's get back to that in a moment. You say, not only is n^1 a product, but so is n^0. If you count n^0 as a product, you have broken the theorem. Because then 1=5^0, 1=10^0, 1=453235^0, etc. So you MUST exclude n^0--n^0 cannot be a "product" for the purposes of this definition. Why are you excluding n^0 as a "product," but then objecting to me wanting to exclude n^1 as a "product"? You've lost whatever principle you were going by. The point of this theorem, as I understand it, is that there are two kinds of whole numbers: prime numbers and composite numbers. The point is that there is no number that is NOT either prime or composite. By not saying every number is either prime or the product of primes, it seems to me you lose the entire point of the theorem. I think a better explanation is that saying "a prime number is a product of itself" is trivial, so it's unimportant either way. Stating it as it is in the video is more concise at the loss perhaps of a little more insight into its point; saying "either prime itself or a product" is more words, hence perhaps less clear even though perhaps it points more to the meaning of the theorem. But either way, the statement that a prime is a product of itself is trivial, so it is the author's choice. As I said in my original post, I would have chosen the more words with the deeper meaning.
So it kinda just comes down to the choice of whether you would rather say, "all positive whole numbers can be expressed as a unique product of prime numbers (excluding 1)" or "prime numbers are numbers which are only divisible by 1 and themselves (excluding 1)" Admittedly saying the definition of a prime is a number that has exactly 2 unique divisors gets around that.
Brady: “what if it’s just a stupid theorem?” That was classic.
But it's a very valid question! Most of maths has had these seemingly arbitrary choices in the past (0! and 0^0 are some others) and we choose the option that gives us the most useful tools to work with. In this case the primes are more useful to us without 1 and that also gives us this theorem which is useful in itself.
@@piderman871 Yeah, but that's fine. Math is not something that is hardcoded into reality and we have to obey it by any means. It's a language and we can agree on the words which go into it. The universe has no concept of prime numbers; it doesn't care whether we consider 1 a prime number or not. It has, however, a concept of unique representations of positive integers as products of members of a specific set of numbers. It's up to us if we want to call this set P or P\{1}.
In this sense the theorem is more fundamental than the notion.
@@iwatchyoutube544 I think he meant hardwired.
@@iwatchyoutube544 Sounds like a "mathematics is invented, not discovered," kinda thing.
@@iwatchyoutube544 Math is our made up language for describing nature and the universe, we need everyone on earth to speak the same language without variations in order to seemlessly explain physical phenomina to every human on earth. What he means is, just like with any other language it isn't set in stone, we add and remove words from our dictionaries all the time. In math we do the same except we need some proof or some new discovery before we add or remove parts of that language
Don't worry 1, Pluto got excluded from his old group of friends too.
I want this on a shirt
+bingahgread Just one planet was exluded... ONE PLUTO, what a sad couple
Ahahaha. You guys. XD
To soon ,you cant bring this up, the wounds arn't healed yet :,(
But Pluto got accepted into a new, more fitting family where he's not alone but has a bunch of new friends! 1 on the other hand is really alone. Poor 1.
So I ordered this one product online and when the order arrived, there was nothing in it. So I called them to complain for having sent me an empty product when I had expressly ordered one. They replied that empty product equals one. Damn mathematicians!
LOL!
Did you order it from Amazon *Prime?*
Surely You Jestin 🤦🏻♂️
domino's pizzas on the way.mam..........
ihujuk
"You can tell it is a very important theory because it has a pompous name." - Love the argument :-D
cry about it Harmen
The number 1 is excluded from the primes. Likewise the name Harmen Breedeveld is excluded from "importants" despite being a pompous name.
@@brianmacker1288 Man, did you get with your wrong leg out of bed this morning? I simply loved his funny remark.
Lighten up :-)
Take care!
@@harmenbreedeveld8026 I thought you "love the argument" so I thought to start the argument with the joke. Is working? Ok, I see, no like the joke? Ok, I tell another.
What is the difference between vacation spots Dubai and Abu Dahbi?
You don't know, because you infidel, so I tell you.
The people of Dubai do not like the Flintstones. But the people of Abu Dahbi do.
“Why did you name me Pompous?”
“Because you are important!”
0:28 _"a prime number is a number that can only _*_de bevided_*_ by"_
All this mathy stuff and you catch this. Thanks for making me laugh to tears.
LOOOOOL
π x 5 = ?
@@memerboi69.0 15.70753.......
Well ex plained
'you can tell it's really important theorem because it has a pompous name'
it's ironic because "prime" kind of means "one", or more precisely "first".
That's just an idiomatic coincidence, in Spanish "Prime" is "Primo" which also means "to have come on top of a conflict or situation or "cousin", so by extension our cousins come first? well, it does apply to people in Alabama lol.
@@eleSDSU SWEET HOME ALABAMA!
I'm Brazilian and "prime number" is "(número) primo" and "primo" also means cousin but indeed it still has some relation because first is "primeiro"
@@eduardoxenofonte4004 Actually all of those are derivations of "prae" the Latin preposition meaning "before", from that we get the superlative form "primus" meaning "first"and their vulgarizations which eventually got to Latin based languages and Portuguese that is not a real language. Now "prime" from "prime numbers" is used in the sense of "first", so they are the "first numbers" as Euclid called them since they are not divisible by others meaning that all other positive integers are measured by primes, but primes are measured only by units making primes the "first numbers". My comment was just a joke but I see there is interest in the topic so it's time for beauty.
Larry Lewinsohn: ...and also the prefix "pre-"
4:26 taught me why anything to the power of 0 equals 1.
1 is the multiplicative identity. So doing no multiplication should give you the identity
@@ster2600 You mean 1 is multiplicative identity like 0 is additive identity. So if you add up nothing you get 0 and if you multiply nothing you get 1.
@@HanakoSeishin thanks, I've corrected it
Another proof : We know a^m÷a^n =a^m-n
So 2/2=2^1-1 2/2=2^0
@@ayantayyab7259 i.. that doesn't look quite right. I'm relatively certain it's only when dividing that you deduct the exponents from one another, not when substracting
Technically speaking, one could use a slightly different definition for prime number:
"A prime number is a number that has only two unique divisors"
Exactly, they must be different caus’ we could say that 1 can be divided by 1 and itself but they’re just the same so....
This is what my teachers always told me
Continuing that because that's only part of the definition I learned at school, "...which is 1 and itself."
This is wrong though. It can be a product of more than 2 products. Eg 2x3x5 is 30
This is wrong though. It can be a product of more than 2 products. Eg 2x3x5 is 30.
" maybe it's just a stupid theorem" I've never actually laughed at a Numberphile video before
You must not have watched enough of them, lol.
3:44 the laugh then the straight face immediately after...
He's a mathematician, he almost certainly has to act out all shows of emotion in order to interact in a socially normal manner.
bro took that personally
The other question: Why isn't 1 considered as a composite number?
If you think about the prime factorisation from any number as a infinite product of all existing primes, but with different exponents (like 15 is 2^0 * 3^1 * 5^1 * 7^0 * 11^0...), 1 also has a unique factorisation: every exponent has to be 0.
PS: I know I'm 9 years too late, hopefully someone sees this comment by chance.
Sheesh...
I'll take a stab at this. A composite number is a product of prime numbers. Since 1 is not a prime, and 1's only unique factor is only 1, it is not a composite number since it is not a product of prime numbers. Also, multiplying 1 by itself infinitely many times is pretty pointless and does nothing fruitful. Because if you were to allow 1 to be prime, every number would have infinitely many prime factors. That's why in the video, Dr. Grime said it has its own category.
@@jpoy21 Awesome explanation.
@@brahmbandyopadhyay Thanks.
@@jpoy21 you're most welcome!
3:43
Lol he laughed and then paused in a second
That would make an awesome GIF.
Gif has been made
Well don't link it here then.
Ok.
can i get a link please?
*Society of Demoted Things*
Current members: One, and Pluto
One, Pluto and floppy disk?
Pluto has just got excluded
pluto has been excluded cause they found about 200 simular siced bodys in sun orbit and it is easier to exclude 1 than to include 200 new planets
Not really, One is on a category by itself while Pluto is in a category that has hundreds if not thousands of other elements. In any case, if you feel for Pluto imagine Triton, demoted from Dwarf Planet to Moon xD
1 is special.
So you kicked out 1 because he didn't fit in. That's mean. :(
+Cellkist watch the video about the infinite prime numbers, 1 would destroy this proof. :)
+Cellkist Bwhaha, I love your comment given the backdrop of your picture :D
+Heinzriko I must admit that I have not watched it, but if they use Euclides' theorem, 1 doesn't destroy the proof. the definition is that a prime number can only be divided by itself and 1, so you don't have to divide P1*P2*P3....*Pn+1 by 1 to proof this
+Cellkist
1 truly is the loneliest number.
+Heinzriko No it would not.
I always justified this by primes have exactly 2 factors, 1 and itself, but 1 only has one factor.
That seems like a reasonable way to think about it.
I think so too
That’s why square numbers (1, 4, 9, 16, 25...) have an odd number of factors.
That’s a nice definition
Only squares have an odd number of divisors
Excluding 1 as a prime doesn't change the absolute, undeniable fact that, Optimus is a Prime. That's because he is unique. He is the product of 1 Matrix and Orion Pax, no other combination will result in Optimus Prime.
Lol
Brilliant
When he kills he is Optimus Crime
Til all are one!
1 is not a prime because it can't clear its own orbit.
But is it spherical? (Answer: only in quaternions).
I love you.
What about 2? 3 is right next to it, and is prime.
Awful analogy. 1 better fits the Sun, which isn't a planet, but way better.
@@myownmeadow1320What about it?
"You can tell it's a very important theorem because it has a pompous name"
love this guy
So glad someone finally took the time to explain this.
Really? It is a matter of convention. That's it. Explained with 6 words.
@@samueldeandrade8535 True. The convention could also easily be changed or reformulated. This video makes it seem like it’s as set in stone, as the Fundamental Theorem of Arithmetic.
"You can tell its an important theorem because it has a name. And you can tell it's a very important theorem because it has a pompous name" - grime
So "1" is the "Pluto" of prime numbers...
not really there are many dwarf planets in our solar system so Pluto isn't alone, i think there might be more dwarf planets than regular but don't quote me.
***** Yep - I was referring to Pluto because it had been (quite rightly) relegated from the list of planets.
Not quite. A more apt comparison would be to say that 1 is to primes as the Sun is to planets.
Wlerin Poetic. I like that analogy.
Referring to 1 as the Pluto of primes isn't a reference to the fact that its not a prime but because historically it was considered a prime.
1 and Pluto need to be friends.
4:14 that question blew my mind. For a moment I thought oops... what now. Empty product for the rescue
If you *had* to use primes, yes. But in reality, 1 = 1 is just fine.
@@moondust2365
8=2×2×2×1
4=2×2×1
2=2×1
1=1
Just make 1 a prime number and retry, "empty product" is a delusion
So the empty product counts as a unique product of primes? That seems like the shiftiest part of the whole deal!
Empty products. Isn't that why anything to the 0 power is 1?
yeeup, it's the same as the multiplicative identity
+Soupy If that's the rationale, then which prime number is it to the power of 0 that makes 1? The idea is that there must be a unique set, right?
+ncooty All of the prime number have to be raised to the power zero: only 1*1*1*1*... = 1. If any of the primes is raised to a power greater than 0, the result is no longer 1.
The fact that this is a unique way to write it follows from the fact that there's only one way to write an empty product: namely, no prime numbers :)
+Klapaucius Fitzpatrick Thank you. I've always wondered about that :)
+Soupy It's similar.
First Pluto, now the number 1. Whats next?
Beer
pluto is back
dslxksanta not quite yet. its still a dwarf planet and still doesn't fall within the official definition of a planet.
• Is in orbit around the sun.
• Is round or nearly round.
• Has "cleared the neighborhood" around its orbit, meaning it is not surrounded by objects of similar size and characteristics.
That last part is the problem.
but it is coming back. isnt there an asteoid int he belt that is a proto planet? it still has the label of "planet"
dslxksanta there are a number of "sub" planet categories such as proto and dwarf. Not a planet like earth or saturn, but something else.
Empty product = 1 sounds like one of the most made-up, arbitrary, things in mathematics. Sounds like it deserves it's own entire video!
I mean, literally all of mathematics is making up rules and using them consistently and seeing where it takes you.
Hahaha yes it is. But again, no harm done. That definition is there for the same reason 1 is not a prime. Its just there to make the wording of everything more convenient.
You've probably already seen something like this before, he didn't explicit it that well though.
16 = 2*2*2*2 = 2^4
8 = 2^3
4 = 2^2
2 = 2^1
1 = 2^0
Indeed, every number power 0 is 1.
Also the real definition of prime number is :
N be an integer. N is prime if, and only if, it can be divided by two *distinct* integers, that are itself and 1.
This definition already excludes 1.
@@notkamui9749 Well hang on a minute, since any number raised to the 0 power is 1, then surely 1 would still have to be excluded because 2^0 and 3^0 both result in 1. The fundamental theorem of arithmetic says that there has to be a unique product.
@@kirablagoev8534 2^0 and 3^0 represent the same product: a product of no numbers. So the product is indeed unique.
Funny thing is
My school exams still have this line when doing a prime test: "excluding the number 1"
For those asking about prime numbers, and how they fit, this is a theorem of ARITHMETIC, not ALL of mathematics. The theorem is about the prime factorization of numbers. For example, it is very explicit to say that the prime factorization of 20 = 2 x 2 x 5. But the prime factorization of eleven, you would explicitly say 11= 11. Arithmetic, however seems to be so fundamental to mathematics, that it is difficult to see a theorem that applies to just the manipulation of numbers rather than on the existence of those numbers.
Going along with what Dr Grimes said at 3:50, imagine if 1 were counted as a prime; we were constantly having to say "for all primes (except 1)" in our proofs and theorems. Mathematicians would, as they always do, come up with a new name for the set of primes not including 1, since it would have been a set that was used frequently. What would they call it? Imagine they called it the Polkadot Set, or whatever.
At that point, the only thing that changed at all is the name of the set for the primes as we defined them now to the "Polkadot Set." And, like he said, prime is just a word. Who cares what they're called? So, whether or not we call 1 prime, the set not including 1 would be used more frequently, and have some name.
omg now i understand why (number)^0 is 1
except for 0^0. they made a video on it. This answer is what you call an indeterminate form
Mind. Blown.
there is another simple explanation :
(number)^0=(number)^(x-x)=(number^x)/(number^x)=1
Fabian Herzog yes
Fabian Herzog: Bro, doesn't that introduce a new problem? Because 1 can be written as empty product of any prime. 1 = 2^0 X 3^0 is equally valid as 1 = 13^0 or for that matter 1 = 17^0 X 5^0 X 97^0. So isn't 1 evading fundamental theorem of algebra again?
1, *excluded from a category*
Pluto, "First time?"
1: "Yeah, but unlike you, I won't be included in a larger category. I'm just stuck, alone."
Still don't get why some people don't know about dwarf planets...
@@moondust2365 1 is included in many larger categories. It's included in "the integers", for instance.
well, it IS the loneliest number...
Plz more likes. Need to tell mom
First prime
That last one feels like cheating...
ComandanteJ It feels like that, but if you look at a bunch of other stuff in maths you'll find that defining the empty product as "1" fits perfectly in every case you can possibly use it in. It's really interesting to study various formulas and models and you discover that defining the empty product as 1 just makes all of them work flawlessly. It really seems to be a universal truth for some reason, even if it might be difficult to understand why.
MasterOfTheChainsaw , I think the last one "feels" like cheating also partly because you are taking the special case of considering the reduction of powers of two; rather than being more general and taking other cases as well (such as those composites which are multiples of more than one prime such as 6 and 12) to give a more fufilling description of why it is an empty product... which interestingly would require multiplying by all of the P_i primes to the power N_j
ComandanteJ It's not though. Other way of seeing it:
16 = 2^4
8 = 2^3
4 = 2^2
2 = 2^1
1 = 2^0
ComandanteJ The fact is that 1 has a special place as the neutral element for the product ; it is only natural that the empty product is 1, exactly the same way that an empty addition ("let's add nothing") is 0, aka the neutral element for the addition.
Kar Thanks , I was just about to ask that
3:43 omg when he suddenly stopped smiling XDDDDDDDD it was ike U Fokin Wot M8?!
A example illustrating that the empty product = 1 (using 2, 3, 5 and 7 as sample multipliers):
product(2, 3, 5, 7)
= product(2, 3, 5) x 7
= product(2, 3) x 5 x 7
= product(2) x 3 x 5 x 7
= product() x 2 x 3 x 5 x 7
... so the remaining empty product on the left must = 1.
The empty sum can be shown to be zero by similar means.
Well then your number can still be factorized into product() x product() x product() x product().... x product() x 2x3x5x7, which still breaks the theorem.
2 = 2 is a product? I see no multiplication going on, which is in the definition of "product". This seems to indicate a major problem in the argument suggested.
Well, no problem if 1 is prime..
2 = 2(x1)
But apparently we should not do that because then we would not got an imaginary empty product.
Its so funny because dividing by 2 in that list is like saying; "we now multiple the rest with 1"
But all of a sudden we should not follow this pattern lol
Yeah, I wondered that too! So is 2 also not a whole number?!
I'm just sharing with the RUclips community, much to my amazement, that I'm really enjoying this series of videos. Thank you for sharing them - you may have a convert...
4:45
Question: how can you call something a product if no multiplication has taken place?
+ColdsideRamrod I know right. That's what I'm trying to figure out right now.
+ColdsideRamrod Just multiply all the previous and the current identity by 1 (or by 1x1, or by 1x1x1, keeping the same number of ones in each).
+ColdsideRamrod 1 can be the product of multiple factors (1x1=1), just not any prime factors. It is an empty product because it is not a product of prime numbers.
+ColdsideRamrod I find it helpful to write out what we're saying when writing a number as a product of primes. A generic formula might be:
n = 2^(something) * 3^(something) * 5^(something) * 7^(something) *...
The exponent of each prime number tells how many times that number is in the total product. The 'unique' part of the theorem is shown in the fact that there can only be one value for each specific exponent.
Now when all these exponents are 0, we get the formula:
n = 2^0 * 3^0 *5^0 *...
n = 1 * 1 * 1 * ...
n = 1
So there is no prime number in 1 in that sense, so it is an empty product, yet still equal to 1.
+ColdsideRamrod He says that to emphasize that there was no multiplication of primes taking place, therefore one is not a prime.
Doesn't the fact that 1 is divisible only by itself and 1, and contains zero primes, make it a super prime number?
Sir, I’m always fascinated by math and numbers just as I am by music and language. I think my children have learned more from watching repetitive clips that are short from you then they have in a school setting or classroom with number of students. Please continue, maybe simplify a little more but they enjoy as do I you’re Classes.
5:10 and that's why 0! is 1.
Wouldn't it be the reason for 2^0 = 1?
@@daisybrain9423 it's actually both
@@daisybrain9423 it's the reason for both.
I like to define primes as all numbers with exactly two factors. Which in all cases would be one and itself, exept for one.
Fun one! How about "exactly two *unique* factors", that would eliminate the 1 without need for an exception, right?
Tom Solstol Yep, that works.
N Knight 27 isn´t prime
Makes sense to me. Maybe that's why I like it.
That's what I learnt in school, and that's the correct definition.
so you made a teorem with an empty product???
1:08 "Oh poor one, one is the loneliest number you you'll ever do" hmm I think I got that reference
Two can be as bad as one.
@@RonJohn63 It's the loneliest number since the number one.
It's lonely being the boss
Rest in peace Robbie Rotten! You were always my number 1.
Ok, but there is still a problem because the decomposition of 1 in primes number is not unique:
1 = 2^0 = 2^0*3^0 = 2^0*3^0*5^0 = etc...
So 1 does NOT have a unique decomposition in prime numbers... THerefore the theorem is false, isn't it ?
2^0 is not a prime number
3^0 is not a prime number
prime^n is not a prime number by definition(for n >1)
Sure, you've just defined a "bag" or "multiset" with that definition.
Also, you may want to say "$0$ for all but finitely many $n \in \mathbb{N}$". The term "almost everywhere" could mean something different (e.g., on a set with zero density).
I would not say that it is well established, in most papers they would say "for all but finitely many..." to avoid ambiguity.
And yes, precisely. So you defined a multiset over the primes, as I said.
3:44. me when someone jokes about me
Korny *OOOF*
Reminds me of Sheldon Cooper
Hmmm.. So if every positive whole number can be written as a unique product of primes, and 1 is not a prime how is 2 a whole number? What am I missing?
2, by itself, is the product. It, like all numbers, can be multiplied by the “empty product” (described at the end of the video) which is simply 1.
"A prime number is a number that can only de bivided by..." had to rewind to make sure my ears weren't dyslexic. (And yes, I know ears can't be dyslexic.)
I read that correctly, got confused, and reread it xD
4:43 i personally think you are wrong here, because the definition specifies a PRODUCT of prime numbers, and in 2=2 there is no product if 1 is not included as a prime number
ElTito Alberto
Mathematicians use the symbol ∑ for a sum of a (finite or infinite) sequence of numbers, and the symbol ∏ for a product. For example, ∑(1,2,3)=1+2+3=6, and ∏(1,2,3)=1*2*3=6.
Suppose that I have one number. What is the sum of all numbers I have? The number itself: ∑(5)=5. Now suppose that I have no numbers. What is the sum now? Zero: ∑()=0; see that ∑()+5=∑(5).
Now similarly, a product of only one number is the number itself, ∏(5)=5, and product of no numbers is defined to be 1 so that ∏()*5=∏(5).
MikeRosoftJH Well if you only have one number, how do you define if it is a sum, or a product? :P I agree it´s not very much better than just saying: "All prime no., except one...", but I guess it gives the mathematicians less to argue about, and more time to do actual work ;) hehe
MikeRosoftJH a product of no numbers isn't just defined to be 1, it is mathematically shown to be 1. It isn't arbitrary, it is a consequence of the mathematical concepts of multiplication and division.
I have a question. If every positive whole number is a unique PRODUCT of primes, and 1 is discounted as a prime, how do prime numbers fit into this theory as you cannot use 1 as a prime number, what would produce, say, 11? --> 1 x 11? but 1 is not prime... so how does this fit?
I really like how this guy talks about numbahs
1:44 "You can tell it's an important theorem because it has a name. You can tell it's a very important theorem because it has a pompous name."
I always heard "prime" defined as having exactly two factors (1, and itself), meaning that 1 could not be prime. Whether that is an accurate definition, I don't know, but because of that I have never considered 1 prime.
That is, actually, a valid reason for not including 1, as a prime; because it appeals to the definition of prime numbers.
I just think of primes as having exactly two _different_ factors.
Never knew 15 was a positive number. That's kind of cool.
What if we changed the definition of "prime" to "A number (or integer) that has exactly two dividers" and no need for all the hassle
That's what many people (particularly elementary school textbooks) have done. But this video is still useful, because it explains why someone would change the definition to that in the first place. Because you can't get around the fact that the definition of prime has _changed_ from its original definition over time.
I will say this, though. When you look for prime elements in different settings (mathematical structures which act similarly to the natural numbers), the number of divisors concept does not generalize well for a definition of prime. So that's something to (potentially) keep in mind.
Ha-ha. But using this definition the 1 should be excluded from prime numbers as well. Because it devided by 1 and itself, which is 1. But this is same and only number :)
I prefer traditional definition. Whole number can be devided by 1 and it self.
1 is not a prime. did you watch the video
+Любомудр Назарьев That was my whole point, a definition that _excludes one_ but not the others. That was literally my goal since we don't need one in it, but all the others. One is not a prime.
But by the original definition of the word 'prime' 1 is the prime to all numbers due to the fact that it is the unit of reference......to change the definition of what a unit is, you would have to use that theorem mentioned above to conclude that a unit is not a positive whole number, which it is axiomatically.....in fact, a unit is required to even understand what a 'whole number' is....
In Vietnam, we have a definition of prime that a prime is a number which has only 2 divisors, 1 and itself. So we can exclude 1 because it has just 1 divisor
3:43 THATS A MEME MATERIAL HE TAKES DRAMATIC PAUSE WHILE LAUGHING
The explanation for how to make 1 out of prime numbers is just as ridiculous as the theorem. This needed to be presented totally differently, and as a poorly defined theorem.
For the fundamental theorem of arithmetic, I prefer the definition "that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique". Without the "prime itself", you end up pretending that the primes only have one factor, but then it's only a product if you multiply by 1, which is against the definition you've used.
ccax74 He's using a more general definition of "product." Strictly speaking, multiplication is a binary operation, meaning it takes on exactly 2 inputs. No more and no less. But clearly this is not the meaning of "product" used in the Fundamental Theorem of Arithmetic (otherwise 8 and 12 and infinitely many more numbers would not be the product of primes!).
Seeing that the multiplication of whole numbers is both associative an commutative, you can generalize the definition of product to take on any finite number of inputs. Usually, when we think of this, we think of allowing more than 2 inputs, but there is a consistent way to allow 0 or 1 inputs as well.
When we have 1 input into our multiplication function, the multiplication function acts as the identity function, giving back the input. Why is this the reason? Well, if we say that our multiplication function is "Prod" then Prod(a,b) = ab.
Now, let's divide both sides by b. We then should get Prod(a) on the left hand side, and we certainly get a on the right hand side. So it makes sense to define Prod(a) = a.
By the same argument, if you divide both sides by a, you get Prod() = 1.
And so, we generalize the definition of "product" to be 1 if there are zero inputs, the identity function if there is 1 input, the standard technical definition of a product for 2 inputs, and iterated standard technical products for more than 2 (but still finitely many) inputs.
This is the sense in which he is using the word "product" in the theorem.
MuffinsAPlenty Well first of all I could point out that your Prod(a,b) logic to arrive at Prod(a) or Prod() doesn't hold true if either of them is zero (and since it's perfectly fine to multiply by zero, that's not obvious), but that's not my main point. If you just google for the fundamental theorem of arithmetic, the vast majority of definitions you'll find will include something like "is either prime or the product of primes" or point out that there could only be one prime factor. He should have just used one of those definitions for the video.
ccax74 Yes, he probably should have used that version of the Fundamental Theorem of Arithmetic for this video, as it seems to be people's biggest contention with this video.
Trabber Shir I like the main idea of your argument because certainly all positive integers can be written uniquely as a linear combination of primes (where vector addition is integer multiplication and scalar multiplication is exponentiation). And thinking about it this way, it certainly makes sense why 1 should be excluded from the primes (it ruins linear independence)...
However, the positive integers do not form a vector space (or a module for that matter). In particular, no vector (besides 1) has a negative vector. For example, if we wanted 2 * x = 1, x = 1/2, which is not an integer, let alone a positive integer. (Though the positive real numbers _do_ form a vector space)
Also, the only way you could ever hope to have a unique basis for a vector space is if your scalars form the field of cardinality 2 (Z/2Z). i.e., the only scalars are 0 and 1. This is because given any nonzero scalar c, if {x1, x2, ..., x(n-1), xn, x(n+1), ...} is a basis, then {x1, x2, ..., x(n-1), cxn, x(n+1), ...} is also a basis. (It certainly spans, since if you would have the coefficient of xn be a to get a vector v, then you can make the coefficient of cxn be a/c, and the set is certainly linearly independent as well since a(cxn) = (ac)xn, and then we can apply the linear independence of the original basis.)
So it looks like you are maybe thinking of the wrong thing. If you do happen to remember, I would love to hear it (since this is certainly a wonderful explanation if it can be salvaged!).
Trabber Shir This is probably not what you were thinking of, but I figured out something similar...
The positive rational numbers (under multiplication) can be represented as a Z-module (a module is like a vector space, but your nonzero scalars don't have to have multiplicative inverses: in this case, the scalars are just the integers). Again, scalar multiplication is exponentiation.
The prime numbers form a Z-module basis for the positive rational numbers. And of course, 1 cannot be a part of this basis, otherwise it would be linearly dependent.
(A point about uniqueness of bases: my previous argument that bases are not unique in a vector space relied on the fact that nonzero scalars had multiplicative inverses, so the same argument does not work here. However, this is still not a unique basis since -1 does have a multiplicative inverse, itself, so you can have reciprocals of primes as basis elements as well, so this basis is still not unique.)
Again, it seems like it isn't exactly what you were thinking of, but is in the same vein.
So what about the number zero? Is that a prime?
Nope. Even if you want to extend the definition to non-positive numbers, you can divide zero by any whole number and get zero again.
This is one of the best math videos I have ever seen! I’ve wondered this since fifth grade.
(Empty product )*5*3 =15
But the empty product isn't a prime ;)
Edit: Oh wait, that doesn't work either, since then 1 wouldn't be a product of primes.
In the end we can just define that this is the same thing as 5*3. We could also just say that 1*3*5 is also the same to get uniqueness but with 1 as a prime, but in the end this way looks nicer
Michael Vogel No, that would not work. The empty product is not a prime number, and the empty product IS a product of primes: an empty product of primes. There is nothing controversial about this.
If you want the theorem to work for any number including 1, you have to consider the number as a list of power of every prime number.
For example, you will say that :
15 = 2^0 * 3^1 * 5^1 * 7^0 * 11^0...
The notation is still unique (but infinite).
So, for 1, you have a unique notation too, which is :
1 = 2^0 * 3^0 * 5^0 * 7^0...
@@philippeklein7876 Correct.
1: They say I'm no longer a Prime 😢
Pluto: I know just how you feel.
Pluto was never a prime number.
@@TheBartomon Joke went a bit over your head, did it?
@@KnightOfGaea What joke? Besides 1 and Pluto can't even talk.
@@TheBartomon 1 isn't a prime the same way Pluto isn't a planet.
@@KnightOfGaea uh muh gawd. I'm messing with you. I was trying to do a joke that was even drier than yours. Woosh.
1 is to multiplication what 0 is to addition.
It doesn't do anything.
Thanks, Newton
it wastes time though
Is Optimus a prime?
Diego Jouaucon no, but in the sexy primes video, James invents the term "octimus primes."
For anyone trying to make the empty product make sense you can think of it as equivalent multiplication.
Multiplying by 6 is the same as multiplying by 2 and then 3. Multiplying by 1 is the same as multiplying by _nothing_.
It's almost like you never did any multiplication in the first place
So 1 is basically The Number-Pluto. Poor thing :(
It's a Grime number.
I taught myself (a bit) differently when, in programming, I wanted to find as many prime numbers as possible. A prime number is not divisible by any other product of prime numbers. In other words, if you had 11, it would not be divisible by lower prime numbers. If 1 were a prime number, then 11/1=11 and thus is not a prime number.
robotica actually, a prime number is divisible by one and itself only. And that is the defenition. Yours doesn't make much sence, and I speak as a programmer
Walwalkn Wewnrkl As i said, the code does work, but since the formal definition of a prime number is: "A number only divisible by itself and 1", logically speaking you can't say that one isn't a prime because you use another definition. Basically, 1 isn't a prime because of is definition exists, but rather his definition works because 1 isn't a prime. As they said in the video, the only reason 1 isn't a prime is because it makes other stuff in maths easier, such as his program. Hope that helps :)
m1g4s It didn't mean to state a definition, I wanted to state an explanation. You can't really say that "a prime number is a number which is not divisible by any lesser prime number", but using examples you can get another perspective to why 1 really isn't a prime number, because it follows another definition.
robotica I know what you meant, but as I explained, your code works because one is not a prime. 1 being a prime is for a different reason, hence why the video proved your code rather than your code proving the video right
m1g4s "hence why the video proved your code rather than your code proving the video right" It's actually both. The video inspired to make a program which would require the first prime number, and when 2 was the first prime number which would make the code work would help people also give an "argument" for 1 not being prime.
Why is 1 the result of an empty product? Here's why:
2^3 = 2x2x2 = 8
2^2 = 2x2 = 4
2^1 = 2 = 2
2^0 = __ = 1
3^0 = __ = 1
n^0 = __ = 1 (any number multiplied by itself 0 times is always 1 -- that was just decided some time, and I'm sure it's since been thoroughly proven, however counter intuitive it may be).
I wish they included the P^0 part in the video, because that would have helped clear up why the empty product is 1.
Even 0 can be excluded by this definition, because while in theory it can be divided by itself (0/0 = 0) and 1 (0/1 = 0), there is no product of primes that results in 0, even the empty product stops at 1.
im not sure about the last part with 0. As you said n^0 = 1. But at the same time 0^n is always 0. What happens when those rules collide?
Gustav Mårdby It depends on what model you use - 0/0 limits on 1 for x/x, but on 0 for 0/x. Like most situations that involve dividing by zero, it's nonfunctional - the result will change depending on the context.
Gustav Mårdby They explained it in some other wideo. Look for it.
the way i explain it:
x^5 / x^5 = x^(5-5) = x^0
any number divided by itself = 1
Hence x^0 = 1
0/0 is not 0. 0/0 is undefined.
Very strange theorem:
Now please apply this theorem , and give me a solution for
1. Write 1 as a product or prime number.
2. Write any prime number as product a product of other primes.
1. It's in the video, it's the empty product. A product of zero primes.
2. it's just the number itself. Of course you can't wrtite it as a product of other primes, that would conradict the theorem. And yes, "2" by itself can be considered a product. If you accept that products might consist of one number, or no numbers at all, everything will make sense.
Well, of course you can argue these definitions and you can disagree with them, so you can say that a product is at least two numbers, and thus prime numbers themselves can't be written as a product of primes. But that doesn't contradict the theorem - you had to rephrase it, sure, but the actual mathematical meaning would stay the same.
Rajesh VP Easily. 1 = 2^0*3^0*5^0*••• There, you have a unique factorization. Now, p(n) = 2^0*3^0*•••*p(n)^1*••• There, you have a unique factorization. Happy?
Math Wiz Yes, because by definition a prime has only 2 factors: itself and 1. If you can get there by multiplying other numbers, it’s not a prime!
But (qua the broader theme of this thread) it’s still silly to say that a prime number is the product of only itself, since ‘product’ implies a generative process and by that definition primes just... *are* !
@@angelmendez-rivera351 Thank you for proving my point.
feuilletoniste *it's still silly to say that a prime number is the product of only itself, since 'product' implies a generative process...*
No, it does not. In fact, mathematically speaking, the idea of a generative process you are trying to appeal to is vague and meaningless. On the other hand, 'product' has a precise mathematical definition, and that definition accounts for unary products just fine. So no, it is not silly, it is simply the way math works. Maybe you should try studying some group theory before you make blatantly false statements such as this one.
Whoa. Empty Product. I like the way it sounds.
But you can't say 2 is a product of primes if its only 2, because you have to multiply it by something (1) to have a product
Maybe this will clear it up: For every prime p, p^2 (its square) is semiprime; 2(prime)^2 = 4(semiprime); 11(prime)^2=121(semiprime), etc.
Assume 1 is prime. 1^2 would be semiprime. 1^2 = 1. So if 1 is prime, 1 is semiprime. See how that doesn't work? So either accept that 1 is not prime, or go back to saying "except 1" every time you talk about primes. I know which I prefer.
Since it has FEWER factors than a prime, I'm sticking with my suggested desgination for 1 as "SUBprime", mortgage scandals be damned.
But if so, the Theorem is wrong... 2 is a whole number and can not be described as a product. Also 1 is a whole number and has the same situation, except that it does if you include one. So... they added: "greater than one and can be represented by itself number". Smh, seems like they took that on their own favor.
On what you remark, you could take same card and said "For every prime p greater than 1, p^2 (its square) is semiprime".
I see one as a real prime.
1's special category is a unit: anything with a multiplicative inverse in its own set (closed).
1^-1=1 an integer. 2^-1, 3^-1, etc are not integers. So for positive integers 1 is the only unit.
You used negatives. Once negatives are involved, 1 and -1 are both unuts.
Sooooo 1 is sort-of prime in a similar way that a free proton is sort-of a Hydrogen atom?
ion*
+Superiorform cation
+Deven McKee No, As an official chemistry nerd that is simply not possible, a proton by itself is extremely unstable and cannot occur naturally. Hydrogen can only exist as a per of atoms in nature. (H2, Not H)
+Sonny Zheng Old comment ik, but aren't all cations ions so we are both correct?
As a chemistry nerd you only consider elements in the presence of and interacting with other elements chemically, where, yeah, it's gonna push HARD for equilibrium. As a physics nerd you'd know how, where, and why you would be chucking around naked protons.
So 1 is a product of [null]x[null]?
My brain cannot handle this.
No, it isn't. It is a product of .
I mean, it's a product of []. There is nothing in brackets. It's just an _empty_ product.
Same problem here... for me 0 x 0 would mean you have "zero times nothing".... but wouldn't that mean it can be everything? Or infinity? Every thought goes a different way, but none of them ends up at 1
1 is a product of no primes, which means it can be a product of 1, because it is not a prime. 1 x 1 = 1, but 1 x 1 has no primes in it, so the result, 1, is a product of no primes.
You're over-thinking it. We altered the definition of a prime number slightly just to make it more convenient for us. It's just a shortcut that doesn't affect the physical laws of the universe. It's like deciding whether to use the 12-hour or 24-hour clock. Time is not affected by our choice, only the way in which we describe it.
I think null and 0 are the same thing, so a more accurate way to describe it is essentially like in the video, as 2^0
And here I was thinking there were no politics in math. Get out of here one, you're ruining our agenda even though you are perfectly valid in every other way.
For the same reason, 2 should be excluded as well, because there are a lot of theorems involving "odd primes". Maybe 2 is related to the "big primes" or it is such a nice guy so that the other primes don't want to exclude it...
Joseph Douglas If you think of primes as integers with exactly two factors, then that makes it work. Or another way of thinking about it is that in purposes like the fundamental theorem, multiplying by one does nothing, so it can basically be removed from the factorization.
@@KnakuanaRka but they teach a prime as a number which can only be divided by 1 and itself, and then the smartass kid says "so is 1 a prime" and the teacher says "no because it's 1 and itself" "yah and you said 1 and itself and 1 is itself" "okay, primes are numbers divisible by 1 and 1 other non-1 number." "but what happened to your other definition" "shut up". why can't they just let 1 be prime and be done with it? And there's nothing really that cool about primes except that they don't have a lot of factors.
No politics in maths? Tell that to Hippasus.
Peter Senior You really don't know anything about applications of number theory if you think prime numbers are not important. And there are many other reasons 1 is not a prime.
Thank you so much, I wuz so confused and puzzled thinking why it isn't a prime number.
0 and 1 are two great numbers and there's a real secret behind them!
With the Fundamental Theorem of Arithmetic being the way that it is, doesn't that mean that ultimately prime numbers are the only actual numbers? And all other numbers are creations of primes?
That’s what he said when he was talking about atoms
If 1 and 0 were prime numbers, they'd be the only primes that are perfect squares. All other perfect squares are not prime because in addition to one and themselves, they have a factor that, multiplied by itself, equals the perfect square (e.g. 3 is prime, but 3x3=9, and 9 isn't prime because it's a multiple of 3).
I'm still not convinced 1 is consistent with the theorem. You're saying that 1 is the product of zero prime numbers and is thus consistent with the theorem. However, is this empty product really unique? The example shown was essentially 2^0 = 1. The product of 0 2s is 1. But that can be applied to any prime number. p^0 = 1 where p is prime.
Prime numbers have 2 devisors while number 1 has only one devisor
number one, you have the bridge!
Now that 0 and 1 are out, 2 should be worried.
2 is not a Gaussuan prime.
2 = (1+i)(1-i).
A number is prime or not with respect to a set of numbers. In the integers, 2 is prime. Gaussian integers are a+bi where a & b are integers.
In the rational numbers (or any field) nothing is prime. Rvery value would be zero or a unit (values with a multiplicative inverse in the set of numbers). By definition neither are prime.
By real definition, not grade school definition.
Undefind - in that case let’s cut to the chase: All primes less than 23 should be worried; they’re not as much fun as their big brothers.
2 doesn't have anything to worry about. Nobody thinks 2 isn't a prime.
Angel Mendez-Rivera 😂
@@angelmendez-rivera351 wait until people come complaining that 2 isn't a prime cos 2 is even :^)
Wow - what a nifty idea. Redefine primes so the theorem is correct. I bet you could prove a lot of theorems using this approach.
The theorem is correct for all primes, but not 1 and has thousands of logical consequences. (It also happens to be the only reason your bank account is safe from cyberattacks.) so excluding 1 doesn’t change anything about the theorem, just that the wording can be cleaner.
right ! 1 is Prime its literally the first prime number, and it does fit the definition of what a prime is. Prima materia first matter prima primero first , 2 is not first lol its second secunda segunda matteria second matter , so why in the world would prime numbers start at 2. prime numbers are extremely important. i know theres a why it was changed, the real reason behind it but are we ever going to find out the why ? i can prove 1 is prime already visually without any fancy formulas. geometry is all you need to observe the fractions and what they become
@@JOELRODRIGUEZ-lk9gu *1 is prime, its literally the first prime number,...*
No, it is not. 2 is the smallest positive prime number (with -2 being the greatest negative prime number).
*...and it does fit the definition of what a prime number is.*
No, it does not, and I would bet real money that you do not know what the definition of a prime number is. You probably think the definition of a prime number is "a positive integer whose only positive divisors are 1 and itself." If you do believe this, then you are in the wrong, because this is not the definition of a prime number. It has not been the definition of a prime number for at least a century now. A prime number is an integer whose only positive proper divisor is 1. The number 1 has no proper divisors at all, so it is not a prime number, and for the same reason, neither is -1.
*Prima materia = first matter = prima = primero = first*
I am glad you know the etymology of the name "prime number," but ultimately, the etymology is irrelevant to the mathematical definitions. My name is Ángel, and the etymology of my name is the Greek word άγγελος, which means "messenger." I am, however, not a mailman, nor a courier. The etymology of my name has nothing to do with the characteristics that define me as a person. Similarly, the etymology of a word used as an English name for a mathematical concept has nothing to do with the definition of the mathematical concept.
*2 is not first lol, its secunda materia, second matter*
Numbers are not made of matter. Numbers are abstract concepts which exist only in the mind of sentient beings which are capable of doing mathematics. That being said, the reason the prime numbers were named after 'prima materia' as a metaphor is that they took the atomic theory of matter as an inspiration: atoms, also called first matter, were considered to be fundamental, elementary particles comprising all matter. Any particle of matter could eventually be reduced to atoms, and atoms built all other particles. The prime numbers play a similar role with the integers: the prime numbers are analogous to atoms, in that they cannot be multiplicatively composed of other numbers, and in turn, the integers are multiplicatively generated by them. -1 and 1 are not prime numbers, because they do not fit that role: they actually do not generate any integers at all, since 1•(-1) = (-1)•1 = -1, 1•1 =
(-1)•(-1) = 1. -1 and 1 are trapped in their own bubble, incapable of generating the other integers. The same happens with 0. The prime numbers 2, -2, 3, -3, 5, -5, etc., are the ones we need to generate integers. 1 is the empty product of primes, and then we have 2, 3, 2•2, 5, 2•3, 7, 2•2•2, 3•3, 2•5, 11, 2•2•3, 13, 2•7, etc. In the other direction, we have -1, with (-1)•(-1), -2, -3, -2•2, -5, etc.
*Prime numbers are extremely important.*
Yes.
*I know there is a why it was changed, the real reason behind it, but will we ever find out the why?*
I strongly recommend you watch Another Roof's video on the topic, because the truth is, for most of history, 1 was not considered to be a prime number. Prior to 1600s, it was not considered to be a prime number. It began to be considered a prime number due to a misunderstanding of the concept, but in the 1800s, they stopped considering it a prime number again. Furthermore, we know exactly why both things happened, and we have known ever since they happened. There is absolutely no mystery at all. I am sure this has actually been explained to you before, but you just were not paying attention.
*I can prove 1 is a prime already visually, without any fancy formulas.*
False. I know you cannot do that, because the definition of a prime number is given by a formula, not by any geometric concepts. You _think_ you can prove it, but this is how like children _think_ Santa Claus is real.
@@angelmendez-rivera351-2 is not a prime? 1, -1, 2, -2... that's 4 divisors
@@shaurryabaheti All prime numbers have exactly 4 divisors. Both -2 and 2 have exactly 4 divisors (they have the same divisors). Therefore, if 2 is a prime number, then so is -2, and vice versa.
1:34 Every positive whole number can be written as a unique product of primes... wouldn't that exclude 2, 3, 5, 7, or any prime number for that matter? I'll continue the video now.
MrDajdawg 2=2
3=3
And so on.... these numbers are already prime so the product is just themselves
@@adamzeno6014 A product is a result of multiplication right? What are you multiplying 3 by to get 3?
I think that the theorem states that every integer greater than one either is a prime or a unique product of primes, but I could be wrong
Knicksboncho Nothing. A multiplication can be done with a single factor. In fact, it can be done with 0 factors. That is exactly what x^0 stands for. And we know x^0 = 1. So there you have it.
@@angelmendez-rivera351 Multiply 3 by a non-one number to get 3
1:20 The first time he said make me confuse because he didn't say unique which is VERY IMPORTANT word (2:15) ?
Both are true. It's much easier to prove the case where we don't insist on uniqueness.
Wait, if "every positive whole number can be written as a unique product of primes", then 1 must not be a positive whole number, regardless of whether it is considered prime or not. I don't think that theorem is written down correctly.
amgemin It is a whole number. It can be written as a unique product of primes. They explained it in the video. It is called the empty product.
@@angelmendez-rivera351 - that "empty product" idea is really powerful and whoever realized it is amazingly smart
Corpsie Corpsie Not at all. The notion of the empty product is completely sensical and consistent with the rest of mathematics. The idea even pre-exists the fundamental theorem of arithmetic. The idea dates all the way back to the invention of algebra itself and the discovery of division.
@@angelmendez-rivera351 - what do you mean?
Corpsie Corpsie What are you confused about?
always just assumed it was because prime numbers had to have two separate factors of one and itself, whereas 1 just has one factor. Really interesting video!
Well that actually is it you’re just looking at it from a different angle
Semantically the base definition implies that "the number" ISN'T one. Such ambiguity could have been avoided; however, I submit the prime word of the definition as the "real" issue: EVERY does include 1. Thus it should also be divisible by two distinct WHOLE numbers, which is false-if your interpretation of the definition… is "rigorous". It's my opinion the word "and" precludes one from the possible values of "ITSELF".
We can reformulate the fundamental theorem as "all whole number have a unique factorisation to prime numbers up to multiplication by unit " to avoid such randomly things ...
Yes, you can. But the theory is arguably much cleaner when allowing empty products/unary products. Reformulating the definition of prime to allow for arbitrary finite products (rather than solely for binary products), units _legitimately do not satisfy_ the definition of prime. So we don't have to "randomly" exclude them from the definition of prime.
Poor One! 😭 It's the Pluto of math!
I know a more excactly definition of primes: A prime is a number, which can be divided by ONLY 2 different numbers (1 and itself). Then the problem would be solved
(divided by 1 = 1/1
divided by itself = 1/1 -> not 2 different numbers)
100 % agree ! if 1 is a prime number, 1-1=0 make Euler product : multiple of (p-1)/p equal to 0, make R.O.S.E(realization of sieve of Eratosthenes) impossible, for example : pi(4)=4*(2-1)/2+0/2+1-1=2, pi(9)=9*((2-1)/2)*((3-1)/3)+1/2-3/6+0/3+2-1=4, pi(25)=25*((2-1)/2)*((3-1)/3)*((5-1)/5)+1/2-1/6-5/10+25/30+1/3-10/15+0/5+3-1=9, pi(x) is exact prime counting function, formula is : x time reverse of Euler product (p-1)/p plus total of mod(x,Po)/Po+ total of p-1.(p is prime number less and equal to x^(1/2), Po is all possible combination of p)
enlong chiou boi wtf u sayin
Ashutosh
How much brown paper do you have?
Numberphile : Yes.
😂
Since the number one is so special, could we call it a "prim" number (or *the* prim number)?
I like the comment that calls it "the origin number", especially since it really is that. All whole numbers come from 1! (They're literally just groups of that!) :D
so is this why if you give a number the power of 0 the product will always be 1?
Yes.
cool, I learned something
As long as you don't try 0^0, which is undefined. (Numberphile probably has a video on this somewhere.)
***** No it's not. It's undefined.
0^0 is top-down defined as 1 in the sense of multiplicative identity. Follows from the definition that n^-n = 1/n^n
A non-empty symbol for empty no-thing is of course logically very problematic. Peano axioms start from defining 0, and whole lot of problems follow from that. 0 is nice tool for making some arithmetical operations easier, but as logical foundation of number theory... less than ideal.
I don't like this formulation of the theorem, it means that "2" is a "product" of one number. I think the theorem is better formulated that every number is either PRIME ITSELF or the product of primes.
But then you'd have to include 1, which they have decided is just in the way of most calculations using primes as a set. So when you want 1 removed from the set in the majority of the cases, that's a strong case for it not belonging there...
athox No, you're missing my point. 1 fits with the definition here too. Whether 1 should or should not be included in the definition is a separate question. I'm saying the definition James gives here forgets to mention that prime numbers should not be included. (That is the definition I have seen, I'm saying, I do not doubt James has some justification for writing it as he did.)
greg55666 Why shouldn't prime numbers be included? The prime numbers have a prime product too.
You can agree that n^2 is the same as n x n, right? As in, n multiplied by itself twice.
Equally, n multiplied by itself once, must also be a valid product, namely n.
n = n^1 = n
See? You cannot get past the n^1 as being a valid multiplication, even if it only has one factor.
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Digression:
Also, n multiplied by itself never, is also a valid product. Which is why n^0 = 1 has no factors, prime or otherwise... If you include 1 in primes then you can do 1^1 = 1, but then we're back to intentionally excluding it for practical purposes.
I think it kinda goes without saying seeing as the *definition* of primes is that they are explicitly not made up of a product of any two other numbers except themselves and 1. Seeing as the theorem includes the use of the idea of prime numbers then it implicitly also excludes primes from needing to comply with its stated definition.
athox n^1 "must" also be a valid product is just circular--I'm questioning whether it should be included.
I think your digression proves my point. I'm saying "n^1" is not a product. Let's get back to that in a moment. You say, not only is n^1 a product, but so is n^0.
If you count n^0 as a product, you have broken the theorem. Because then 1=5^0, 1=10^0, 1=453235^0, etc. So you MUST exclude n^0--n^0 cannot be a "product" for the purposes of this definition.
Why are you excluding n^0 as a "product," but then objecting to me wanting to exclude n^1 as a "product"? You've lost whatever principle you were going by.
The point of this theorem, as I understand it, is that there are two kinds of whole numbers: prime numbers and composite numbers. The point is that there is no number that is NOT either prime or composite. By not saying every number is either prime or the product of primes, it seems to me you lose the entire point of the theorem.
I think a better explanation is that saying "a prime number is a product of itself" is trivial, so it's unimportant either way. Stating it as it is in the video is more concise at the loss perhaps of a little more insight into its point; saying "either prime itself or a product" is more words, hence perhaps less clear even though perhaps it points more to the meaning of the theorem. But either way, the statement that a prime is a product of itself is trivial, so it is the author's choice.
As I said in my original post, I would have chosen the more words with the deeper meaning.
So it kinda just comes down to the choice of whether you would rather say, "all positive whole numbers can be expressed as a unique product of prime numbers (excluding 1)" or "prime numbers are numbers which are only divisible by 1 and themselves (excluding 1)"
Admittedly saying the definition of a prime is a number that has exactly 2 unique divisors gets around that.