Proving sqrt(2) is irrational! (Classic Proof)

Here's the classic proof that square root of 2 is irrational.
Proof by contradiction!

blackpenredpen,
math for fun

[sichelsam]

Really good example of the contradiction concept. I like the macabre comparison to the flying human 👍

[blackpenredpen]

If you can prove "humans cannot fly", then you will understand "proofs by contradiction"

[knochentrocken]

Everyone is talking about that the proof of humans cannot fly is wrong but I don't think they understand the intention of this "proof"

[Tracy_AC]

"And that's messed up, right?!" XD

[jacksainthill]

You only proved that humans cannot fly _other than downwards._

[DjVortex-w]

We have been trying to talk sense to sqrt(2), but it just keeps being stubborn and so utterly irrational. It's impossible to talk to it. It won't change its mind no matter what, and no matter how clearly it's shown how wrong it is. A lost case.

[risou-burner-account]

Personally I prefer "Q.E.D".

[rasitcakir]

But humans can fly...
in an airplane.

[faisca]

Thank you so much for this video! I love how you explain things with an enormous joy and passion for math. Thank you again, keep up the good work!!

[ericmiller]

The story is told that this proof was originally discovered by one of the students in Pythagoras's school. But one of the most strongly held Pythagorean beliefs -- a religious belief, not a rational* one, obviously! -- was that all numbers were rational, that the whole universe consisted of "harmonious ratios". The story continues that they had the student killed, in order to suppress his proof that this wasn't the case. (But I think they had him drowned at sea, rather than throwing him off a building.)

* So, an irrational belief in the rationality of the universe? ;-)

[AnatoArchives]

Humans can't fly but flies can human.

[terryendicott]

"Humans can't fly" -- Well it looks like one stick figure was dying to prove you wrong.

[chair]

Suppose root 2 has rationality
To find a contradiction we must try
to write our root as fraction a on b
but show we can't, for something goes awry

By squaring our equation on both sides
then multiply; a squared is 2b squared
That 2 divides a squared is thus implied
and therefore 4 (squared factors come in pairs)

But be a squared a multiple of 4
Then half that (b squared) must divide by 2
By using the same logic as before
Then 4 divides b squared, it must be true

To share no factors, a and b are bound
But 2 they share, a contradiction found

[iabervon]

I think the best way to really understand this is to also prove that sqrt(3) is irrational and that sqrt(4) is irrational, and notice where that last proof doesn't work. For 3 and 5, there are several cases in the lemma, but they still prove that the only way a^2 can be a multiple is if a is a multiple. With 4, another case works when it wasn't supposed to, which is how we can have perfect squares.

[justmathsbymike]

Thanks a lot sir. I have been searching for this.

An easy (although less formally strict) way to prove that if a² is even, then a is even: The prime factor of 2 has to come from somewhere, it cannot be a product of multiple prime factors in a×a (because then it would not be prime), so each a must have at least one prime factor 2.
And an alternative second blackboard:
√2×b=2k
b=√2×k
b=a/b×k
1=a×k
1=2k²
0.5=k²
0.25=k

[mathcanbeeasy]

Very important the explanation of why gcd(a, b)=1. And a lots of teachers skip that part. Practically the contradiction is that, if sqrt(2) is rational than his fraction can be simplified infinitely by 2.

[MikeRosoftJH]

This proof implicitly depends on the following statement: "Every non-empty set of natural numbers has a minimum". This statement happens to be true, but it's not obvious. (It can be proven using the axiom of induction; I'll leave this as an exercise for the reader. Hint: Try to prove an equivalent statement: if the set has no minimum, then it is empty.)

[dyer]

Can you do one proving the irrationality of e!

[nicoleleung]

Why we have to assume a/b is in its simplest form? I believe its not part of the definition of a rational number?