I am having some trouble with this proof.

Part of it is that I have to prove it for every integer. Does this mean I have an inductive step that goes for $P(k+1)$ and $P(k-1)$? Assuming my base case is $n=0$.

Prove $n^2 \geq n$ for every integer.

Starting with the base case $n=0$:


We assume $P(k)$ is true for $k=n$.

Want to show $P(k+1)$ holds, that is

$(k+1)^2 \geq k+1$

Case 1: $P(k+1)$, for integers $\geq 0$

$k^2 \geq k$

$k^2+1 \geq k+1$

$(k^2+1)^2 \geq (k+1)^2$

$k^4+2k^2+1 \geq k^2+2k+1$

$k^2(k^2+1)+(k^2+1) \geq (k^2+k)+(k+1)$



$n^2-n=n(n-1)$. If $n>0$, $n-1 \geqslant 0$, so their product is 0 or positive. If $n \leqslant 0$, $n-1<0$, so again their product is 0 or positive.

To summarise, $n^2-n=n(n-1) \geqslant0\Rightarrow n^2\geqslant n$.


A simple approach would be to write : $$(k+1)^2=k^2+2k+1$$ Since $k^2 \geq k$, you have $(k+1)^2 \geq 3k+1\geq2k +k+1\geq k+1$

Then you have to prove that it is also true for negative integers, you notice that $\forall k>0,(-k)^2=k^2\geq k \geq -k$. So the inequality is also true true for negative integers, so it is true on $\mathbb{Z}$.

  • $\begingroup$ This is very helpful thank you. But does this show this is true for every integer, or simply all the positive, non-zero integers? $\endgroup$ – J00S Jun 4 '16 at 7:10
  • $\begingroup$ I just edited about that point :) $\endgroup$ – Jennifer Jun 4 '16 at 7:10
  • $\begingroup$ I messed with the inequalities it is fixed now $\endgroup$ – Jennifer Jun 4 '16 at 7:13

Hint: Let $x \in \mathbb R$. $x^2-x \ge 0 \Leftrightarrow x \le 0 $ or $x \ge1$

$\mathbb Z \subset \mathbb R$


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