Wow!! You are so good at teaching! Thank you everybody!HallsofIvy said:"Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. This in turn, would mean that "b" isn't a natural number."
How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that [itex]\sqrt{2}[/itex] is irrational.
Better to note that if [itex]\frac{a}{b}= \sqrt{2}[/itex] then, squaring both sides, [itex]\frac{a^2}{b^2}= 2[/itex] so that a^{2}= 2b^{2} showing that a^{2} is even.
Crucial point: the square of an odd integer is always odd:
If p is an odd integer, then it can be written 2n+ 1 where n is any integer.
p^{2}= (2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+2n)+1.
Since 2n^{2}+ 2n is an integer, p^{2} is of the form 2m+1 (m= 2n^{2}+2n) and so is odd.
Do you see why knowing that tells us that a must be even?