Irrationality of $\sqrt{2m}$

Question

I have this problem:

If $\sqrt{m}$ is irrational, and $m$ is odd, so:

Is $\sqrt{2m}$ always irrational ?

My thought was:

I have a $m$ number, that can be of the form $(2n \pm 1)$, that is square is irrational.

To find a counterexample, I need this:

$2m = x^2$,

then I need an integer, that have perfect square, divided in $2$ is an odd number, that is equal to:

Lets $x$ any integer:

$\frac{x^2}{2} = k$

Then, $k$ must be an integer, since it is divisible by 2 and must be a number of form $(2n \pm1)$ that is $m$.

With these conclusions, I can reform my problem to:

Is there an perfect square, which is divisible by $2$, and this division is an odd number?

I do not ask this directly, because there may be an easier approach, but still this form seems very interesting to me.

PS: I have tested it in programming between $[1, 1000]$ and there is no perfect square, with these characteristics.

Answer 1

Suppose $a^2 = 2k$ for some odd $k$.

If $a$ is odd, then $a^2$ is odd too, so that can't be the case.

However if $a$ is even, then $a=2n$ for some $n$, and $(2n)^2 = 4n^2 = 2k$, so $k=2n^2$ and was not odd after all. So this is also impossible.

Since neither even nor odd integers can be roots of your perfect square, it cannot exist.

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In general one can prove that if a prime $p$ divides a perfect square $a^2$, then $p$ also divides $a^2/p$.

Answer 2

If $2m $ is a perfect square,

$$2m=n^2=(2^{a_1}3^{a_2}5^{a_3}...p^{a_k})^2$$

thus $$a_1\ge 1$$ and

$$m=2 (2^{a_1-1}3^{a_2}...p^{a_k})^2$$ is even.

Answer 3

Let's take the scenic route. Remember that $\sqrt{ab} = \sqrt a \sqrt b$. This means that $\sqrt{2m} = \sqrt 2 \sqrt m$. So $\sqrt{2m}$ is the product of two irrational numbers. It could still be a rational number, though. Okay, sorry, that was more of a dead end than a scenic route.

You say you're looking for $2m = x^2$, where $x$ is an integer. Clearly $x$ must be even. And it turns out that $m$ must also be even, because all even squares are divisible by 4, but this contradicts your stipulation that $m$ is odd.

Or what if instead $m$ is a multiple of $\sqrt 2$ but not of 2? Then $m$ is not a regular integer (though it is an algebraic integer) and neither is $x$. But $2 \sqrt 2$ is not an integer in the usual sense.

Or what if $m$ is simply half $\sqrt{2m}$? For example, suppose that 6 is a square. Then in $2m = x^2$, $$m = \frac{\sqrt 6}{2}.$$ From this it follows that $$m^2 = \frac{6}{4} = \frac{3}{2}.$$

We can slice this thing a thousand different ways, they will all fail to deliver the number you're looking for, because it doesn't exist to be found. No amount of brute force number crunching can find what doesn't exist.

Answer 4

That is equivalent to: There are no square numbers of the form $2m$, where $m$ is odd.

Proof: Take any number $x$, if it is odd then $x^2$ is odd. If it is even, then $x^2$ is divisible by $4$. That means that if $2m = x^2$, $m$ must be divisible by $2$, which contradicts the statement $m$ is odd.