4.2 Direct proof

As we discussed in Section 1.4 What is proof in mathematics?, deductive reasoning is fundamental to mathematics. Deductive reasoning means starting with some assumptions and making a correct argument that leads to a conclusion.

A direct proof of a statement is a chain of deductive reasoning that concludes with the claim of the statement. That is, we prove the claim directly. Examples 3.3 and 3.6 included direct proofs. A calculation, such as an algebraic rearrangement, is a form of direct proof.

Direct proofs very often start by applying a formal definition. For example, if we assume that $a$ is even then it is natural to apply the definition and deduce that there exists an integer $k$ for which $a=2k$ . Direct proofs very often end by using a formal definition. For example we might conclude ‘…which equals $2k$ , and so it is even’. Many direct proofs are therefore a ‘definition chase’ from definitions used in the hypothesis to definitions used in the conclusion.

Example 4.9.

Here is a proof that the product of two even integers is even.

Proof. By Definition 2.6, we can write $a=2k$ and $b=2l$ for some integers $k$ and $l$ . Then their product $ab$ can be expressed as

	$\displaystyle ab$	$\displaystyle=2k\times 2l$
		$\displaystyle=4kl$
		$\displaystyle=2\times(2kl).$

Since $2$ , $k$ and $l$ are integers then so is their product $(2kl)$ . Therefore $ab$ fits the definition of an even integer. Hence the product of two even integers is even. ∎

Exercise 4.10.

Prove that the product of an even integer and an odd integer is even.

Solution.This proof is similar to the previous example.

Proof. Using the definitions of even and odd from Section 2.2.2, we can write $a=2k$ and $b=2l+1$ for some integers $k$ and $l$ . Their product is then

$\displaystyle ab$ $\displaystyle=2k\times(2l+1)$

$\displaystyle=4kl+2k$

$\displaystyle=2\times(2kl+k).$

As $2$ , $k$ and $l$ are integers, so is $(2kl+k)$ . Thus $ab$ is an even integer. Hence the product of an even integer and odd integer is even. ∎

Example 4.11.

What can be deduced about the product of two odd integers? [Take a few minutes to think about this yourself.]

We shall prove that the product of two odd integers is odd.

Proof. Let $a=2k+1$ and $b=2l+1$ for some integers $k$ and $l$ . Then,

	$\displaystyle ab$	$\displaystyle=(2k+1)\times(2l+1)$
		$\displaystyle=4kl+2k+2l+1$
		$\displaystyle=2\times(2kl+k+l)+1.$

Note that $(2kl+k+l)$ is an integer as $2,k$ and $l$ are integers. Therefore $ab$ is an odd integer and we have proven our claim. ∎

Exercise 4.12.

Let $n$ be an integer. Prove that $n^{2}+n$ is even.

Solution.Suppose $n$ is an integer. Factorising gives $n^{2}+n=n(n+1)$ , and so $n^{2}+n$ is a product of two consecutive integers. Any pair of consecutive integers must consist of one even and one odd integer. By Exercise 4.10, the product of an even and an odd integer is even. Therefore $n^{2}+n$ must be even.

4.2.1 Proofs of three familiar results

We end this section by proving what should be three familiar results, just in case you have never seen proofs of them before. Remember from Section 1.4 that proofs sometime leave out details that you should be able to fill in yourself. As you read the proofs below, you may find it necessary to convince yourself of certain deductions. We have given you some hints to help.

When you write your own proofs as a student, you should aim to convince your tutor or lecturer that you understand your argument deeply. Therefore, you should not leave out important details.

Proposition 4.13 (Dividing fractions).

If $a,b,c$ and $d$ are integers with $b,c,d\neq 0$ , then $\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$ .

Proof.

Suppose $a,b,c$ and $d$ are integers and $b,c,d\neq 0$ . Then

$\displaystyle\frac{a}{b}\div\frac{c}{d}$	$\displaystyle=\frac{\frac{a}{b}}{\frac{c}{d}}$
	$\displaystyle=\frac{\frac{a}{b}\times\frac{d}{c}}{\frac{c}{d}\times\frac{d}{c}}$	(muliplying top and bottom by $\tfrac{d}{c}$ )
	$\displaystyle=\frac{a}{b}\times\frac{d}{c}$	(since the denominator is $\tfrac{c}{d}\times\tfrac{d}{c}=1$ ).	∎

Note that we really do need to assume that $b,c,d\neq 0$ to avoid division by zero. Can you see precisely why each assumption is needed?

Theorem 4.14 (The Pythagorean theorem).

Suppose that a planar triangle has sides of length $a$ , $b$ , and $c$ . If the angle between sides $a$ and $b$ is a right angle then $a^{2}+b^{2}=c^{2}$ .

Proof.

Suppose that a triangle has sides of length $a$ , $b$ , and $c$ . Assume that the angle between sides $a$ and $b$ is a right angle. Then we can arrange four copies of the triangle as follows to form a large square of side length $a+b$ .

[How can you be sure that the figure formed is actually a square? And what would happen if $b$ was larger than $a$ ?]

The shaded shape is also a square [can you see why?], and we can calculate its area $A$ in two ways:

First, $A=c^{2}$ because $A$ is a square with side length $c$ .

Second, the area of the large surrounding square is $(a+b)^{2}=a^{2}+2ab+b^{2}$ . The area of the original right-angled triangle is $\frac{1}{2}ab$ . So the area $A$ of the shaded square is

	$\displaystyle A$	$\displaystyle=\bigl{(}a^{2}+2ab+b^{2}\bigr{)}-4\bigl{(}\tfrac{1}{2}ab\bigr{)}$
		$\displaystyle=a^{2}+2ab+b^{2}-2ab$
		$\displaystyle=a^{2}+b^{2}.$

We have calculated $A$ in two ways and so these must give the same value. Therefore $c^{2}=a^{2}+b^{2}$ . ∎

Theorem 4.15 (Converse of the Pythagorean theorem).

Suppose planar a triangle has sides of length $a$ , $b$ , and $c$ . If $a^{2}+b^{2}=c^{2}$ then the angle between sides $a$ and $b$ is a right angle.

Proof.

Suppose a triangle has sides of length $a$ , $b$ , and $c$ . Assume that $a^{2}+b^{2}=c^{2}$ .

Form a new triangle with side lengths $a$ and $b$ at right angles. Then, by the Pythagorean theorem, the third side of this new triangle has length $\sqrt{a^{2}+b^{2}}$ . But $a^{2}+b^{2}=c^{2}$ and so $\sqrt{a^{2}+b^{2}}=\sqrt{c^{2}}=c$ , since $c$ is positive (because it’s the side length of a triangle).

Thus we have two triangles with the same side lengths: the original one from the statement and the new one we have just constructed. Because they have the same side lengths, they must have the same angles [why is this?]. Therefore the angle between sides $a$ and $b$ in the original triangle must also be a right angle. ∎

	$\displaystyle ab$	$\displaystyle=2k\times(2l+1)$
		$\displaystyle=4kl+2k$
		$\displaystyle=2\times(2kl+k).$