$\begingroup$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?
$\endgroup$83 Answers
$\begingroup$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.
$\endgroup$$\begingroup$Unnecesarily sophisticated proof: $$ 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. $$
$\endgroup$$\begingroup$Yes, because you are going round in circle's if you don't say the extra step. It is like proving that cats are better than dogs, and assuming 'if cats are not better than dogs' but this is wrong because 'blah blah blah' and then ending your proof. You have not said anything much, except a>b so b
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