How abut the following proofs that sqrt{2} is irrational.
1. If 2=a^2/b^2 with a and b relatively prime then a^2=2b^2 so a is even, and, letting a=2c, b is also even a contradiction.
2. Use the lemma that a positive real r is rational if and only if there is a positive integer b such that br is an integer. (Proof left to the reader.) if sqrt{2} is ration then there is an integer b such that bsqrt{2}=a for some integer a. Let b be the smallest such integer. Then, if c=b(sqrt{2}-1) then c is smaller than b, c=bsqrt{2}-b is an integer, and sqrt{2}c=sqrt{2}b(sqrt{2}-1)=2b-b*sqrt{2} is an integer, a contradiction (this uses infinite descent).
3. Use the theorem that if x, y, and n are positive integers such that x^2-ny^2=1 then sqrt{n} is irrational, and apply with n=2, x=3, y=2.
Proof of theorem.
a. If x^2-ny^2=1, then using (x^2-ny^2)^2=(x^2+ny^2)^2-n(2xy)^2 to show that there are arbitrarily large solutions to x^2-ny^2=1.
b. If n=a^2/b^2 then 1=x^2-(a^2/b^2)y^2 so b^2=b^2x^2-a^2y^2=(bx+ay)(bx-ay)>=bx+ay>bx so b>x but this contradicts the existence of arbitrarily large x.
1. If 2=a^2/b^2 with a and b relatively prime then a^2=2b^2 so a is even, and, letting a=2c, b is also even a contradiction.
2. Use the lemma that a positive real r is rational if and only if there is a positive integer b such that br is an integer. (Proof left to the reader.) if sqrt{2} is ration then there is an integer b such that bsqrt{2}=a for some integer a. Let b be the smallest such integer. Then, if c=b(sqrt{2}-1) then c is smaller than b, c=bsqrt{2}-b is an integer, and sqrt{2}c=sqrt{2}b(sqrt{2}-1)=2b-b*sqrt{2} is an integer, a contradiction (this uses infinite descent).
3. Use the theorem that if x, y, and n are positive integers such that x^2-ny^2=1 then sqrt{n} is irrational, and apply with n=2, x=3, y=2. Proof of theorem. a. If x^2-ny^2=1, then using (x^2-ny^2)^2=(x^2+ny^2)^2-n(2xy)^2 to show that there are arbitrarily large solutions to x^2-ny^2=1. b. If n=a^2/b^2 then 1=x^2-(a^2/b^2)y^2 so b^2=b^2x^2-a^2y^2=(bx+ay)(bx-ay)>=bx+ay>bx so b>x but this contradicts the existence of arbitrarily large x.
How are any of these the same as any other?