Why does algebra work?

Why does algebra work?
In middle school, our teachers taught us the mechanics of algebra: isolate the variable by adding, subtracting, multiplying, and dividing on both sides. And then in later years, our toolboxes expanded to include logarithms, exponentials and other, fancier operations.
However, I think it’s easy to get lost in the mechanics and not realize why it all works out. It’s one crucial piece of logic, the basis of all algebra, but it’s easily overlooked, despite becoming important in more complicated problems.
Let’s say we have the equation $2x + 1 = 5$ , and we’re trying to find all possible values of $x$ . We solve it by subtracting $1$ from both sides and then dividing both sides by $2$ :
$2x + 1 = 5,$
$2x + 1 - 1 = 5 - 1,$
$2x = 4,$
$\tfrac{2x}{2} = \tfrac{4}{2},$
$x=2.$
The key to this working? These clauses are linked by “if and only if”s.
$2x + 1$ equals $5$ if and only if $2x$ equals $4$ , which is true if and only if $x = 2$ . Therefore, we can conclude that $2x + 1 = 5$ if and only if $x = 2$ .
That is, one way to look at it is that we’re replacing the original equation with equivalent forms. $2x + 1 = 5$ is true exactly in all the cases that $x = 2$ is true. That’s really the whole idea of algebra (and logic) — replacing one statement with another exactly equivalent until it’s obvious what your result is.
That’s why, in general, it’s not a productive idea to multiply both sides by zero: $2x + 1 = 5$ is not exactly equivalent to $0 \cdot (2x + 1) = 0 \cdot 5$ because the latter is true no matter what $x$ is.
Another tricky case where remembering this concept helps is when squaring numbers. Remember that $x^2 = a$ is not exactly equivalent to $x = \sqrt{a}$ , since $x$ can be $-\sqrt{a}$ too.
“If and only if” is abbreviated $\Leftrightarrow$ , and I think using it generously can help keep algebra tidy and correct.
Originally written January 31, 2013.