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

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Discussion Overview

The discussion revolves around the proof of the irrationality of √2, specifically focusing on the assumption that the fraction a/b representing √2 is in lowest terms. Participants explore the implications of this assumption and its necessity in the proof by contradiction.

Discussion Character

  • Exploratory, Debate/contested, Conceptual clarification

Main Points Raised

  • One participant questions the necessity of assuming that a/b is in lowest terms, suggesting that it could have been in a non-reduced form from the beginning.
  • Another participant argues that assuming a/b is in lowest terms allows the proof to demonstrate that √2 cannot equal any reduced fraction, implying that it also covers non-reduced fractions.
  • A later reply expresses understanding of the argument, indicating that the proof's contradiction shows that √2 cannot be expressed as a fraction at all, thus establishing its irrationality.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of the assumption that a/b is in lowest terms. The discussion remains unresolved regarding the broader applicability of the proof to non-reduced fractions.

Contextual Notes

Participants highlight the assumption that a/b is in lowest terms as a critical point in the proof, but there is no consensus on whether this assumption is necessary or if it limits the proof's applicability.

bonfire09
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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.
 
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oops I think I posted it in the wrong section.
 
bonfire09 said:
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.
 
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.
 

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