Prove: if x is odd, then sqrt(x) is odd. If $x$ is odd, then $\sqrt{x}$ is odd, where $x$ is an integer.
Any hints welcome and preferred. Thank you!
 A: 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).
A: 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.
A: 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.).
A: 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.
A: 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.
