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I recognize that a rational number a/b means that b does not divide a and also that a and b are relatively prime. Does that mean rational numbers are always expressed in their simplest form? Or is my statement about coprime wrong?

I have tried using Euclid's Algorithm and Bezout's Identity, since that is the section that the question comes from but I have not gotten it to work out.

Please Help.

Eric Wofsey
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Tikwen
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  • This is a huge FAQ, e.g. see the linked dupe (and its links) for many methods of proving this well-known theorem that if $\sqrt{m},$ is rational then it is an integer (or, equivalently, if $m$ is not an exact square then $,\sqrt{m},$ is irrational), e.g. it is a special case of the Rational Root Test. – Bill Dubuque Sep 18 '20 at 22:45

1 Answers1

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Hint:

Write your rational number as an irreducible fraction ($\gcd(r,s)=1$), use that $$\Bigl(\frac{r}{s}\Bigr)^2=m \iff r^2=ms^2$$ and deduce with Gauß' lemma that $s=1$.

Bernard
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  • I do not know what Gauss' Lemma is, we have not covered it in class. I have looked it up too but I think my proof is suppose to use Bezout's Identity, Euclid's Algorithm and the division theorem. – Tikwen Sep 18 '20 at 23:10
  • Gauß' lemma asserts that if $n$ divides $ab$ and $n$ is coprime to $a$, then $n$ divides $b$. And precisely, its proof uses Bézout's identity. – Bernard Sep 19 '20 at 07:37