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Click Here For Image To Proof

With reference to the above image, why is it for this proof that sqrt(2) is irrational, after making the first assumption that sqrt(2) is rational, we can also make what seem like an additional assumption - that the fraction (m/n) is in its lowest form / irreducible (which is later used to produce a contradiction).

It seems like we are making more than one assumption for this proof by contradiction which begs the following 2 questions:

  • How do we know when such additional assumptions can be made? - My classmate have shared that this assumption is fairly made due to a "Proof by infinite descent" - I will like to still hear this community's thought on this / any resources which will point me to a similar direction etc. for a deeper understanding will be greatly appreciated !
  • How can we know which assumption is responsible for the contradiction?
Bill Dubuque
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rustlecho
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    Yes, we may blame any assumption we want. But one is the "guilty suspect", because we do not know if it is true or not. Thus, if from the assumption that it is true we produce a contradiction, then... – Mauro ALLEGRANZA Mar 13 '23 at 14:09
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    In your second line you have irrational where you should have rational. The statement that a rational can be put in lowest forms should be familiar. You can use the Euclidean algorithm to find the GCD of the numerator and denominator and divide them both by it. – Ross Millikan Mar 13 '23 at 14:10
  • @RossMillikan Okay I have made a typo mistake there. I have edited the question already. Thank you – rustlecho Mar 13 '23 at 14:12
  • The idea of the proof is clearer if as here we reformulate it to deduce a parity contradiction in the (unique) power of $2$ that divides a natural – Bill Dubuque Mar 13 '23 at 15:15

1 Answers1

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Actually we implicitly use here another statement: if $q$ is rational number, then there exist integer numbers $m$ and $n$ s.t. $q = m / n$ and $m / n$ is irreducible.

So, better formulation will be: assume $\sqrt{2}$ is rational, then there exist $m$ and $n$ s.t. $m / n$ is irreducible and $\sqrt{2} = m/n$. Then we use this numbers in the rest of proof.

Or, in other words, we don't need to assume that $m / n$ is irreducible, we can choose it to be irreducible.

mihaild
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  • One can also prove sqrt(2) is irrational with the Rational Root Theorem for the polynomial x^2 - 2 = 0, which is much quicker. – Nate Mar 14 '23 at 05:21