This Will Be Your Favorite Simple Problem
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- Опубликовано: 4 окт 2024
- Here's a result that, at first, is quite hard to believe. However, a little bit of logic can go a long way.
This is one of my favorite simple math problems in the world. Enjoy!
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
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First comment
i love when you pick sqrt(2) every single time for every number, it reminds me of that one video where it's like "that's right, it goes in the square hole!" for *every single shape*
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Ah yes... The classic square hole😌😂
Lmao
That’s such a niche reference but it’s perfect lmao
LOOOOL I WAS THINKING THE SAME THING
i mean e^ln2=2, wgich is much more straightforward, assuming we know irrationality of e and ln2
but in such questions we would not be given that ln(2) is irrational as you have assumed. We would have to prove that ln(2) is irrational
@@p_square you can refer to standard mathematical results in such problems. in particular, here the lindemann-weierstrass theorem applies.
@@p_square while true in same way you need to prove that square root of 2 is irrational
@@TrimutiusToo to be fair, sqrt2 is much simpler to prove than e or ln2, the latter being a real killer in terms of proof's complexity
@@timurpryadilin8830 If we're given that e is transcendental, then ln2 being irrational follows. Concretely, if e^(p/q) = 2 for positive integers p,q, then e is a root of x^p - 2^q = 0 and thus algebraic.
I have now realized I hate the way I say "irrational"
Now is that irrational or rational? 🤔
haha
@loc share what the heck
@@aashsyed1277 How's it going?
@@rubenvela44 good but did you become a member?
Reminded me of my elementary school days when I asked my math teacher the same question. Well, he resolved it by saying e^{ln(2)}=2.
Your explanation is also awesome:)
Thanks so much!
Huh
Never thought of that. Good example
Logarithms in elementary school days???
@@anirbanmallick8502 back in my days when i was in elementary school I got my head burning from thinking what the heck is the result of 2 divided by 4. And now this??
e^πi=-1
My immediate first thought was e^ln3, for example. I feel like exponentials and logs would be the most natural first choice.
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ah, natural
I got no idea what the hell logs even are
@@pxolqopt3597Logs are the inverse of exponents, log 2 with base 8 is basically asking 2^x=8 which is 3. Imagine an animation where the 8 on the bottom going to the right side and the 2 staying there.
@@-petrichor-7263 I must have sent that comment before I finished calculus haha
This has always been my favourite proof, using an undecided (undecidable?) fact to prove another conjecture. I understand that Wiles’ proof of Fernat’s Last Theirem depends on a similar logical partition (ie considering X and not X separately and showing that something is true either way). So elegant! Thank you for making this video.
Very glad you enjoyed it!
"Either A or B, but C in both cases" is also a good trick to aid in solving sudoku puzzles
It feels weird to consider r as a rational xD But anyway, nice video as always :)
Mates Mike aquí :0 Me gusta más tu forma de hacer videos ;)
@@jorge-_-1562 quéee va, somos muy parecidos, y además su estilo me encanta, se hace muy ameno que hable a la cámara :)
@@MatesMike Si, cuando mira a cámara es mejor, pero este video ha sido diferente. El resuelve por lo general problemas (series, integrales, etc.) Tu canal es más completo y divulgativo, dejando atrás el cálculo vas a la esencia
Pero madre mia Mike que haces aqui compañero
@@MatesMike you mean weird to consider r as an *_irrational_* number, surely? :-B
This IS my favorite simple theorem, thank you so much for covering it!!!! :)
You're very welcome! Thanks for watching :)
Mine too, I make sure I teach this whenever I am teaching intro to proofs. I first saw this from a constructivist mathematician who did not believe in the Law of Excluded Middle and so did not take this proof as valid (since we can't know from the proof which is rational)
@@BriTheMathGuywhy can't it be so that √2^√2=√2
√2^√2
= √2^2^1/2
Then by exponent rule, a^b^c = a^bc
Then √2^2^1/2 = √2^(2×1/2) = √2^1
So
For a nice constructive example with an easy proof: (sqrt(2)) ^ x = 3, where x is the log base sqrt(2) of 3. If x were rational, say x=p/q, then sqrt(2)^p = 3^q, and squaring both sides, 2^p = 9^q; contradiction because one side is even and the other is odd.
(Many people have pointed out e^(ln2)=2, but it is quite difficult to prove ln2 is irrational. Numbers of the form log_a(b), where a and b are rational powers of rational numbers, are some of the easiest numbers to prove irrationality for - even easier than sqrt(2) in my opinion - because it just boils down to comparing the prime factorizations of a and b)
Excellent comment.
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You know that a video is good when you feel like it did last ~10% of its actual duration
e^ln(2)=2
😂😂
Now prove e and ln2 are irrational xD
@@kylesheng2365 use sqrt2 and log_2(3)
@@jatloe ya I was replying to the guy above
you would end up proving e is transcendental, which is not trivial at all
Its just like what Matt Parker said on his channel
π^π^π^π could be an integer as far as we know
No, it's nothing like that.
@@leonhardeuler675 prove it
@@leonhardeuler675i beg to differ
How is it different?
@@leonhardeuler675 Prove it.
@@alienbroccoli8296 It's not on me to show they're different. It's on the first guy to show that they are the same.
This is one of my favorite introductory problems in Number Theory and how to think about proving stuff! If A then done, else not A implies B so done
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Everyone is saying e^ln(2)=2,
But you don't even have to be that specific.
x^logBASE(y, x)=y
You have infinitely many answers when x is irrational and y is rational.
This is literally one of the first proofs we are tought about in my introduction to proofs calss. Nice
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I'm a math nut and this is my newest favorite math video!! Thank you lots for posting!!
Very glad you liked it!
I really loved the reveal. You’re right, it’s simply my irrational favorite! I’m going to watch it again.
Glad you enjoyed it!
Math is so amazing, it confuses us not giving a single clue, and the next moment it becomes totally known to us
Him: explaining for 3 minutes
The majority of the comments: e^ln(2)
"Teacher, when are we gonna need this in real life?" "Imagine youre at the grocery store..."
"I want you to imagine, just for a second, that you are a mathematician..."
I am one! :D
"...that loves numbers"
Oh, no. I don't care about numbers. I love manifolds, varieties, and spaces of any kind.
My favourite example of the importance of constructive math.
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I know others have already brought this up, but:
e is irrational, ln(2) is irrational, e^ln(2) = 2
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I also love the infinite irrational tower: sqr(2)^sqr(2)^sqr(2)^sqr(2)^... = 2
I love your content. It makes so much sense. The way you explain things is so much better than how most of these school teachers try and explain stuff! There's some great ideas which I may use for my youtube channel! Thanks!
0:00 Integers and whole numbers are literally the same thing: A whole unit number of something. That’s, what an integer is; and that’s, what a whole number is. The central set should have been called: ”Natural Numbers”. 😅
My favorite example of irrational^irrational = rational is e^ln(rational) = rational.
S'pose b = 1, we have r^q = a. Apply ln on both sides, we get ln r^q = q ln r = ln a -> q = ln a / ln r.
So...for any (yes, yes) irrational number r there exists and exponent q (rational or irrational) that converts it to a natural number a.
idk math words in english, but what about my solution:
square root of a number equals to this number in 1/2 degree. so square root of 2 degree square of 2 is like 2 degree 1/2 degree 2 degree 1/2, which means two degree 1/2. so square root of 2 degree square root of 2 equals square root of 2. I am in eight form, so tell me about my mistakes please.
This is the exact same example I came up with back in high school when faced with the same question. Later on, I realized I couldn’t prove whether sqrt(2)^sqrt(2) was an irrational number in the first place! Classmates kept asking if there were two or three irrational numbers-no two of which were allowed to be opposites or inverses or trivial by definition-whose sum or product was rational. Again, I could come up with sums of roots of third degree polynomials or sums and products of cosines and sines or squares of tangents, etc… which did add up or produce rational or even whole numbers, but there was no way (for me) to prove their irrationality to begin with. In fact, I believe it’s not yet proven if pi^e or pi+e or many combinations of irrationals are irrational; it’s that hard!
Minor correction. It is in fact proven that pi^e and pi+e and a lot of those cominations are irrational. What hasn't been proven is whether they're transcendental which is a subtype of irrational number which are MUCH MUCH MUCH harder to prove than regular irrationality.
If root 2 to the power of root 2 was rational, then it solves the problem
Let i represent the square root of -1. Have you talked about the fact that i, raised to the i power, gives us a real answer?
Assuming a _SET_ of real numbers can be considered "a real answer"... o.O
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Maybe nice follow up: proof sqrt(2)^(sqrt(2)) irrational. Maybe in the general form (Gelfond-Schneider Theorem).
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sqrt(2) raised to itself is equal to the sqrt(2) root of 2. Since exponents and roots are inverse functions, raising the sqrt(2) root of 2 to the power of sqrt(2) "undoes" the root, leaving just 2. I bring this up because if you put sqrt(2)^sqrt(2) = sqrt(2) root of (2) in math notation, it looks funky.
A really neat, but unsatisfying counterexample to the claim that an irrational exponentiated to an irrational is also irrational e^ln(2)=2. We get a natural number out and this example is actually a bit stronger than disapproving the initial claim since both e and ln(2) are transcendental.
EDIT: Wow okay looks like I'm not the first person to note this. Here's a fun related problem. Suppose I give you the "power tower" 2 = x^x^x.... (Yes this is an abuse of notation). We can easily solve for the value of x by claiming 2=x^2 ----> x=√2. But what if instead I claim 3=x^x^x..., then similarly 3=x^3 ----> x=3^(1/3). What's going on here?
Better yet is trying it with 4, then realizing that 2^½ = 4^¼. 3b1b did a video on it.
There is an issue of convergence here. The sequence of power towers will never converge to a real number larger than e. 3 is larger than e.
@@angelmendez-rivera351 Not if you take 3 to be sufficiently small.😂
How is 2=x^2?
@@zebran4 Because (by assumption) 2 = x^x^x^x^... = x^(x^x^x^...) = x^2. It hinges on the fact that adding an x to the bottom of the infinite power tower leaves it exactly the same: an infinite power tower.
Not really relevant to the problem at hand but I messed around with sqrt(2)^sqrt(2) algebraically and found it's equal to 2^(1/(sqrt(2)). In other words it's the "square root of 2th" root of 2, which I think is kind of neat.
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Yeah, I found that it's the quad root of 2
Because by using the properties of exponents you get 2^1/4 since root 2 = 2^1/2
I see that nobody has mentioned that exponentiation is not associative, that (a^b)^c /= a^(b^c). Even more bizarrely, in general (x^x)^x /= x^(x^x)!!! In particular, (√2^√2)^√2 /= √2^(√2^√2)
Well root 2 power root 2 means root2^ 2 ^1/2 which is root 2 power 2/2 (since a^m^n = a^mn) so ans is root2
This has absolutely been my favorite simple problem for like 10 years
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Suppose sqrt(2)^sqrt(2) is algebraic.
(sqrt(2)^sqrt(2))^sqrt(2)=2 which is algebraic => sqrt(2)^sqrt(2)=0 or sqrt(2)^sqrt(2)=1 or sqrt(2) is rational, none of which is true.
Thus, sqrt(2)^sqrt(2) is transcendental.
Although, while I used the contrapositive of the Gelfond-Schneider theorem, the theorem itself also helps proving sqrt(2)^sqrt(2) is transcendental.
Wut xd
I don't get it, why (sqrt(2)^sqrt(2))^sqrt(2) should imply that sqrt(2)^sqrt(2) is 0 or 1?
@@tonaxysam the Gelfond-Schneider theorem states that if a and b are algebraic such that a=/=0 and a=/=1 and b is irrational => a^b is transcendental.
Here, we assumed that sqrt(2)^sqrt(2) is algebraic. Since sqrt(2)^sqrt(2) and sqrt(2) are algebraic and (sqrt(2)^sqrt(2))^sqrt(2)=2 is algebraic, the contrapositive of the Gelfond-Schneider theorem states that a=0 or a=1 or b is rational. But here, a=sqrt(2)^sqrt(2) which is neither 0 nor 1, and b=sqrt(2) being irrational is the poster child of the proof by contradiction examples, so... contradiction.
@@tonaxysam also, I said (assuming sqrt(2)^sqrt(2) is algebraic) (sqrt(2)^(sqrt(2))^sqrt(2) is algebraic=> sqrt(2)^sqrt(2)=0 or sqrt(2)^sqrt(2)=1 or sqrt(2) is rational. Don't cut my sentence mid-way.
I immediately thought e and log(2)
When you speak for a non-native listener it is sometimes difficult to distinguish “a rational number” from “irrational number”
I like how this video can be played at 1.75x speed and I feel like he's talking at a normal pace
not only it is rational, it's natural which is even more crazy
I don’t know if this will be my favorite problem. I do know, though, you’re one of my favorite RUclipsrs.
Thank you! I really appreciate it!!
Well, you’re a very good explainer and very good looking.
Before watching this video, I tried to simplify sqrt(2)^sqrt(2) and what I got was the [sqrt(2)]th root of 2. Clever video, though!
doesn't sqrt(2)^sqrt(2) fulfill the Gelfond-Schneider theorem? In that way, we can prove that sqrt(2)^sqrt(2) is also trancendental.
e^ln(2) would be an easy solution
And thus irrational because it is real...
I invented a number e which is transcendental and another number ln2 that is also transcendental such that e^ln2=2. e=2.7...,ln2=0.69... For more precise values of e and ln2 it approaches 2 like e=2.718...,ln2=0.6931... Since both are transcendental, a^b could be rational even if both a and b are irrational. So i found two irrational numbers such that a^b is rstional
I thought you were going to show that e^iπ=-1, which also answers the question.
ans of √2^√2 = 1.632
we can simply apply log on both side we can solve that
Incredible!
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I have a question, can I make sqrt of 2 ^ sqrt of 2 like this
(Sqrt of 2) ^ 2^1/2 which by properties of exponents means it’s sqrt of 2^ 2*1/2 = sqrt of 2
_proceeds to never state the square root of two to the power of the square root of two_
My friend sent me a question today. It was to find the value of x^x where x = sqrt(i) where i = sqrt(-1)... This video gave me a different way to look that problem. I will try solving it again.
(2^1/2)^(2^1/2)
2^(1/2*2^1/2)
1/2 * 2^1/2 = 2^-1 * 2^1/2 = 2 ^-1/2 = 1/2^1/2
Therefore we get a transcendental number as our result, 2 to the root 2 root.
My argument:
Girl: So where is the part you prove me wrong?
this video is amazing, but i do prefer it when when you show up!!!
I’ll see if I can make an appearance next time 😀
@@BriTheMathGuy ok, i'll see you in that one 😅😅
Oh yeah, as soon as I heard that it is about irrational numbers I knew what he's going to show. I am a mathematician who loves numbers and this is for sure my favourite simple problem.
Irrational: no closure under the field of fractions.
Now, we have the word "transcendental", and that means not a root of any polynomial.
But if we include hyperoperations (i.e. tetration, pentation, hexation, etc.), is the word "transcendental" still appropriate?
What if... pi can be explained with tetrations, or pentations...
you can compute hyperoperations with finite repetitions of the usual operations, so nothing would change in terms of trancendental numbers
if you want a proper extension of, say, algebraic numbers (numbers which are roots of polynomials), i would say the computable numbers are a good next step. those are numbers for which there exist an algorithm that allows their computation to an arbitrary precision. this includes all algebraic numbers, plus some of the trancendental ones, like e or pi, but not all of them. in fact, the set of computable numbers is countable.
it should be a rabbit hole deep enough for you to have fun!
@@intvl ok that cool
the infinite tetration of the square root of 2 is just 2 --- a natural number
Tylko skomentuje. Na miniaturce jest pierwiastek z dwóch drugiego stopnia do potęgi pierwiastek z dwóch drugiego stopnia to można zapisać jako dwa do potęgi jednej drugiej i całość jeszcze raz do potęgi 2 i do potęgi jedna druga. Czyli mamy 2, a reszta potęgi się ze sobą mnoży i wychodzi ostatecznie dwa do potęgi jednej drugiej, czyli pierwiastek drugiego stopnia z dwóch.
I was already looking for a solution to this question. Prove that there exists a pair of irrational numbers such that one to the power of the other is rational and surprisingly this video pops up.
Intersting. If you potentiate the nth root of n n times with the nth root of n your result will always be n: x = n ^ (n ^ (-1) * n ^ (n * 1/n)) = n ^ (n ^ 0) = n.
I just came here to fervently reject the law of excluded middle.
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you can solve this problem in multiple ways
one of the alternatives is:
e^ln 2 = 2
This looked a lot like a short 3b1b video. Nice work!
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e to the ln2 pops to mind immediately for me.
2^(1/2)^(1/2)
e to the power of ln2 is, by logarithm rules, 2. Both are not only irrational but transcendental as well.
Proving that e and log2 are irrational/transcendental is significantly harder than using √2
When I saw sqrt(2)^sqrt(2) I thought "well, sqrt(2) is 2^½ so this is sqrt(2)^2^½ which is 2^½ which is sqrt(2)?" and forgot that order matters for this sort of thing.
Beautiful logic
Glad you think so!
I didn’t even need to watch the video yet, but this was the first proof I ever understood in high school
Well try to understand proofs it will be very much required in your future
@@amiyabora1011 oh I’m a masters math student now, I’m just pointing out that this WAS the first proof I ever understood
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I tried to imagine that I was a mathematician but then my brain imploded
You can do the same thing with sqrt(3)^sqrt(3)^sqrt(3), which becomes 9. Or any repeated thing like that, as long as it's irrational (for example, sqrt4 wouldn't work for this, as it equals 2).
It's 3*sqrt(3)
Could you use trig to simplify (sqrt2)^sqrt2 with special triangles?
You made an error at 0:29 - in English usage not math. You said "And this might beg a few questions...". That is a misusage of the phrase "beg the question". What that expression means is to use an argument that assumes the truth of the very thing one is trying to prove. What you should have said is "raise a few questions".
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🤓
I did notice this before this video and it blown my mind 🤯
🤯
This violates the Gelfond-Schneider Theorem.
If a and b are complex algebraic numbers with a
otin {0,1} and b not rational then a^b is transcendental.
sqrt(2) is algebraic and
otin {0,1} so by the above theorem sqrt(2)^sqrt(2) is transcendental and because it is real, also irrational.
Gelfond-Schneider theorem says √2^√2 is irrational.
i don't know if anyone suggested this, but if the square root is considered as an exponent (1/2) there's a slimmer way to prove this. Go check it out!
Cries in taking log both sided
I fucking like this use of Logic
There are lots of irrational numbers beyond pi and e -
nth root of any prime number p is irrational for any integer n > 1
log of any integer that is not already a rational power of the base is an irrational number - e.g. e^(ln(n)) = n for any positive integer n>1 and irrational log(n).
and so on ...
These irrational numbers can be mixed and matched in so many ways to create rational numbers.
I genuinely muttered "whoa" when i saw the thumb, it's not like i didn't think in something like that before but my mind was like "wtf" when i saw it.
you can also do e^ln2 which is just 2
rationality always leads to such beautiful proofs
You should talk about fractional exponents and its problematic defnition
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What, title was absolutely right
that was pretty cool. Thank you.
"logically, one of two things are true" .. s/b "is true"
√ 2^√ 2 = X (say),take log base X both sides of equationr RHS simplifies to be 1,L.H.S=log base X of √ 2^√ 2 bring the exponent down of the log send x to RHS side equatio then becomes X^1 = √ 2x√ 2 = 2,simple!
I love mathematical proofs where you look into a rabbit hole but don't actually go inside, and then use that to find a solution.
Is there any known case of a rational number taken to an irrational power being rational? It doesn't seen intuitively possible.
2^(log_2(3))=3
Very late, but a few simple examples are:
0^(sqrt2)=0
1^(pi)=1
2^(log2(5))=5
@@aj_style1745 Ah logs, stupidly obvious (ignoring the trivial examples...)
And I suppose ALL such examples are logs.
Either it's rational or is irrational caught me
Finally, a math video I can understand
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sqrt of 2 to the power of sqrt of two is actually equal to the sqrt of 2th root of sqrt of two:
(√2)^√2 = (2^1/2)^(2^1/2) = 2^((2^(√2)/2)= 2^(1/√2)
= the sqrt of 2th root of sqrt of two :)
Root 2^ log7 base2
sqrt(2)^sqrt(2) actually _is_ irrational. But that is a considerably harder proof (strictly speaking, the proof of the irrationality of that particular example is trivial: x = sqrt(2) is algebraic therefore by the Gelfond-Schneider theorem x^x is transcendental. Proving the GS theorem is the hard bit.)
You know a lot more math than I do, but my thought was that this video did not validly prove that an irrational raised to an irrational can be rational. To be a valid proof, we cannot ignore what sqrt2 to the power of sqrt2 is.
@@martinlindenbusch7175
Here, let me rephrase the video a little:
We know there are only two things sqrt(2)^sqrt(2) can be: Irrational or rational. Since we do not know which it is, we write a proof for both:
- Assume sqrt(2)^sqrt(2) is rational, we already have our result since sqrt(2) is irrational and, by assumption, sqrt(2)^sqrt(2) is rational
- Assume sqrt(2)^sqrt(2) is irrational. Let's raise it to another irrational power like sqrt(2).
(sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^sqrt(2)sqrt(2) = sqrt(2)^2 = 4
As you can see, it doesn't matter if sqrt(2)^sqrt(2) is rational or irrational, we can find a result where irrational^irrational = rational.
@@mrocto329 i believe you meant sqrt(2)^2 = 2, not 4
@@robinsparrow1618 oh yeah thanks
what's really fascinating is that these types of proofs were banned in ancient greek mathematics, meaning you had to actually find an example where r^q is rational
Can you post a video calculating this? I did it myself and got the answer as half. Dk if I did it right tho