Prove if 3n^2 + 1 is even, then n is odd Step by Step

Step by step

The question

Question: prove that if 3n² + 1 is even, then n is odd.

1 Turn it around

An if-then statement means the same as its contrapositive: if not Q, then not P.

We saw it with truth tables.

if P, then Q ≡ if not Q, then not P

Turned around, it says: if n is even, then 3n² + 1 is odd.

Starting from n is much easier: we know how to write it with letters.

turned around: if n is even, then 3n² + 1 is odd

Careful!

That swaps the two parts.

It is the converse, a different statement.

Turned around means: swap the parts and put not on both.

Careful: not if n is odd, then 3n² + 1 is even ✗

2 Write it with letters

n is even, so n = 2a, where a is a whole number.

n = 2a

That is what even means.

even = 2 × a whole number

3 Put them in

Now put them in: 3n² + 1 becomes 3(2a)² + 1.

3n² + 1 = 3(2a)² + 1

4 Work it out

(2a)² means the bracket times itself.

= 3(2a)(2a) + 1

Every part times every part: 12a² + 1.

= 12a² + 1

5 Show the form

We take out 2 from all but the 1: 2 × 6a² + 1.

= 2 × 6a² + 1

6a² is a whole number, because whole numbers added and multiplied stay whole.

6a² is a whole number

2 times a whole number, plus 1: that is the form of an odd number.

So 3n² + 1 is odd.

The turned statement is true, so the first one is true too: if 3n² + 1 is even, then n is odd.

The square at the end means: the proof is done.

So: if 3n² + 1 is even, then n is odd. ∎

6 Check

Try n = 1.

Then 3n² + 1 is 4: even.

n = 1: 3n² + 1 = 4, even

And 1 is odd, as the statement says.

Right!

and 1 is odd

Let's look back.

First: turn it around.

Then: write it with letters.

Then: put them in.

Then: work it out.

Then: show the form.

Last: check.

Answer