Product of two odd numbers is odd

Question

How do I prove that the product of odd integers is odd? I know that I'm supposed to use an algebraic equation.

Answer 1

Any odd number $m$ can be written as $m= 2k+1$, where $k$ is any integer.

Take $m = 2k_1+1$, $n=2k_2+1$, and multiply them together.

$$nm = (2k_1+1)(2k_2+1) = \cdots$$

What do you get as a result?

Answer 2

If you are really $14$ I am guessing you haven't had much experience with formal proofs. I'll attempt to make this as simple as possible. Let $n$ and $k$ be a positive integer. That is, we are choosing any two number from the list $1,2,3, \dots$. We know an integer is even if it is divisible by two. If a number is not divisible by two, it is odd. So now we will use a "trick" to give ourselves some odd numbers. No matter what $n,k$ we choose, it should be obvious that $2n, 2k$ are integers that are divisible by two. If we add one to $2n$ and $2k$ then it should also be clear that $2n+1$ and $2k+1$ are NOT divisible by two, due to the remainder of one. In this way, we have chosen two arbitrary odd integers. The arbitrariness is very important, because it means that the final result will be true for any two odd integers we start with. Now let's consider the product of our two odd integers, $$(2n+1)(2k+1)= 2n \cdot 2k +2k +2n +1 \\ = 4nk+2k+2n+1 \\ = 2(2nk+k+n)+1$$ Next you can observe that $(2nk+k+n)$ is some integer, let's say $m = (2nk+k+n)$. Then we can write $$2(2nk+k+n)+1 = 2(m)+1$$ and finally we can conclude that for any two odd integers $2k+1, 2n+1$ that $$(2n+1)(2k+1) = 2m+1$$ for some integer $m$. Hopefully this makes sense! Formal proofs can be challenging, especially when you are used to working with explicit numbers.

Answer 3

It depends a bit how you define an even number, which might seem like a strange thing to say, since even and odd numbers are obvious and familiar. In particular "it is obvious that" an odd number is always of the from $2r+1$ where $r$ is an integer. However, if we define an even number to be a number which is divisible by $2$, and an odd number to be one which isn't even, it may not be immediately obvious how to prove that if $n$ is a number not divisible by $2$, then $n-1$ is even.

One important property of prime numbers is that if a prime $p$ is a factor of $mn$ and $p$ is not a factor of $m$ then $p$ must be a factor of $n$. This is probably the first non-trivial fact which gets proved when we start looking at integers afresh from a rigorous mathematical perspective.

We can use this fact for a proof in your case - a proof by contradiction, which is a little different from the others which have been suggested.

Suppose $m, n$ are odd integers and their product $r=mn$ is even. Then $2$ is a factor of $r$. $2$ is a prime so ... and this contradicts the assumption that ... so the product $r$ cannot be even. A number which is not even is odd.