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The proof of the irrationality of $$\sqrt{2}$$ begins by Let us suppose that $\frac{a}{b}$ is in its lowest terms, which is to say $a$ and $b$ have no common factor.

Why can we suppose this?

Elena Greg
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    If a fraction with value $\sqrt{2}$ exists , we can divide numerator and denominator by their gcd and get a fraction with the same value ($\sqrt{2}$) and coprime integers. – Peter Oct 13 '20 at 11:59
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    Let me turn it around and ask you: can you think of any non-integer rational number expressed as a ratio of two integers, where you cannot reduce it to its lowest terms (the point at which no prime factors are common between numerator and denominator). – Deepak Oct 13 '20 at 12:02
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    Assume that $a,b$ have a common factor $k$. This means that $a=a_1k$ and $b=b_1k$ and thus $\dfrac a b = \dfrac {a_1k}{b_1k}= \dfrac {a_1}{b_1}$. End of the story. – Mauro ALLEGRANZA Oct 13 '20 at 12:15
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    The answer to your question depends on the method. If proof by contradiction, first assumption is that "It were rational and can be expressed as a fraction a/b in lowest terms, with a and b integers, i.e., at least one is odd." – MathArt Oct 13 '20 at 12:23
  • Thank you very much – Elena Greg Oct 13 '20 at 12:23

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