This is a very simple proof, but I know that proofs which are this simple can often have some erroneous assumptions. Is mine okay?

Argument: With the exception of perfect squares, there are no other whole numbers whose square root is a rational number.

Proof. Let $\frac{a}{b}$ (where $b \neq 0$, $1$, or $a$, and $a>b$) be the square root of a certain number $x$, and $a$ and $b$ share no common factors. Therefore, $$x=\frac{(a\cdot a)}{(b \cdot b)}$$ and this is clearly irreducible. Therefore, $x$ cannot be a whole number.

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    $\begingroup$ Behind "clearly" lies many common errors - just as here. Proving that is the essence of the matter. $\endgroup$ – Math Gems Mar 12 '13 at 19:59
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    $\begingroup$ Take a look at the "clearly" part, and make sure that you can actually justify it (if you can't figure out how, then perhaps it is not so clear as all that). $\endgroup$ – Cameron Buie Mar 12 '13 at 20:00
  • $\begingroup$ Well I think it is pretty intuitive and anyway this statement could be pretty easily shown by writing a and b in their prime factorizations and showing that (a^2)/(b^2) also share no common factors. $\endgroup$ – Ovi Mar 12 '13 at 20:18
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    $\begingroup$ The "pretty intuitive" fact you are using is called The Fundamental Theorem of Arithmetic. It may seem obvious, but there are some types of "numbers" (different from the usual integers) for which the corresponding statement is false. $\endgroup$ – Noah Stein Mar 12 '13 at 20:25
  • $\begingroup$ @Ovi That's right, you can easily prove it using existence and uniqueness of prime factorizations. You should explicitly state that instead of "clearly". Otherwise there is no way for the reader to know for sure what proof you have in mind. Throughout history many believed that uniqueness of factorization was a "law of thought" which did not require rigorous proof. Such errors were clearly exposed only when more general number systems were investigated, many of which do not enjoy uniqueness of factorization. $\endgroup$ – Math Gems Mar 12 '13 at 21:00

A good rule of thumb is that any time you must say "clearly" or "obviously" in a proof, you should either omit the statement or justify it further. I know they sound like that in math books all the time, but they are trying to teach with their mathematical writing, whereas you're trying to prove something. (Also, in my opinion, they shouldn't.) For example, your "clearly" part is generally justified by infinite descent when proving $\sqrt{2}$ is irrational, which is a pretty nontrivial technique.

Another heuristic for telling whether a direct proof is good is to turn it into the contrapositive statement and see how well it works. If a step needs more justification in that direction, it probably does in the original direct proof too. Applying this to your proof, the step "if $x$ is a whole number, then $a^2/b^2$ is reducible" is exactly the assertion behind the infinite descent argument, which obviously requires elaboration.

  • $\begingroup$ Thanks for your feedback guys. $\endgroup$ – Ovi Mar 12 '13 at 20:46

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