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If $x$ is odd, then $\sqrt{x}$ is odd, where $x$ is an integer.

Any hints welcome and preferred. Thank you!

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    $\begingroup$ Considering that $\sqrt{x}$ is rarely an integer, this needs more assumptions. Regardless, write down the contrapositive and see what you can do with it. $\endgroup$
    – user61527
    Jun 1, 2014 at 20:34
  • $\begingroup$ @mathh: Hardly. $\sqrt{3}$ isn't odd either. $\endgroup$
    – cHao
    Jun 2, 2014 at 0:15
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    $\begingroup$ This may be better phrased as "Prove: if $n^2$ is odd, then $n$ is odd." $\endgroup$
    – wchargin
    Jun 2, 2014 at 0:44

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Here's the easy way to do it: try proving the contrapositive. If $\sqrt{x}$ is even (not odd), then $x$ is even (not odd).

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As stated in the comments, a necessary assumption is that $\sqrt{x}$ is itself an integer. See what you can get from the following:

  • An odd number times an odd number yields an odd number.
  • An even number times an odd number yields an even number.
  • An even number times an even number yields an even number.

All of these facts can be proven by applying the fundamental theorem of arithmetic. Now think about how these can help you with your problem.

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  • $\begingroup$ Since this seems a very basic question, it may also be useful to remember that $\forall k \in N: 2k + 1$ is an odd number (and 2k is always even). $\endgroup$
    – Voo
    Jun 1, 2014 at 20:41
  • $\begingroup$ Yes! That's probably an easier approach. Likewise, even numbers are of the form $2k$ for some $k \in \mathbb{N}$, and from there, we can simply compute $(2k)^2$ and $(2k+1)^2$ and see what we get for each. $\endgroup$
    – Kaj Hansen
    Jun 1, 2014 at 20:43
  • $\begingroup$ Exactly what I was going for - I just wasn't sure how the modus for these kind of questions is here, i.e. how much help we should provide. $\endgroup$
    – Voo
    Jun 1, 2014 at 20:45
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    $\begingroup$ It's a fine line. Typically, I will try to give more bare-bones hints for easier questions (algebra/precalculus/calculus), and once I encounter more substantial questions from more advanced topics, such as Galois theory, I'll tend to give more detailed responses. The temptation to outright answer a question is always there (and many people do it), but it's not necessarily encouraged as much as giving hints is. Also, it's definitely more acceptable to give a lot away when it's obvious that the question didn't come from homework. Ultimately, it comes down to your personal philosophy. $\endgroup$
    – Kaj Hansen
    Jun 1, 2014 at 20:56
  • $\begingroup$ ^I should add that there are quite a few discussions on this very topic on the Meta board, and there is a wide range of opinions. $\endgroup$
    – Kaj Hansen
    Jun 1, 2014 at 21:01
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Is this proof interesting? ($x$ is an odd number and perfect square.)

First. It is easy to show that if $x$ is odd so is $x^2$.(Take $x=2k+1$ and square.)

Second. Consider the number $$(x-\sqrt x)(x+\sqrt x)=x^2-x$$ The RHS is even so at least one of the factors of LHS is even too. With no loss of generality suppose that $$(x-\sqrt x)$$ is even. Since $x$ is odd $\sqrt x$ must be an odd number(because$(x-\sqrt x)$ is even.).

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    $\begingroup$ love this proof! $\endgroup$
    – Alex
    Jun 2, 2014 at 0:03
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I assume you also mean to assume that $x$ is a square.

Now just think about what odd means. (ie, $n$ is odd if it is not divisible by 2, and even if it is divisible by 2).

Now just look at all the squares you can think of.

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

Of course if the number you're squaring is odd (hence not divisible by 2), its square is also not divisible by 2, and if the number you're squaring is even (ie, divisible by 2), then its square is, since its square is divisible by the number itself.

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An even (positive) integer is divisible by 2.

If any integer is not divisible by 2, then no factor of it is divisible by 2.

So, if any N is odd then all its factors are odd, specifically if N is a square.

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