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The problem I'm having with this proof is that I'm not sure if my proof actually proves the theorem correct or if I'm using circular reasoning.

Theorem:

Prove that the square root of any irrational number is irrational.

Proof:

=> Suppose not. The square root of any irrational number is rational.

=> Let $m$ be some irrational number. It follows that $\sqrt{m}$ is rational.

=> By definition of a rational number, there are two positive integers $p$ and $q$ such that $\sqrt{m} = \dfrac{q}{p}$

=> $m = \dfrac{q^{2}}{p^{2}}$

=> $q^{2}$ and $p^{2}$ are integers, and by definition of a rational number, $\dfrac{q^{2}}{p^{2}}$ is rational

=> $m$ is irrational and is equal to the rational number $\dfrac{q^{2}}{p^{2}}$. This is a contradiction.

=> Thus, the square root of any irrational number is irrational.

I've seen this proof also done with the addition of these steps:

$m = \dfrac{q^{2}}{p^{2}}$

$m \times p^{2} = q^{2}$

Because a an irrational number times a rational number is irrational, we have an irrational number equaling a rational number which is a contradiction.

My question is: Is this step really necessary?

me_ravi_
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HLM
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  • see here http://math.stackexchange.com/questions/499819/a-square-root-of-an-irrational-number – adember May 09 '15 at 06:06
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    Regarding your question of "is this step really necessary?" -- to which step are you referring? Also, in the second proof, you said "an irrational number times a rational number is irrational" -- this isn't always true. Take $\pi$ and $0$. We have $\pi \cdot 0 = 0$, and $0$ is rational. BUT it is true if the rational number you are multiplying by is non-zero. And in your proof, $p^{2}$ is assumed to be non-zero since it was in the denominator of the fraction, so everything is still OK except that you should amend the statement to "irrational times non-zero rational = irrational". – layman May 09 '15 at 06:07
  • I meant the entire step where we state the products of $m$ and $p^{2}$ to be an irrational number equaling a rational number $q^{2}$ and stating that this is a contradiction. Can't we just say that the contradiction is the irrational number $m$ equals the rational number $\dfrac{p^{2}}{q^{2}}$ and be done? Is there any point in taking the products of $m$ and $q^{2}$? – HLM May 09 '15 at 06:12
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    Suppose not. Then the square root of some irrational number is rational. Let $m$ be such a number. Then continue as you did. – André Nicolas May 09 '15 at 06:29
  • Thank you for all the help! And thank you for that catch Andre. – HLM May 09 '15 at 06:49
  • Note that the claim is equivalent to: The square of a rational number is always rational. – Hagen von Eitzen Sep 25 '16 at 15:29
  • Are we assuming that the irrational number is positive? There is a counter example in $\sqrt{-\sqrt{2}}$, which doesn't even have a real root. – Obinna Nwakwue Apr 24 '17 at 23:04
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    Following the words "Suppose not" in your proof: The negation of "All dogs are black" is not "Every dog is non- black". It is "There exists at least one non-black dog". The negation of "All sqrts of irrational are irrational" is not "All sqrts of irrationals are rational". It is "There exists at least one irrational with a rational sqrt". – DanielWainfleet Jul 20 '18 at 20:08

2 Answers2

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Said shortly, $$\left(\frac pq\right)^2=\frac{p^2}{q^2}$$ is rational.

The square of any rational is rational, hence no rational is the square root of an irrational.

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Unnecesarily sophisticated proof: $$ m^2\not\in{\Bbb Q}\implies[{\Bbb Q}(m^2):{\Bbb Q}]>1\implies [{\Bbb Q}(m):{\Bbb Q}] = [{\Bbb Q}(m):{\Bbb Q(m^2)}]\,[{\Bbb Q}(m^2):{\Bbb Q}] > 1. $$

Martín-Blas Pérez Pinilla
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