Writing formal proof

Question

I need to either prove or disprove that if the square root of n is even, then n is even.

I believe the statement is true as there is no such n that contradicts this but I'm not sure about where to start the proof. Thank you.

Answer 1

Let $k$ be a number such that $k^2=n$. We need to show that if $2|k$ then $2| n$.

Since $2|k$ we have $k=2j$ for some integer $j$.

Notice that then $n=(2j)^2=2(2j^2)$. So $2|n$.

Answer 2

You have a couple of ways to go. The first step is identifying your hypotheses and your thesis. In this case, your hypothesis is: "$\sqrt{n}$ is even" and your thesis is "$n$ is even".

Now, you need to show $H \Rightarrow T$.

There's a couple of ways of facing this problem. One way of doing it is starting in $H$ and "manipulating" what you know (in this case $H$ = "$\sqrt{n}$ is even, which can be expressed as $\sqrt{n} = 2k,\ k\in \mathbb{Z}$", for example), to arrive to $T$ (which you could express as $n = 2q, q \in \mathbb{Z}$). How could you go about it? You need to get rid of the square root...

Another way is by means of the contrarreciprocal: showing $H \Rightarrow T$ is equivalent to showing $\neg T \Rightarrow \neg H$, in this case you would say "If $n$ is not even then $\sqrt n$ is not even" (we should clarify that $n$ is a square, though, for this to make sense). I think this is harder for this problem than the other way around, though.

A final strategy is using a proof by contradiction, in which you assume that both $H$ and $\neg T$ are true, and then deduce something outright impossible or absurd (like $1 = 0$). This means that $H$ and $\neg T$ can't be both true, so if $H$ is true then so must be $T$. In your case, we say "Let's assume $\sqrt n$ is even, and $n$ is odd", and then try to derive the contradiction. If you try it this way, you'll probably end up doing something similar to the other two-- sometimes proof by contradiction is the only way to go: this isn't the case.

When one faces this kind of problems one should consider these different alternatives: in this particular case the first one is the correct choice.

I suggest you try to prove it yourself before reading what's next, but for the sake of completeness, here's the proof:

Doing it the first way: $\sqrt{n} = 2k, k \in \mathbb{Z} \Rightarrow \sqrt{n}^2 = (2k)^2 \Rightarrow n = 4k^2 \Rightarrow n = 2(2k^2) \Rightarrow n = 2q, q \in \mathbb{Z} $

Answer 3

You are given that $\sqrt{n} = 2k$ for some integer $k$.

So $n = (2k)^2 = 4k^2 = 2(2k^2)$ so $n$ is even.

I think maybe you meant that you are given $n = 2m$ and $\sqrt{n} = k \in \mathbb Z$ and you need to prove $k$ is even.

You know $k^2 = 2m$ so $\frac {k^2}2 = m$. As $2$ is prime $2$ must be a prime factor of $k^2$. But $k^2$ has the exact same prime factors as $k$. So $2$ is a factor of $k$ and $k$ is even.