How can I prove using an induction technique that an even number plus an odd number is always equals to an odd number?
Thanks in advance!
$\endgroup$ 23 Answers
$\begingroup$No need for induction. Consider $2n$ and $2s+1$, numbers that are even and odd for $n,s\in\Bbb Z$ respectively. Then the sum is given by $2n+2s+1=2(n+s)+1$ which is odd.
$\endgroup$ $\begingroup$Induction is not needed.
An integer $m$ is even iff$m = 2n$for some integer $n$.
Similarly, an integer $m$ is odd iff$m = 2n+1$for some integer $n$.
The sum of an even and odd integer is of the form$2n+2m+1 =2(n+m)+1 $so it is odd.
Exercise: Prove that an integer can not both be even and odd.
$\endgroup$ $\begingroup$The use of induction here is somewhat contrived but here goes . . .
Let $2m$ be an arbitrary even number.
Add $1$. Then $2m+1$ is odd.
Assume $2m+(2n+1)$ is odd for all odd numbers less than or equal to $2n+1$.
Consider the next odd number. We have $$2m+(2(n+1)+1)=2(m+n+1)+1$$ is odd by definition (as it is one away from a multiple of two).
But $2m$ was arbitrarily even. $\square$
$\endgroup$
