Formal writing in math: equations

Question

What is the formally correct way to solve a bunch of equations in math?

Is it \begin{align} 42x = 4324 \\ x = 4324/42 \end{align} or \begin{align} 42x = 4324 \\ \Rightarrow x = 4324/42 \end{align} or \begin{align} 42x = 4324 \\ \Leftrightarrow x = 4324/42 \end{align} or \begin{align} 42x = 4324 \\ \text{and so} \ x = 4324/42 \end{align} or something completely different? I've been using the second listed above (it looks the best), and now my instructor just told me that it's wrong to use that logical symbol in that manner.

Answer 1

The problem your instructor has (I think) is that you state a set of propositions with these symbols, and a proposition is either false or true. In the case of a set of propositions it can be true for some $x$ and for some $x$ it can be false. For example, it is perfectly fine to state $$[42x=4324] \Rightarrow [x=35867879]$$

Which is true for all $x \neq \frac{4324}{42}$.

The same goes actually for the third. This leaves the first and the fourth. Since the first is less writing, I would prefer the first.

Answer 2

There are two very different ideas among your proposals:

1. Implication: $\implies$ or "and so" indicate that the second row follows from the first by some logical inference rule.

2. Equivalence: $\iff$ indicates a double implication, the first row reciprocally follows from the first. This is a stronger statement.

In your example it makes no difference, but it would in an asymmetrical case like $x=3$ vs. $x^2=9$.

When solving an equation, a straight implication $\implies$ ensures that you don't drop a solution, while reverse implication $\impliedby$ ensures that no solution extra solution is introduced. For a correct solution, you need equivalence.

So the notation $\iff$ is bulletproof, while with other variants you should make your meaning explicit.

Answer 3

To use $\implies$ is problematic as it does not really do what you want.

When you have (Eq 1) $\implies$ (Eq 2) and $x$ is a solution of (Eq 2) then it does not follow that $x$ is a solution of (Eq 1).

You only get that if $x$ is a solution of (Eq 1) then it is also a solution of (Eq 2). Thus (Eq 2) gives your restrictions on the solutions of (Eq 1). You then would have to go back and check if the soltutions of (Eq 2) actually are solutions of (Eq 1).

If you do not do this, you did not really solve (Eq 1). To avoid this you should write an equivalence $\Leftrightarrow$.