When you want to prove that
you must construct a proof of arguments that implies each other, such that
With a direct proof, you are supposed to make a logical chain of reasoning. You begin with . The claim needs to imply something, which then implies something else, and so on, until you end up implying , where is what you were supposed to prove from .
When you are constructing a proof, it’s smart to make an expression you can work with. In this case you’ll need to make expressions for and . Since is an even number, it can be written as , where is an integer. The number is an odd number, and can therefore be written as , where is an integer. You can insert these expressions into , and then you get
where is an integer, and so is an odd number.
You wish to prove that is always divisible by 2:
You know that all even numbers have a factor that is even. This is true because all even numbers can be divided by 2 given the definition of even numbers.
So now you see that you can factorize as . Then you see that and are two integers that follow each other on the real number line (like 3 and 4, 4 and 5). You can now argue that no matter what is, one of the two factors has to be an even number, since every second integer on the real number line is even.
If is an even number, then has to be an odd number. If is an odd number, then insert this expression for in . Thus
Mathematically, the proof looks like the one below, where the argument is constructed with implications: