MHB Is There an Easier Method to Prove $n^2>n$ for Negative Integers?

[cbarker1]

Dear Everyone,

Directions: Decide whether the statement is a theorem. If it is a theorem, prove it. if not, give a counterexample.

$$n^2>n$$ for each negative integer n

Examples might work for this inequality

$$n^2-n>0$$

Let n=-1. Then
$$(-1)^2-(-1)>0$$
$$1+1>0$$
$$2>0$$

Let n=-2. Then
$$(-2)^2-(-2)>0$$
$$4+2>0$$
$$6>0$$

Let n=-3. Then
$$(-3)^2-(-3)>0$$
$$9+3>0$$
$$12>0$$

I figure out the pattern of the inequality. So I need to prove it for all cases.

PROOF: Let n be the negative integers. Then,
$$n^2-n>0$$
$$n(n-1)>0$$
$$n>0 \land n>1$$

Here is where I am stuck with my reasoning. Is there better way to prove it?

Thanks
Cbarker1

 

[Greg]

$ n^2 \ge \sqrt{ n^2 } = |n| > n $ if $ n $ is negative.

 

[cbarker1]

$ n^2 \ge \sqrt{ n^2 } = |n| > n $ if $ n $ is negative.

Yes, it is true.

But I think there is an easier method to prove this, right?
So, I will rewrite the statement into a condition statement.

If n is a neg. integer, then $n^2>n$.

Work for this statement and I am stuck in this part upcoming:

Proof: Suppose n is negative integer. Then, $n>0$. Then $n-1>0$. So $n-1>n>0$.

What to do next in this proof?