Why Must a/b Be in Lowest Terms to Prove sqrt(2) Is Irrational?

[bonfire09]

In proofs like prove sqrt(2) is irrational using proof by contradiction it typically goes like-We assume to the contrary sqrt(2) is rational where sqrt(2)=a/b and b≠0 and a/b has been reduced to lowest terms. I understand that at the very end of the arrive we arrive at the conclusion that it has not been reduced to lowest terms which leads to a contradiction. But what allows me assume a\b has been reduced to lowest terms? For all I know a/b could have not been reduced at the very beginning.

 

[bonfire09]

oops I think I posted it in the wrong section.

 

[Michael Redei]

But what allows me assume a\b has been reduced to lowest terms? For all I know a/b could have not been reduced at the very beginning.

You (the writer of the proof) can assume ("demand" might be a better word actually) whatever you like, as long as you can live with the consequences. In this case, assuming that a/b has been reduced means we're only really proving that √2 can't be equal to any reduced fraction. So this well-known proof actually only applies to a very special case of fractions. √2 might still be equal to some non-reduced fraction.

But you see why proving the special case is enough to cover the non-reduced case as well, don't you? This small step ("no reduced fraction, so no non-reduced fraction either") is so obvious that it's not even mentioned. At least I've never seen it being spelt out explicitly.

 

[bonfire09]

Oh wow now I get it. I never thought of it like that. So basically the proof is saying that sqrt(2) is reduced to lowest terms and sqrt(2) is not reduced to lowest terms which means that it rules out any possibility of it being written as a fraction.Thus its irrational.