Proving that there is no negative integer with $n^2+n<0$

Question

I'm trying to prove the statement: There is no negative integer with $n^2+n<0$. For this, I went with proving it by contradiction. Here's what I have so far:

Let us assume there is a negative integer $n$ with $n^2+n<0$. Since $n$ is negative, this means that $n<0$. Then, $n^2\ge 0$, where $n^2=n*n$. However, this is a contradiction because $n^2\ge 0$ and we assumed $n^2+n<0$, $n^2+n$ will at least be $0$ and $0 \nless 0$.

I'm not sure if my proof is correct or even going in the right direction, I think my negated statement may be wrong, which makes my proof wrong too. Any feedback or help is appreciated.

Answer 1

In your argument it is not clear how you conclude that $n^{2}+n $ is at least $0$.

Let $m =-n$. Them $m >0$ and $n^{2}+n <0$ gives $m^{2} <m$. But then you can cancel $m$ and get $m <1$. there is no positive integer $m$ such that $m <1$ so we have a contradiction.

Answer 2

$$n(n+1)<0$$

If $n>0,n+1<0\iff n<-1$ which is untenable

If $n<0,n+1>0\iff n>-1\iff -1<n<0$

Answer 3

Your proof is faulty.

You claim that $n<0$, which is true, and that $n^2\geq 0$, which is true, and then claim that this is a contradiction. Why? There are plenty of numbers $x$ for which $x<0$ and $x^2\geq 0$, there is no contradiction there.

You also claim that "$n^2+n$ will be at least $0$". How do you know this? Note, if you can prove this, then yes, indeed, you can claim that you reached a contradiction. But you didn't prove this yet.

Answer 4

It is not enough. Because $-n\gt 0$ and if $\exists n,0\leq n^2\lt -n$ then $n^2+n\lt 0$

The proof goes as follows $n^2+n=n(n+1)$.

If $n\lt -1$ then $n$ and $n+1$ are both negative and their product is positive.

If $n=-1$ then $n^2+n=0$

Answer 5

Let $n<-1$;

Then

$n^2>-n$; or

$n^2+n >0$; Contradiction.