1. ## Understanding proof by contradiction

Hello guys,
I have problems understanding proofs by contradiction.

This is one of my exercises :

Give a proof by contradiction that if 2n^9 + 5n^7 - 6n^4 - n^2 - 27 = 0 where n is an integer, then n is a positive integer (n > 0)

Obviously, I have to assume that n is negative integer. What I don't know is how to proceed with the contradiction. I try to substitute n in the equation with different positive and negative integers, but it is never equal to zero

2. Hi,
Originally Posted by cyb3r
Give a proof by contradiction that if 2n^9 + 5n^7 - 6n^4 - n^2 - 27 = 0 where n is an integer, then n is a positive integer (n > 0)

Obviously, I have to assume that n is negative integer. What I don't know is how to proceed with the contradiction. I try to substitute n in the equation with different positive and negative integers, but it is never equal to zero
Let's assume $n\leq 0$.

• Is $2n^9$ positive ( $>0$) or non positive ?
• Is $5n^7$ positive or non positive ?
• Is $-6n^4$ positive or non positive ?
• Is $-n^2$ positive or non positive ?
• Is $2n^9 + 5n^7 - 6n^4 - n^2$ positive or non positive ?

Do you see why $2n^9 + 5n^7 - 6n^4 - n^2-27=0$ isn't possible if $n\leq0$ ?

3. So $2n^9$, $5n^7$, $-6n^4$, $-n^2$ are going to be all non-positive, and the whole equation is going to be non-positive, and Zero is not a non-positive number and we make the conclusion that $n$ is positive ?

4. Originally Posted by cyb3r
So $2n^9$, $5n^7$, $-6n^4$, $-n^2$ are going to be all non-positive
Yes.
and the whole equation is going to be non-positive
Yes, $2n^9 + 5n^7 - 6n^4 - n^2 \leq 0$ so $2n^9 + 5n^7 - 6n^4 - n^2 {\color{blue}- 27} < 0$. In other words, $2n^9 + 5n^7 - 6n^4 - n^2 {\color{blue}- 27}$ is negative.
and Zero is not a non-positive number
Yes but what's interesting here is that 0 is non negative. As $2n^9 + 5n^7 - 6n^4 - n^2- 27$ is negative, we've found the contradiction we were looking for.
and we make the conclusion that $n$ is positive ?
Yes we do.