Common errors in proofs

There are lots of pitfalls in setting out proofs and you should start to collect a set of examples of where proofs can go wrong and look out for these sorts of errors and misunderstandings in your own work and in proofs that you are given to study. In this section we shall look at a few examples of the sorts of mistakes and errors that can occur in proofs; but, be warned, this is not an exhaustive list and there are many errors that mathematicians can make when setting out proofs.

Square roots and squaring equations:

In this specification we shall take it that $\sqrt{x}$ means the positive number $y$ such that $y^2 = x$; this is standard in mathematics. Generally, we need to be careful with equations when we square them. We need to be careful in case we generate extra solutions to the equation. Here are two examples:

Dealing with inequality signs:

When students first meet inequality signs, they naturally assume they behave the same way as equals signs: they don’t. For an equals sign, the general rule that students go by is that “whatever you do to one side of the equals sign you must do to the other” and this is usually fine for equals signs. However, if you use this rule within inequality signs then it might be the case that the inequality is no longer preserved. Here are some pitfalls that you need to watch out for:

Squaring both sides:

$-5 < 4$ is correct but on squaring we obtain the false $25 < 16$

Multiplying both sides by a negative number:

$1 < 2$ is true, but on multiplying by $-1$ we obtain the false $-1 < -2$

Taking some function of both sides:

$\frac{\pi}{4} < \frac{\pi}{3}$ is true but on taking the cosine of both sides we obtain the false $\cos\frac{\pi}{4} < \cos\frac{\pi}{3}$

Dividing or multiplying by zero:

Dividing both sides of an expression by a second expression that is equal to zero can cause problems. Generally, we cannot divide by zero as it can generate nonsense. For instance, we know $7 \times 0 = 5 \times 0$ but we cannot divide both sides by $0$ to give $7 = 5$. This issue extends to examples that contain algebra. Here is a classic proof that commits this error [can you spot exactly where the error occurs?]:

Let $x$ and $y$ be non-zero numbers such that $x = y$
Then we can write $x^2 = xy$
Subtract $y^2$ from both sides: $x^2 - y^2 = xy - y^2$
So $(x + y)(x - y) = y(x - y)$
Dividing by $(x - y)$: $x + y = y$
As $x = y$, we have: $2y = y$
Then dividing by the non-zero number $y$: $2 = 1$
Subtracting 1 from both sides: $1 = 0$
Therefore $1 = 0$