[buzzin_]

What is going on in the third example?

"Let a,b,m,n be integers, and suppose that b^2 = 2a^2 ..."

So, (b^2)/(a^2)=2 b/a = sqrt(2)

or, in other words, sqrt(2) is rational. then it goes and uses this in the rest of the example.

[lupire]

You are referring to https://hrmacbeth.github.io/math2001/01_Proofs_by_Calculatio...

That's hilarious. It's the brick joke, in math. https://plantsarethestrangestpeople.blogspot.com/2012/01/bri... It's a great example, made terrible (for teaching) because the punchline is never explained.

Your excerpt left out key information, which is the secondary hypothesis that am+bn=1

Hint: what are a and b? I won't wait for you to find values that satisfy the hypotheses.

It's the same flavor as: assume a=b+1, and b=a+1. Then a-b=b-a You can prove that p=>q, sometimes, even if p is false. (Especially if p is false!)

A later chapter provides the punchline.

https://hrmacbeth.github.io/math2001/07_Number_Theory.html#t...

[dullcrisp]

The point is that in order to prove something by contradiction, you must assume something that turns out to be false and see what you can derive from that assumption (ideally a contradiction). The last link in the sibling comments does indeed contain proofs that the given premise is impossible. Think about how you know that the square root of 2 is irrational.

[bmacho]

It is correct, true, and funny. Funny in the way that it creates a seeming contradiction and your brain feels good when it solves it.

Nothing special is going on there, but it may be distracting if the joke doesn't land.