Writing proofs of A = B

Section 4.1 Writing proofs of \(A = B\)

Starting out with proof writing, many proofs that we write will either consist of or be heavily centred around showing some equality \(A = B\text{.}\) Or possibly it will be an inequality \(A \le B\) for which this advice will also apply.

Let us start by looking at the kind of proof many people write when they begin writing proofs.

Example 4.1.1. To show that \(\ln(a) + \ln(b) = \ln(ab)\text{,}\) beginning proof.

\begin{align*} \ln(a) + \ln(b) \amp = \ln(ab)\\ e^{\ln(a) + \ln(b)} \amp = e^{\ln(ab)}\\ e^{\ln(a) + \ln(b)} \amp = ab\\ e^{\ln(a)} \cdot e^{\ln(b)} \amp = ab\\ a \cdot b \amp = ab\\ \text{true} \end{align*}

Therefore \(\ln(a) + \ln(b) = \ln(ab)\text{,}\) QED.

As you progress with proof writing, and reading your professors' and textbooks' proofs, you'll see that this style is discouraged. If you ever end up teaching proof writing in any capacity, you'll inevitably have to communicate what exactly is wrong with this proof.

There are two main problems here. The first is that these chains of implications need to be reversible (if and only if) statements. This is important because it's not true that if \(A = B\) implies \(C = D\) and we can see that \(C = D\) is true, then we cannot necessarily reverse that and claim that \(A = B\) is true. Such non-reversibility is the basis of many a false proof. E.g.

\begin{align*} x \amp = 1\\ x \cdot x \amp = x \cdot 1\\ x^2 - x \amp = 0\\ x(x - 1) \amp = 0\\ x \amp = 0\\ 0 \amp = 0 \end{align*}

Here where we went wrong is that \(x(x - 1) = 0\) is not the same as \(x = 0\text{.}\)

I feel however that this point gets harped on disproportionally and to the extent that all it teaches people to do is to insert some \(\iff\) symbols and maybe reverse the order of the statements at the end as in the next example.

Example 4.1.2. To show that \(\ln(a) + \ln(b) = \ln(ab)\text{,}\) intermediate proof.

\begin{align*} \amp\amp a \cdot b \amp = ab\\ \amp\iff\amp e^{\ln(a) + \ln(b)} \amp = ab\\ \amp\iff\amp e^{\ln(a)} \cdot e^{\ln(b)} \amp = ab\\ \amp\iff\amp e^{\ln(a) + \ln(b)} \amp = e^{\ln(ab)}\\ \amp\iff\amp \ln(a) + \ln(b) \amp = \ln(ab) \end{align*}

Therefore \(\ln(a) + \ln(b) = \ln(ab)\text{,}\) QED.

Now, I won't say this isn't better, but it does bring us to problem number two. The problem for which I feel is the most important of the two, namely this: proofs are about communication. Specifically, a proof should appeal to definitions and previous facts (theorems), as well as explain each step, in order to convince the reading that what they are reading follows logically.

Let me first show an example of how I would write this proof and then discuss.

Example 4.1.3. To show that \(\ln(a) + \ln(b) = \ln(ab)\text{,}\) my proof.

First, recall that \(\ln(ab)\) is defined to be the quantity \(t\) for which \(e^{t} = ab\text{.}\) Therefore, we need to show that \(t = \ln(a) + \ln(b)\) is such a quantity. That is, that \(e^t = ab\text{.}\)

To that end, we calculate

\begin{equation*} e^{\ln(a) + \ln(b)} = e^{\ln(a)} \cdot e^{\ln(b)} = a \cdot b \end{equation*}

since again, by definition, \(e^{\ln(x)} = x\text{.}\)

Therefore, \(\ln(a) + \ln(b) = \ln(ab)\text{.}\)