Irrational Root 2!

In this video, we introduce and prove the "Carpets Theorem" and then utilize the theorem to prove that both the square root of two and the square root of three are irrational by a visual infinite descent argument.

This animation is based on an argument from Stanley Tennebaum. If you are interested, I recommend this article from John Conway and Joseph Shipman about the irrationality of square root of two (and others):

#irrationalnumbers #realnumbers​ #manim​ #math​ #mtbos​ ​ #animation​ #theorem​​ #visualproof​ #proof​ #iteachmath #mathematics #irrational #carpetstheorem #proofbycontradiction #root2 #algebra #infinitedescent

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[jan-timolobner]

I finally understood one of your videos.
I might not understand every video, but I am determined to learn, so thank you for your hard work on these awesome videos.

[Emily-fmpt]

I appreciate the elegance of the proof! Especially the last part I feel is more intuitive than the usual “you can always divide by two, ” because in this case it’s clear that you’ll eventually just run out of positive integers to work with, which causes the contradiction

[holyek]

These videos make me so happy and mad. Where were they when I needed them in geometry and calculus? They make everything so clear and comprehensible.

[Ninja]

I really like these geometric proofs because I was only ever taught algebraic proofs for proving this such as the classic proof, or other things like the FTA or rational zero theorem. Thank you for teaching me new things

[nargacugalover]

Luckily 2! is still 2. So the title is still right in a way

[dragansantrac]

Wow. One of the best proofs of irrationality of sqrt(2)

[tweep]

This is the best explanation of this I've ever seen

[matthewbay]

I love that proof, proving root two is irrational is part of my favorite proof which proves you can raise an irrational number to an irrational number and get a rational number.

[FLS]

Nice proof, I've haven't seen a visual one of this before! If someone's still not getting it:

If √2 were rational, there would have to exist (finite) integers a and b so that a/b = √2.

a/b = √2, square both sides
a²/b² = 2, multiply by b²
a² = 2b²

So proving the irrationality of √2 is equivalent to proving there is no square of an integer, that is double the square of another integer. The proof by contradiction in the video assumes the opposite. There are no finite a and b to satisfy these equations, so √2 has to be irrational.

Correct me if I'm wrong, thanks!

[hidude]

awesome! love how the well ordering axiom can prop up in fun ways

[MathFromAlphaToOmega]

If you go in the other direction with increasing pairs, you'll get better and better approximations to sqrt(2). Those should correspond to solutions to Pell's equation a^2-2b^2=±1.

[tamaz]

This would have been a perfect video to be made into a loop, but looks like we didn’t get to see that

[jakobr_]

I feel like there’s a missing step where it’s explained that b is between a and a/2 so that the small carpets can overlap like that

[orisphera]

I prefer the following instead of infinite descent:
Take the smallest counterexample. By the step, there's a smaller one. This is a contradiction.
You can note that this is essentially a special case of a reasoning that under a slightly different formulation is known as transfinite induction. It's similar to induction, but more powerful. It's also proven differently. Normal induction is proven from the definition of natural numbers (and 0) as ones reachable by the operations. (By “reachable”, I mean present in every set closed under them.) Transfinite induction is essentially the well-orderedness (existence of the smallest in every subset) of the natural numbers (or whatever set you're using), although for N it may be easier formulated as the normal induction over prefixes

[noahblack]

Took me a minute to figure out why a^2=2b^2 means √2 is rational.

If √2 is rational, then there exists integers a, b where a/b = √2. Multiply both sides by b and then square them.

[chixenlegjo]

Nice proof, but I’ll propose my personal favorite:
The exponents in the prime factorization of a perfect square are always even. When you multiply this perfect square by a prime number, the exponent in that prime factor becomes odd. This means that you cannot multiply a perfect square by any prime number to get a perfect square. Therefore, the square root of any prime number is irrational.

[stephenliao]

simplify sqrt to r
a/b = r(2) for some a, b inZ+, gcd(a, b)=1
a=r(2)b
aa=2bb
lhs has factor 2 but there are two a so 4 | a
Thus 4 | 2bb and this imply 2 | bb
Which means 2 is factor of b
So gcd(a, b) does not equal to one. Therefore, no such coprime integer pair (a, b) exists
I think this is the same to the video but in different way of recursively decreasing a and b

If we use the well ordered property of integer and state (a, b) is the least element satisfying the equation then we do not have to state the recursive part

[Yam-fjmh]

My whole YouTube recommended is just various ways to show the irrationality of root 2

[kexcz]

Wow, cool, yet I know I will forgot that lin like next 10 mins... 😂

[narfharder]

For a long time something vaguely bothered me about this particular proof by contradiction, now I've finally put my finger on it. Thanks for the epiphany! (Surely this has been thought of before, I'm just an armchair amateur who wouldn't even know what book to read about it.)

Consider this is really a proof that √2 is not rational. So, either "irrational" is merely defined as not rational, meaning we have only proven a contradiction by contradiction; or, if "irrational" has a positive definition, then we must first prove that "rational" and "irrational" are in fact mutually exclusive.

Now I wonder if these two ways of defining "irrational" are themselves mutually exclusive - so, what obvious thing have I missed?