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Prove n^2 + n is divisible by 2 by induction

Step by step, with the reason for every step · Induction · Discrete Math

Step by step

The question

Question: prove by induction that n² + n is divisible by 2, for every whole number n.

n2 + n is divisible by 2

Like dominoes: show it for n = 1, then show that k passes it on to k + 1.

1 Check n = 1

For n = 1: 2, and that is 2 × 1.

Divisible by 2: the first domino falls.

12 + 1 = 2 = 2 × 1

2 Assume it works for k

Suppose it works for k: k² + k is a multiple of 2.

So it is 2 times some whole number.

Call it m.

k2 + k = 2m
m is a whole number

3 What we must show

We must show the same for k + 1: every n becomes k + 1.

= (k + 1)2 + (k + 1) is divisible by 2

Careful!

A bracket to a power is the bracket times itself.

Every part times every part.

Careful: not (k + 1)³ = k³ + 1 ✗

4 Use the assumption

Each power of the bracket is the bracket times itself.

(k + 1)2 + (k + 1) = (k + 1)(k + 1) + (k + 1)

Open the brackets.

= k2 + k + k + 1 + k + 1

Like terms together.

= k2 + 3k + 2

Look for the old one inside: k² + k.

The rest is left over.

= (k2 + k) + 2k + 2

By our assumption, the old one is 2m.

= 2m + 2k + 2

5 Take out 2

Every term has a 2 in it.

Take out 2: 2(k + m + 1).

= 2(k + m + 1)
k + m + 1 is a whole number

So it works for k + 1 too.

True for n = 1, and passed on each time: true for every whole number n.

The proof is done.

divisible by 2 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 12.

n = 3: 32 + 3 = 12

12 is 2 times 6.

Divisible by 2.

Right!

12 = 2 × 6

Let's look back.

First: check n = 1.

Then: assume it works for k.

Then: what we must show.

Then: use the assumption.

Then: take out 2.

Last: check.

Answer
k + m + 1 is a whole number

Your turn

Try this one yourself, then check:

n3 − n is divisible by 3
Show the answer
3(k2 + k + m)

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