1.2 Proof by deduction — Practise with help

Work each proof on paper first, closed book. Keep the general letter all the way through, and write a sentence at the end.

Warm-up: remember

Expand $(x + 3)(x - 2)$ .
Factorise $n^2 + 3n$ .
Write an odd whole number in the form $2n + 1$ , and write an even whole number in the form $2m$ .

Answers

$x^2 + x - 6$ .
$n(n + 3)$ .
An odd number is $2n + 1$ and an even number is $2m$ , for whole numbers $n$ and $m$ .

Practise with help

1. Prove that the sum of any two even numbers is even.

Hint

Name the two numbers so that the factor $2$ is visible in each one. Call them $2m$ and $2n$ .

Next step

Add the two expressions, then take out the common factor $2$ .

Full working

Let the two even numbers be $2m$ and $2n$ , where $m$ and $n$ are whole numbers. Their sum is

2m + 2n = 2(m + n).

The number $m + n$ is a whole number, so the sum is a multiple of $2$ . Therefore the sum of two even numbers is even.

2. Prove that the product of two consecutive whole numbers is even.

Hint

Two consecutive whole numbers are $n$ and $n + 1$ . One of any two consecutive numbers is even.

Next step

Split the proof into the two cases: $n$ even, and $n$ odd.

Full working

Let the consecutive whole numbers be $n$ and $n + 1$ .

If $n$ is even, it can be written as $2k$ , so

n(n + 1) = 2k(n + 1),

which is a multiple of $2$ .

If $n$ is odd, then $n + 1$ is even, so $n + 1 = 2k$ and

n(n + 1) = 2kn,

which is a multiple of $2$ .

In both cases the product is even. The two cases cover every whole number $n$ , so the product of two consecutive whole numbers is even.

3. Prove that the square of an odd number is odd.

Hint

Write the odd number as $2n + 1$ and square it. The result should be one more than a multiple of $2$ .

Next step

Expand $(2n + 1)^2$ , then factor a $2$ out of the first two terms.

Full working

Let the odd number be $2n + 1$ . Its square is

(2n + 1)^2 = 4n^2 + 4n + 1.

Factorise the first two terms:

4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1.

This is one more than a multiple of $2$ , so it is odd. Therefore the square of an odd number is odd.

4. Prove that the difference between the squares of two consecutive whole numbers is odd.

Hint

The two numbers are $n$ and $n + 1$ . Write down the larger square first, so the difference is positive.

Next step

Expand $(n + 1)^2 - n^2$ and simplify.

Full working

The two consecutive whole numbers are $n$ and $n + 1$ . The difference of their squares is

(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2.

The $n^2$ terms cancel:

n^2 + 2n + 1 - n^2 = 2n + 1.

The result $2n + 1$ is odd, so the difference between consecutive squares is odd.

5. Prove that $x^2 + 2x + 5$ is positive for every real number $x$ .

Hint

Complete the square. Half of $2$ is $1$ , so start with $(x + 1)^2$ .

Next step

Write $x^2 + 2x + 5 = (x + 1)^2 + c$ and find $c$ by comparing the constant terms.

Full working

Complete the square:

x^2 + 2x + 5 = (x + 1)^2 + 4.

A square is never negative, so $(x + 1)^2 \geq 0$ . Therefore

(x + 1)^2 + 4 \geq 4.

Since $4 > 0$ , the value is always positive.

6. Prove that the product of any three consecutive whole numbers is a multiple of $3$ .

Hint

Three consecutive whole numbers contain exactly one multiple of $3$ . Write them as $n$ , $n + 1$ and $n + 2$ .

Next step

Consider the remainder when $n$ is divided by $3$ . There are three cases: remainder $0$ , remainder $1$ and remainder $2$ .

Full working

Let the three consecutive whole numbers be $n$ , $n + 1$ and $n + 2$ . Every whole number leaves remainder $0$ , $1$ or $2$ on division by $3$ .

If $n = 3k$ , then $n$ is a multiple of $3$ .
If $n = 3k + 1$ , then $n + 2 = 3k + 3 = 3(k + 1)$ is a multiple of $3$ .
If $n = 3k + 2$ , then $n + 1 = 3k + 3 = 3(k + 1)$ is a multiple of $3$ .

In every case one of the three numbers is a multiple of $3$ , so their product has a factor $3$ . Therefore the product is a multiple of $3$ .