Fantastic proof. I also really appreciate how you replied to a lot of the questions in the comments. If it wasn't for your replies, I wouldn't have understood it. Thanks a ton!
I understood that. So either it was a very clear and lucid explanation or I am a genius. I am going to go with the latter. Seriously though, great video.
Both can be true! Thanks for watching! You might also like this video where we prove square root 3 is irrational: ruclips.net/video/sLaAU49zBRo/видео.html&pp=ygUkc3F1YXJlIHJvb3Qgb2YgMyBpcyBpcnJhdGlvbmFsIHByb29m
I don't understand, aren't we assuming two things:- 1. Square root of 2 can be expressed as a ratio of two integers, a and b 2. The ratio is in lowest terms Why does it have to be in lowest terms, wouldn't the number still remain the same even if the fraction is not in lowest terms?
Thanks for watching! Yes, we are assuming root two is rational, since this is a contradiction proof we begin with that assumption, so that we can show it leads to a contradiction - thus the assumption mustn't be true. When we work with an unknown fraction, we have to name its numerator and denominator, a and b in this case. However, like you say - fractions don't have unique representations, they can be expressed in infinitely many ways but still have the same value. But, since every fraction can be uniquely expressed in lowest terms, it is essentially extra information we get for free if we specify that is the expression we are using. If I say r is a rational number and r = p/q, then I don't know anything about p and q except they are integers and q is nonzero. However, if r is rational then it can be expressed as a ratio in lowest terms, so I could just as easily say r is a rational number, expressed in lowest terms as p/q. Then I know p and q are integers, q is nonzero, AND p and q have no common factors. The fully reduced fraction is like the bare essential form of the rational number. Working with any other form involves unnecessary fluff. In this particular proof, showing a and b have a factor of 2 in common isn't a contradiction unless we specify originally (which we did) that they are representing root 2 in lowest terms, and so cannot have any common factors. Hope that helps!
@mohamedelzoheiry1413 0 seconds ago the solution of square root 2 in nearest fraction is 1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796 the solution of square root 2 is 25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136 1.4142136 = (2)^(1/2) the solution of square root 2 is 11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136 1.4142136 = (2)^(1/2)
@@WrathofMath @mohamedelzoheiry1413 0 seconds ago the solution of square root 2 in nearest fraction is 1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796 the solution of square root 2 is 25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136 1.4142136 = (2)^(1/2) the solution of square root 2 is 11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136 1.4142136 = (2)^(1/2)
1. Sqrt(2) being a ratio of two integers is the definition of being rational so proving sqrt(2) cannot be expressed this way is precisely what it means to prove that sqrt(2) is irrational. 2. The reason we can assume that the expression is in lowest terms is because if the fraction were not in lowest terms I can simply force it to be. Let p/q be a fraction not in lowest terms, then let n = gcd(p, q). So, p/q = (p/n) / (q/n). Notice that the right hand side is always in lowest terms. So when say that sqrt(2) = p/q, we are allowed to assume that p/q is in lowest terms because I can always just pick (p/gcd(p,q)) / (q/gcd(p,q)) which is in lowest terms by definition. However I’d rather just write p/q and assume gcd(p,q) = 1 because that is far less cumbersome. The point is that this is the same thing
Nice. I've watched other proofs of this idea, but those other proofs didn't go into the idea of 'k'. By doing that, you created at a compelling argument that is easy to follow, and which shows that the original assumption (a and b being relatively prime) must be false. Well done!!!
We don't. The proof hinges on a and be not both being even. We do that because we assume a/b is the lowest form of the fraction. And Wrath says "assume," but I think it would be less confusing to say "define". Every rational fraction has a lowest form, one where a and b share no common factors, and we are defining a and b such that a/b is the most reduced form of the fraction.
The most general version of this proof is one which shows that every natural number that is not a perfect square has an irrational square root. This obviously includes 5. Let n be a natural number such that n is not a perfect square (more precisely, there does not exist an integer p such that p^2 = n). We show that sqrt(n) is irrational. Assume for contradiction that sqrt(n) is rational. We know that sqrt(n) is positive by the definition of the square root function. Since sqrt(n) is rational and positive, by definition there exist natural numbers, p and q, such that sqrt(n) = p/q. We assume without loss of generality that p and q are relatively prime (meaning gcd(p,q) = 1, i.e p and q have no common factors other then 1). If p and q were not relatively prime then p/gcd(p,q) and q/gcd(p,q) satisfy the prior conditions. Therefore we can assume that p and q meet the aforementioned criteria. So sqrt(n) = p/q -> n = p^2 / q^2. But if q^2 divides p^2 then trivially gcd(p^2, q^2) = q^2. But gcd(p^2, q^2) = (gcd(p, q))^2 = 1. So q^2 = 1 so clearly q = 1. But that means n = p^2 and that’s a contradiction.
Maybe I misunderstood something. But the initial hypotheis was it is a fraction AND in its lowest terms. You derived a contradiction. This means the initial statement is false ie it is not a fraction OR not in its lowest terms OR both. It doesn't necessarily mean it is not a fraction - it may simply not be in its lowest terms. I'm sure I got something wrong ...
Nope. He clearly stated that the fraction IS in it's lowest terms. That means that a and b should not have any common divisors. BUT since we arrive at the fact that they do have common divisors that must mean a fraction actually cannot be put in it's lowest term which means it's not a fraction because any fraction can be put in it's lowest terms and if it can't then it's not a fraction. P.S.: Sorry for bad wording, English is my 3rd language
@@user-mo8in1by7h "A fraction is said to be written in the lowest term if its numerator and denominator are relatively prime numbers, i.e. they have no common factors other than 1. In other words, a fraction is said to be in its lowest terms or lowest form, if the HCF of the numerator and denominator is 1." Just learn how to google, please. Lowest term of 4/2 will be 2/1.
Hello! Is it generalizing things too much if I just thought that if a^2/b^2 = 2 is true, then the absolute value of a has to be exactly 2 times the absolute value of b for the fraction to be two. So then we have 4b^2=2b^2 which is false. Your proof is great, leaves nothing unexplained but I suppose I was on the right tracks at least, I kind of proved that they must have a common factor of 2 but I was not clever enough to think of it that way! I will definitely watch more of your stuff and subscribe, my studies are not that math heavy but I suck at it since I didn't take high school seriously so I am trying to learn some proof writing on my own. Thanks for the video!
The bit about being in lowest terms bothers me. I mean given the fact that there are an infinitude of pairs of values to check, how can we be so certain that we have lowest terms?
The reasoning is that every fraction CAN be expressed in lowest terms, and so if we assume we can write sqrt(2) as a fraction, let's not just say it equals x/y - let's go a step further, since we can, and take the sqrt(2) fraction in lowest terms, say a/b. It's legitimate to do this because if sqrt(2) does have a fraction representation, which we're assuming it does, then it also has a fraction representation in lowest terms.
Here's a simpler way. ✓2 = p/q => 2 = (p/q)^2 => 2q = p^2/q Since 2q is natural And there's no common factor between p and q, Therefore p/q or p^2/q can't be natural. Hence, 2q = p^2/q can't be true.
Why exactly can't a and b have any common factors? I tried the proof with n instead of 2 and it seems to work, "proving" that there are no rational numbers at all... My math is most certainly wrong, but please tell me how
Thanks for watching and good question! In the contradiction assumption, we assumed sqrt(2) could be written as the ratio of two integers, say a/b. Then certainly, either a/b is already a fully reduced fraction (meaning a and b have no common factors) or it could be reduced further if a and b do have common factors. So, let's say a/b has been fully reduced. Does that make sense? We know if sqrt(2) is a fraction, it can be written as a fully reduced fraction, and so that's how we'd choose to write it. The proof would not be able to apply to just any number, because one of the key steps uses the fact that a/b = sqrt(2). So if sqrt(2) were just n, the rest of the proof wouldn't work. Hope that helps! Edit: Here is a link to a similar proof if you're interested: ruclips.net/video/sLaAU49zBRo/видео.html
@@WrathofMath Thanks for answering. The problem I see is that nobody says "let's just assume this real quick". If a and b would be allowed to share factors, the proof wouldn't work. It's kind of a big deal, but I still found no logical reason for that
i wanna question the man who created this que did he really didnt know root 2 is irrational why does he need to create such an illogical question, nice explation btw
Do you mean what we are accomplishing by putting it there? Or do you mean why k instead of some other letter? k happens to be a common choice for arbitrary integers. n for example wouldn't be a great choice as it's generally reserved for positive integers, and x is typically a real number. So just sticking with pre established convention.
Did we just prove that the square root of 2 cannot be a rational number? I watched some other video that I didn't quite understand that show that the square root of 2 exists using some kind of bounded theorem. Can you od this proof so we can understand? Thank you
You said "Can you od this proof so we can understand?" Is 'od' a typo? I guess you probably mean 'do'. Yes, the square root of 2 is irrational but it DOES exist by the completeness axiom, which is probably what the other video was talking about. The square root of 2 is the least upper bound of a certain set of rationals, so the real numbers contain it.
The proof starts with three important statement. We assume for contradiction that sqrt(2) IS rational. This means we can write it as a ratio of two integers a/b. But if we can write it as a ratio of integers, we can also write it as a fully reduced ratio of integers, so we assume it is written that way and thus a and b cannot have any factors in common otherwise we could reduce the fraction further. If it were 4/8 for example, this wouldn't be possible because we assumed it was written in "lowest terms" and so the common factor of 4 would have already been cancelled out.
The contradiction is that we took a/b to be in lowest terms; so if it turns out it isn't in lowest terms - that is a contradiction. The sequence is like this; if a number is rational then it can certainly be expressed as a rational number in lowest terms, so we assume for the sake of contradiction that root 2 is rational, and thus it equals a/b in lowest terms. But assuming this leads to a/b being further reducible, and thus a/b couldn't have been in lowest terms, and so root 2 couldn't have been rational. Another way to think of it is that assuming root 2 is rational leads to an infinitely reducible fraction, which is not possible.
I understand the contradiction that it’s not in its simplest terms but a/b would still be rational even if it had a common factor. I can do it but can’t get my head around this 🤯
But we aren't proving a/b isn't rational. a/b is certainly rational, it was constructed precisely to be rational. We're proving that if sqrt(2) = a/b, a contradiction is found.
I understand there’s a contradiction because a/b is not fully simplified but a/b is still rational simplified or not and therefore we are still showing root 2 as rational.
Every rational fraction has a form where the numerator and denominator do not have factors in common. 3 and 6 have common factors, but 3/6 can be reduced to 1/2. a/b is defined as the √2 fraction such that a and b are both integers with no common factors. If √2 was rational, such a fraction would need to exist.
seems like an old video but I would very much like to understand this but i just cant get it is there any way we could chat on discord or something and you could help me understand this ?
and know imagine ancient alexandria with Euclid as teacher, proving this for the first time to a stunned audience. other civilizations did such math also ofc, but irrational numbers were a very new thing back then. our dear trigonomy guy Pythagoras hated the idea btw..... yep, that stuff is >2000 years old, the proof is valid without improvement since ~300bc
The idea is that, when we assumed we could write sqrt(2) as a/b, we assumed a/b was in lowest terms. This means it cannot be reduced, and so a and b mustn't have common factors otherwise a/b COULD be reduced. It's fine to assume the fraction is in lowest terms because IF sqrt(2) can be written as a fraction, then it can be written as a fully reduced fraction, so we just assume we're working with that fully reduced fraction.
If a was odd, a^2 would necessarily be odd. Here is a proof of the result about evens: ruclips.net/video/FewsjiKug8Q/видео.htmlsi=7moxfkJXMNfUfQJy If a is odd, it must not have a factor of 2, in which case a^2 doesn't have a factor of 2 either, in which case a^2 would be odd. So if a^2 isn't odd, then a isn't odd and so a is even.
I hear you say "k". But I see you write... actually I do not know how to describe what you write. It's not an American Alphabetical Letter. That crazy symbol you use SCARES ME!!! :0
@@DeVirgilWouldn't the ∈ symbol actually indicate that k is an element of the set of all integers? If he wanted to write that k is a subset of the integers, then he would have written {k} ⊆ Z.
If √4 is irrational, then √4 != a/b Let us Assume that: √4 = a/b Then: 4 = a^2/b^2 4b^2 = a^2 2b = a Substituting 2b = a into 4 = a^2/b^2: 4 = [(2b)^2]/b^2 4b^2 = 4b^2 So √4 =a/b, therefore √4 is not irrational (it is rational)
@mohamedelzoheiry1413 0 seconds ago the solution of square root 2 in nearest fraction is 1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796 the solution of square root 2 is 25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136 1.4142136 = (2)^(1/2) the solution of square root 2 is 11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136 1.4142136 = (2)^(1/2)
@mohamedelzoheiry1413 0 seconds ago the solution of square root 2 in nearest fraction is 1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796 the solution of square root 2 is 25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136 1.4142136 = (2)^(1/2) the solution of square root 2 is 11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136 1.4142136 = (2)^(1/2)
Fantastic proof. I also really appreciate how you replied to a lot of the questions in the comments. If it wasn't for your replies, I wouldn't have understood it. Thanks a ton!
niko oneshot :o
@@jfinxindid Yessir :3
Proof so good it got Hippasus killed by Pythagoras himself
yup
I understood that. So either it was a very clear and lucid explanation or I am a genius. I am going to go with the latter. Seriously though, great video.
Both can be true! Thanks for watching! You might also like this video where we prove square root 3 is irrational: ruclips.net/video/sLaAU49zBRo/видео.html&pp=ygUkc3F1YXJlIHJvb3Qgb2YgMyBpcyBpcnJhdGlvbmFsIHByb29m
I don't understand, aren't we assuming two things:-
1. Square root of 2 can be expressed as a ratio of two integers, a and b
2. The ratio is in lowest terms
Why does it have to be in lowest terms, wouldn't the number still remain the same even if the fraction is not in lowest terms?
Thanks for watching! Yes, we are assuming root two is rational, since this is a contradiction proof we begin with that assumption, so that we can show it leads to a contradiction - thus the assumption mustn't be true.
When we work with an unknown fraction, we have to name its numerator and denominator, a and b in this case. However, like you say - fractions don't have unique representations, they can be expressed in infinitely many ways but still have the same value. But, since every fraction can be uniquely expressed in lowest terms, it is essentially extra information we get for free if we specify that is the expression we are using.
If I say r is a rational number and r = p/q, then I don't know anything about p and q except they are integers and q is nonzero. However, if r is rational then it can be expressed as a ratio in lowest terms, so I could just as easily say r is a rational number, expressed in lowest terms as p/q. Then I know p and q are integers, q is nonzero, AND p and q have no common factors. The fully reduced fraction is like the bare essential form of the rational number. Working with any other form involves unnecessary fluff. In this particular proof, showing a and b have a factor of 2 in common isn't a contradiction unless we specify originally (which we did) that they are representing root 2 in lowest terms, and so cannot have any common factors. Hope that helps!
@@WrathofMath Ohh, I get it now, thanks a lot
@mohamedelzoheiry1413
0 seconds ago
the solution of square root 2 in nearest fraction is
1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796
the solution of square root 2 is
25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136
1.4142136 = (2)^(1/2)
the solution of square root 2 is
11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136
1.4142136 = (2)^(1/2)
@@WrathofMath
@mohamedelzoheiry1413
0 seconds ago
the solution of square root 2 in nearest fraction is
1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796
the solution of square root 2 is
25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136
1.4142136 = (2)^(1/2)
the solution of square root 2 is
11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136
1.4142136 = (2)^(1/2)
1. Sqrt(2) being a ratio of two integers is the definition of being rational so proving sqrt(2) cannot be expressed this way is precisely what it means to prove that sqrt(2) is irrational.
2. The reason we can assume that the expression is in lowest terms is because if the fraction were not in lowest terms I can simply force it to be.
Let p/q be a fraction not in lowest terms, then let n = gcd(p, q). So, p/q = (p/n) / (q/n). Notice that the right hand side is always in lowest terms. So when say that sqrt(2) = p/q, we are allowed to assume that p/q is in lowest terms because I can always just pick (p/gcd(p,q)) / (q/gcd(p,q)) which is in lowest terms by definition. However I’d rather just write p/q and assume gcd(p,q) = 1 because that is far less cumbersome. The point is that this is the same thing
Nice. I've watched other proofs of this idea, but those other proofs didn't go into the idea of 'k'. By doing that, you created at a compelling argument that is easy to follow, and which shows that the original assumption (a and b being relatively prime) must be false. Well done!!!
Thank you!
After watching many videos， finally there is one that I can understand. 😭
Tq for not assuming we know anything to begin with. 🙏
I can't understand something there,why do we assume that a/b is an even number? can you explain that please
in squares a=b if we talk about geometry
We don't.
The proof hinges on a and be not both being even. We do that because we assume a/b is the lowest form of the fraction. And Wrath says "assume," but I think it would be less confusing to say "define". Every rational fraction has a lowest form, one where a and b share no common factors, and we are defining a and b such that a/b is the most reduced form of the fraction.
What if you have been given √5 prove that it's irrational?
The most general version of this proof is one which shows that every natural number that is not a perfect square has an irrational square root. This obviously includes 5.
Let n be a natural number such that n is not a perfect square (more precisely, there does not exist an integer p such that p^2 = n).
We show that sqrt(n) is irrational. Assume for contradiction that sqrt(n) is rational. We know that sqrt(n) is positive by the definition of the square root function. Since sqrt(n) is rational and positive, by definition there exist natural numbers, p and q, such that sqrt(n) = p/q. We assume without loss of generality that p and q are relatively prime (meaning gcd(p,q) = 1, i.e p and q have no common factors other then 1). If p and q were not relatively prime then p/gcd(p,q) and q/gcd(p,q) satisfy the prior conditions. Therefore we can assume that p and q meet the aforementioned criteria.
So sqrt(n) = p/q -> n = p^2 / q^2. But if q^2 divides p^2 then trivially gcd(p^2, q^2) = q^2. But gcd(p^2, q^2) = (gcd(p, q))^2 = 1. So q^2 = 1 so clearly q = 1. But that means n = p^2 and that’s a contradiction.
Maybe I misunderstood something. But the initial hypotheis was it is a fraction AND in its lowest terms. You derived a contradiction. This means the initial statement is false ie it is not a fraction OR not in its lowest terms OR both. It doesn't necessarily mean it is not a fraction - it may simply not be in its lowest terms. I'm sure I got something wrong ...
Nope. He clearly stated that the fraction IS in it's lowest terms. That means that a and b should not have any common divisors. BUT since we arrive at the fact that they do have common divisors that must mean a fraction actually cannot be put in it's lowest term which means it's not a fraction because any fraction can be put in it's lowest terms and if it can't then it's not a fraction.
P.S.: Sorry for bad wording, English is my 3rd language
@@user-mo8in1by7h "A fraction is said to be written in the lowest term if its numerator and denominator are relatively prime numbers, i.e. they have no common factors other than 1. In other words, a fraction is said to be in its lowest terms or lowest form, if the HCF of the numerator and denominator is 1." Just learn how to google, please. Lowest term of 4/2 will be 2/1.
Hello!
Is it generalizing things too much if I just thought that if a^2/b^2 = 2 is true, then the absolute value of a has to be exactly 2 times the absolute value of b for the fraction to be two.
So then we have 4b^2=2b^2 which is false.
Your proof is great, leaves nothing unexplained but I suppose I was on the right tracks at least, I kind of proved that they must have a common factor of 2 but I was not clever enough to think of it that way!
I will definitely watch more of your stuff and subscribe, my studies are not that math heavy but I suck at it since I didn't take high school seriously so I am trying to learn some proof writing on my own. Thanks for the video!
How did you arrive at “then the absolute value of a has to be exactly 2 times the absolute value of b for the fraction to be two”?
The bit about being in lowest terms bothers me. I mean given the fact that there are an infinitude of pairs of values to check, how can we be so certain that we have lowest terms?
The reasoning is that every fraction CAN be expressed in lowest terms, and so if we assume we can write sqrt(2) as a fraction, let's not just say it equals x/y - let's go a step further, since we can, and take the sqrt(2) fraction in lowest terms, say a/b. It's legitimate to do this because if sqrt(2) does have a fraction representation, which we're assuming it does, then it also has a fraction representation in lowest terms.
@@WrathofMathBeautifully explained. Many thanks to both of you
Here's a simpler way.
✓2 = p/q
=> 2 = (p/q)^2
=> 2q = p^2/q
Since 2q is natural
And there's no common factor between p and q,
Therefore p/q or p^2/q can't be natural.
Hence, 2q = p^2/q can't be true.
Correct as well, I suppose.
Why exactly can't a and b have any common factors? I tried the proof with n instead of 2 and it seems to work, "proving" that there are no rational numbers at all... My math is most certainly wrong, but please tell me how
Thanks for watching and good question! In the contradiction assumption, we assumed sqrt(2) could be written as the ratio of two integers, say a/b. Then certainly, either a/b is already a fully reduced fraction (meaning a and b have no common factors) or it could be reduced further if a and b do have common factors. So, let's say a/b has been fully reduced. Does that make sense?
We know if sqrt(2) is a fraction, it can be written as a fully reduced fraction, and so that's how we'd choose to write it. The proof would not be able to apply to just any number, because one of the key steps uses the fact that a/b = sqrt(2). So if sqrt(2) were just n, the rest of the proof wouldn't work. Hope that helps!
Edit: Here is a link to a similar proof if you're interested: ruclips.net/video/sLaAU49zBRo/видео.html
@@WrathofMath Thanks for answering. The problem I see is that nobody says "let's just assume this real quick". If a and b would be allowed to share factors, the proof wouldn't work. It's kind of a big deal, but I still found no logical reason for that
Perhaps because n is for all natural numbers, and it follows that natural numbers are all rational, thus it works.
i wanna question the man who created this que did he really didnt know root 2 is irrational why does he need to create such an illogical question, nice explation btw
long time ago but anyone know why K is used specifically?
Do you mean what we are accomplishing by putting it there? Or do you mean why k instead of some other letter? k happens to be a common choice for arbitrary integers. n for example wouldn't be a great choice as it's generally reserved for positive integers, and x is typically a real number. So just sticking with pre established convention.
Did we just prove that the square root of 2 cannot be a rational number? I watched some other video that I didn't quite understand that show that the square root of 2 exists using some kind of bounded theorem. Can you od this proof so we can understand? Thank you
You said "Can you od this proof so we can understand?"
Is 'od' a typo? I guess you probably mean 'do'. Yes, the square root of 2 is irrational but it DOES exist by the completeness axiom, which is probably what the other video was talking about. The square root of 2 is the least upper bound of a certain set of rationals, so the real numbers contain it.
I still don’t understand why a and b can’t be even and can’t share the same factor?would you mind tell? I’m new
The proof starts with three important statement. We assume for contradiction that sqrt(2) IS rational. This means we can write it as a ratio of two integers a/b. But if we can write it as a ratio of integers, we can also write it as a fully reduced ratio of integers, so we assume it is written that way and thus a and b cannot have any factors in common otherwise we could reduce the fraction further. If it were 4/8 for example, this wouldn't be possible because we assumed it was written in "lowest terms" and so the common factor of 4 would have already been cancelled out.
Read about this in Stephen hawking’s book but didn’t understand, thank you
Glad to help, thanks for watching!
Which book?
@@dolmathesimp4570 God created the integer
I don't understand. If a/b is not in it's lowest terms at the end it's still rational so there's no contradiction. Can you explain this? Thanks
The contradiction is that we took a/b to be in lowest terms; so if it turns out it isn't in lowest terms - that is a contradiction.
The sequence is like this; if a number is rational then it can certainly be expressed as a rational number in lowest terms, so we assume for the sake of contradiction that root 2 is rational, and thus it equals a/b in lowest terms. But assuming this leads to a/b being further reducible, and thus a/b couldn't have been in lowest terms, and so root 2 couldn't have been rational. Another way to think of it is that assuming root 2 is rational leads to an infinitely reducible fraction, which is not possible.
I understand the contradiction that it’s not in its simplest terms but a/b would still be rational even if it had a common factor. I can do it but can’t get my head around this 🤯
But we aren't proving a/b isn't rational. a/b is certainly rational, it was constructed precisely to be rational. We're proving that if sqrt(2) = a/b, a contradiction is found.
I understand there’s a contradiction because a/b is not fully simplified but a/b is still rational simplified or not and therefore we are still showing root 2 as rational.
@@paullloyd4443 If √2 is rational, there must be a fraction a/b which is both fully simplified and equal to √2. a/b is defined as that fraction.
AMAZING!!! THANKS
My pleasure, thanks for watching! Here is a similar lesson on square root of 3 if you're interested: ruclips.net/video/sLaAU49zBRo/видео.html
how come both the numerator and the denominator has to have no common factor in order to prove it to be rational? thanks
Every rational fraction has a form where the numerator and denominator do not have factors in common. 3 and 6 have common factors, but 3/6 can be reduced to 1/2.
a/b is defined as the √2 fraction such that a and b are both integers with no common factors. If √2 was rational, such a fraction would need to exist.
But Y are you using 2k
'k' here is used to represent a number, which, when multiplied by 2, becomes an even number.
Thanks alot 💗
Can you please prove if the square of 2 - 1 is irrational
Thanks for watching! I'm not sure what you mean. Are you asking about sqrt(2) - sqrt(1)?
@@WrathofMath i think he meant sqrt(2-1)
U make me understand
Glad to help - thanks for watching!
Just found this vid, amazing!
Thank you!
thank you
Wow thanks 😭😭
Hope it helped!
seems like an old video but I would very much like to understand this but i just cant get it is there any way we could chat on discord or something and you could help me understand this ?
If you can time stamp the part(s) you don't understand and explain some of your confusion I may be able to help
@WrathofMath thank you ill do that today.
By that logic isnt 2/4 also irrational?
Please explain it in another way I can't understand
Thank you sir
and know imagine ancient alexandria with Euclid as teacher, proving this for the first time to a stunned audience.
other civilizations did such math also ofc, but irrational numbers were a very new thing back then. our dear trigonomy guy Pythagoras hated the idea btw.....
yep, that stuff is >2000 years old, the proof is valid without improvement since ~300bc
Why can't a and b share common factors?
The idea is that, when we assumed we could write sqrt(2) as a/b, we assumed a/b was in lowest terms. This means it cannot be reduced, and so a and b mustn't have common factors otherwise a/b COULD be reduced. It's fine to assume the fraction is in lowest terms because IF sqrt(2) can be written as a fraction, then it can be written as a fully reduced fraction, so we just assume we're working with that fully reduced fraction.
Many thanks for explanation@@WrathofMath
Alexandrea Ports
wtf, how you pass from a^2 being even, to a being even. That is not that simple.
If a was odd, a^2 would necessarily be odd. Here is a proof of the result about evens: ruclips.net/video/FewsjiKug8Q/видео.htmlsi=7moxfkJXMNfUfQJy
If a is odd, it must not have a factor of 2, in which case a^2 doesn't have a factor of 2 either, in which case a^2 would be odd. So if a^2 isn't odd, then a isn't odd and so a is even.
I hear you say "k". But I see you write... actually I do not know how to describe what you write. It's not an American Alphabetical Letter. That crazy symbol you use SCARES ME!!! :0
He actually wrote K is a subset of Z. Z being he set of all integers and that symbol is epsilon, which denotes 'belong to' when talked about sets.
@@DeVirgilWouldn't the ∈ symbol actually indicate that k is an element of the set of all integers? If he wanted to write that k is a subset of the integers, then he would have written {k} ⊆ Z.
Cool
Got it in my math textbook when I was in 8 th grade.
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If you use the square root of 2, then this statement will be true: 2x = x+x = x^(3.
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Very true
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Now prove root 4 is rational by this method root 4 is also irrational which is not
If √4 is irrational, then √4 != a/b
Let us Assume that:
√4 = a/b
Then:
4 = a^2/b^2
4b^2 = a^2
2b = a
Substituting 2b = a into
4 = a^2/b^2:
4 = [(2b)^2]/b^2
4b^2 = 4b^2
So √4 =a/b, therefore √4 is not irrational (it is rational)
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if anyone that understands this would contact me i would very much appreciate it. id very much like to be able to understand this
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😮
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Not really helping.
@mohamedelzoheiry1413
0 seconds ago
the solution of square root 2 in nearest fraction is
1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796
the solution of square root 2 is
25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136
1.4142136 = (2)^(1/2)
the solution of square root 2 is
11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136
1.4142136 = (2)^(1/2)
thank you
@mohamedelzoheiry1413
0 seconds ago
the solution of square root 2 in nearest fraction is
1871/1323 or 1970/1393 or 3124/2209 or 3363/2378 or 5234/3701 or 7105/5024 or 8119/5741 or 9611/6796
the solution of square root 2 is
25/18 + 1/42 + 1/660 = 19601/13860 = 1.4142136
1.4142136 = (2)^(1/2)
the solution of square root 2 is
11/8 + 3/83 + 6/1955 = 1835819/1298120 = 1.4142136
1.4142136 = (2)^(1/2)