Question on even numbers I came along a question asking to define an even number.  My definition was any number divisible by $2$. However, I was thinking that $-2$ is divisible by $2$, even though I never thought of it as an even number. 
Therefore, my question is: Are negative numbers like $-2$ and $-4$ even numbers?
Thank you.
 A: Yes, $-2$ and $-4$ are both even since they can be written as $2(-1)$ and $2(-2)$ respectively. Note that $0$ is also even, since $0 = 2(0)$.
A: Absolutely. Let's look at this from two different angles.
First, the angle of algebraic number theory: let's make sure we agree on what "divisible" means. The sum of an integer to another integer is also an integer. The product of integers is also an integer. We say that the ring of integers is "closed" under addition and under multiplication.
But an integer divided by another integer may or may not be an integer. For example, 22 divided by 7 is approximately 3.142857. So 22 is not divisible by 7. But 21 is divisible by 7, since $3 \times 7 = 21$.
And so we have that 22 is also divisible by 2, whereas 21 is not. So we say one integer is divisible by another if the result of the division is also an integer.
The same goes for negative numbers: $-22$ is divisible by 2 because $$\frac{-22}{2} = -11,$$ whereas $-21$ is not, because 10.5 is not an integer.
In short, if $n$ is an even integer, then $-n$ is also an even integer.
And secondly, look at it from the angle of computer programming. You probably know that almost all computers use binary numbers internally. What you might not know is that almost all computers use two's complement to store negative numbers.
For example, in a signed byte, 2 is 00000010 (no surprise there). And 1 is 00000001 and 0 is 00000000. But $-1$ is 11111111. And $-2$ is 11111110, $-3$ is 11111101, $-4$ is 11111100, etc.
I'm not going to explain two's complement here. It might seem like a needlessly complicated system. Why not just agree that the first bit is the sign bit? Then $-1$ is 10000001 rather than 11111111, $-2$ is 10000010 rather than 11111110, etc.
Of course I'm hardly the first or the last to suggest this. But the big problem with this system is that you have two different representations for 0. That's a bigger problem than the asymmetry between positive and negative integers in two's complement.
But in either system, odd integers have a least significant bit that's on regardless of the sign bit, while all even integers have a least significant bit that's off, whether they be positive, 0 or negative.
