What is the best way to prove: Prove every x is Q.

Question

The example question is "prove that every integer that is divisible by 10 must be an even integer".

However this can be generalized to prove that every x is Q, where is a statement such as "must be an even integer". So what is the best way to approach this proof? Contradiction, contrapositive, direct?

My thinking was:

Question translates to $\forall A(10k=A\rightarrow A = 2c)$, such that $k\in\mathbb Z,c\in\mathbb Z$ (also can someone tell me if this way of translating the question is correct)

From there:

Let B be some number that is divisible by 10. In other words, $B=10K$, where $K\in\mathbb Z$

$B = 2(5K)\\B=2P$

Therefore, B is even because for some B, divisible by 10, B must be even by the definition of even numbers.

Is this form the best way to prove these types of questions? Does this count as a proof by contradiction?

Edit: Or would a proof by induction work best here? In that case how would this question be proven (I have an idea; just need a confirmation)?

Answer 1

So what is the best way to approach this proof? Contradiction, contrapositive, direct?

I think there is no "best way" to approach a proof. But most of the time a direct proof should be tried first, because they are most of the time more clear and have other advantages over a proof by contradiction, for example.

Like that a direct proof can give you an algorithm which you can use or implement.

Your translation of the question seems fine to me. Your proof also looks good, but it is not a proof by contradiction. That would be a direct proof.

A proof is generaly build like this. You assume, that $A$ holds and then you conclude $B$. $A\Rightarrow B$. $A$ is here the statement, that you have an integer $k$, which is divisable by $10$. And you want to conclude $B$, which is that $k$ has to be divisable by $2$.

In a proof by contradiction you assume, that not $B$ holds and conclude that not $A$ holds. $\neg B\Rightarrow \neg A$.

It is $A\Rightarrow B\leftrightarrow \neg B\Rightarrow \neg A$.

EDIT: (For the following note the answer of Arsen Berk)

[A proof by induction does not work here. First of all you are working over the integers, but induction is for statements over the natural numbers. Also I can not imagine how the inductive step should work out here. There is no link to this task and induction.]

Answer 2

I think your proof does not count as proof by contradiction, it is more like direct way of proving it and if you can think the direct way, in my opinion, it is clear enough (but it is a matter of style so I can't say this is the "best" way). Induction would also work here, but in your direct way, the argument is simple and clear enough to be understood so I don't think an inductive argument is needed for this kind of proof.

Since you asked it, proof by contradiction would be as the following:

Suppose, for a contradiction, that $x \in \mathbb{Z}$ and $x$ is divisible by $10$ but $x$ is not even. But since $10 = 2\cdot 5$, we need to have $2|x$ and $5|x$. But if $2|x$, then $x$ is an even integer, which is a contradiction as required. Therefore, $x$ must be an even integer.