Square Root of 2 is Irrational

Square root of 2 is irrational

In this video, I present the classic proof that square root of 2 is irrational, by using a proof of contradiction. Featured is also a special fluffy guest :)

[beatoriche]

A little food for thought for all those who might encounter this classical proof for the first time:

Why does this line of argumentation actually work? I mean, if _p_ and _q_ were just arbitrary real numbers, we could easily find infinitely many fractions _p/q_ that are equal to the square root of 2. What makes this proof work is that the natural numbers have a property the reals do not have – they have a smallest element, namely 1. And this becomes crucial when looking for two natural numbers whose ratio is the square root of 2 – the line of argumentation in the video implies that, if these two natural numbers exist, they are both even, meaning they can both be divided by 2, yielding two more natural numbers whose ratio is the square root of 2. But then these two are both divisible by 2 again by the same argument, and those are divisible by 2 as well, and so on and so forth. For real numbers, this would be perfectly fine – you can divide them by 2 again and again and again. But it's different with the natural numbers – you can't just generate smaller and smaller natural numbers since, at some point, you'd arrive at 1; and at that point, it's over.

This is a _really_ common line of argument in number theory, and it even gets a special name – it's called _proof by infinite descent, _ a special type of proof by contradiction where you exploit the special properties of the natural numbers. You may imagine a proof of this type as consisting of a long staircase with you on one of the steps, and if the statement in question were true, you could go downstairs for a bit, at which point you can go downstairs again and again and again. But the staircase has to start at some point, so you'd have to stop at that point at the latest. In other words, the statement cannot be true.

Anyway, just wanted to put this out there to shed some more light on this proof.

[aaditolkar]

This man is putting out great math content even during a pandemic. Respect!

[CliffStamp]

This is wonderfully clear.

BUT, for most students introduced to this type of Math they often walk away thinking this is some kind of witchcraft, for how would you ever know to do those steps, where does that Math come from in proofs. This is what can cause a lot of people to think Math is impossibly hard because you have to essentially memorize vast amounts of proofs in order to do non-trivial deductions.

[akshataggarwal]

Not Thinking about this stuff is irrational, isn't it?

[harrywotton]

Let me give you another proof.
Suppose 1:√2 = m:n
Notice that m and n have no common factor (except 1). In other words, m and n are the smallest possible natural numbers.

Since 1:√2 = √2-1:2-√2,
m:n = n-m:2m-n
1:√2 = n-m:2m-n

Since m < n < 2m,
0 < n-m <m
This means n-m is a natural number that is smaller than m. However, as I said, m is the smallest possible natural number. Therefore,
1:√2 = m:n = n-m:2m-n
can't be true.

[thomasthedankengine]

Is there another proof? I always see this by one by contradiction but maybe there is a COOL way to prove this.

[SeeTv.]

I know the proof already but now I finally understand why you suppose p and q have no factors i common, thanks!

[MathemaTeach]

Ohhh very cute pet! Very organized in your proof, that is a great help to our students to better understand our Math language. Good Job

[thomasborgsmidt]

What I mean is: It is the then simple to show, that ANY prime number r of any s root where p/q is SUPPOSED to be equal to r^(1/s).
Now there is NO way that r=r^s (times some arbitrary number without r as prime factor)

[gauravgautam]

Sir can you prove it using any other method??

[ricardoguzman]

Super simple proof: p^2 = 2q^2. Now consider the prime factorization of both sides of the equation. The left side has an even number of factors of 2 while the right side has an odd number of factors of 2 so the two sides cannot equal each other.

[ZipplyZane]

I tried to do this one but got stuck, but then had a thought about the more general concept, and wound up proving that the nth-roots of all integers are either integers or irrational numbers. (Well, technically, I proved it for square roots and realized that it worked for all roots.)

It happened a week ago, but I just saw this video now, and decided I had to tell someone who might appreciate it.

[aneeshsrinivas]

is it a good idea to end my proofs by contradiction with "or else i'm gonna need to contact god or his tech support team to fix this bug in the next version of the universe ".

[thomasborgsmidt]

What I mean by being unnessarily restrictive: The rational numbers are the SPCIAL case! NOT the irrational.
There is however a class of numbers that can be described by the product of prime numbers (all of them - only some of them might be raised to the power of 0 - generally the majority have the power of 0).
where the power of the prime numbers are rational. I. e. 27 ^ (5/2).
If You write a paper on it - please include my name as co-author.

[gurindersinghkiom]

Good... In India it's in 10th standard. U explored every aspect wonderfully. Thanks

[szerszen]

Excellent! Dr Peyam, maybe you can make a video about number e, many facts and lemmas about this number. I will be greatful!

[aneeshsrinivas]

in fact in general if p1, p2, p3, ..., pn are distinct prime numbers then √(p_1*p_2*...*p_n) is irrational.

This is one way to prove the fact that any natural number which isn't a perfect square has an irrational square root.

[thomasborgsmidt]

The problem is that this proof is FAR from being general enough. In other words it is unessesarily constrained.
Now write both p and q as prime factors. Then fold q into p; but with q having negative powers. If p and q have a common factor it will mean that p contributes with a positive power and q contributes with a negative power. This means that you can subtract the of q from the power of p.

[squareheadc]

Thanks Dr Peyam for another lecture. Some things that come to mind when I see that.
1) Are you kidding when you are saying that the proof is very important, or is it actually?
2) I guess that this proof, more than proving sqrt(2) to be not rational, suggests a definition of irrational numbers, i.e., that they can't be written as a ratio of two integers. But we haven't for instance ruled out that it can't be written as a ratio of two real numbers (I guess it can be proven easily that turning to real numbers doesn't change anything but this proof doesn't claim that).
3) For a computer scientist, it is a bit hard to grasp what you mean with drawing a random real number. In a computer you would by necessity always grab a real number with a limited number of digits, and hence all samples would be rational by design. I guess that you refer to something along the line of: for any pair of real rational numbers that you can grab on the real line, e.g., 0.111...11 and 0.111...12 you will find infinitely many real irrational numbers and hence the probability of selecting any of the rational numbers will be zero. I guess it works fine in theory but it seems harder to produce such random samples in practice.

[garyhuntress]

Very interesting proof. You opened by saying it is one of the most important proofs in all of math. Can you explain briefly why it's important?