Induction $n\ge 0, n^2-n$ is even

Question

I am stuck trying to get the inductive step done on the following proof...

For all $n \ge 0, n^2 - n$ is even.

I have established that the problem is even by setting it equal to $2k$ and then plugging in $n+1$ but after that I am stuck.

Proof: Proceed using induction. Base case: Set $n = 0.$ $0^2 - 0 = 0$ which is even.

Inductive step: Set $n\ge0$. Suppose that $n^2-n$ is even. Then there exists integer $k$ such that $n^2-n = 2k$.

Then $(n+1)^2-(n+1) = 2k$

I recommend reading this answer. It lays out nicely how to do an induction proof. — Ross Millikan, Nov 15 '17 at 18:37
Here is a MathJax Tutorial that'll help you typeset formulas so they become a good bit more readable. I went ahead and did some basic typsetting of your post already. Edit your post to see what I did. — Bram28, Nov 15 '17 at 18:39
You're close. Expand your $n+1$ step out. Can you replace any bit of your expression with your assumption? What's left of it? Is the remaining part of it even? — thesmallprint, Nov 15 '17 at 18:40
You could also do the inductive step as moving from $n(n-1)$ to $n(n+1)$ and $n(n+1)-n(n-1)=n(n+1-n+1)=2n$ because you can take out the common factor $n$. — Mark Bennet, Nov 15 '17 at 19:37

score 1 · Accepted Answer · answered Nov 15 '17 at 18:39

1

Hint

\begin{aligned} (n+1)^2 -(n+1) &= n^2+2n+1-n-1 \\ &= n^2+n \\ &= (n^2-n)+2n \\ &= 2k+2n \\ &= 2(k+n). \end{aligned}

answered Nov 15 '17 at 18:39

Eff

score 0 · Answer 2 · answered Nov 15 '17 at 18:40

0

It's just $n(n-1)$. One of those terms is even. Unless you are being forced to induct, then ignore me.

answered Nov 15 '17 at 18:40

Randall

score 0 · Answer 3 · answered Nov 15 '17 at 19:02

$n \ge 0 $, $n^2 -n$ is even.

$n=0$: $n^2-n =0$; ok.

Hypothesis:

Assume valid for $n$, I.e $n^2- n $ even.

Step:

Consider: $(n+1)^2- (n+1) = $

$n^2 +2n+1-n-1= $

$(n^2-n) +2n.$

Since by hypothesis $(n^2-n)$ is even,

adding an even number $2n $,

gives an even number.

Hence:

If $n^2-n$ is even then

$ (n+1)^2 -(n+1) $ is even.

3 Answers3