8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ see here math.stackexchange.com/questions/499819/… $\endgroup$ – adember May 9 '15 at 6:06
  • 1
    $\begingroup$ 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". $\endgroup$ – layman May 9 '15 at 6:07
  • $\begingroup$ 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 $q^{2}$? $\endgroup$ – HLM May 9 '15 at 6:12
  • 4
    $\begingroup$ Suppose not. Then the square root of some irrational number is rational. Let $m$ be such a number. Then continue as you did. $\endgroup$ – André Nicolas May 9 '15 at 6:29
  • 1
    $\begingroup$ 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". $\endgroup$ – DanielWainfleet Jul 20 '18 at 20:08
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
-2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.