# Why do we do the same thing to both sides?

The purpose of this article is to provide students with a deeper understanding of why the rules for solving equations and inequalities work.

## Equations

To begin, notice that if \(a = b,\) then \(f(a) = f(b),\) no matter the function \(f(x).\) More verbosely, putting \(a\) into \(f\) yields the same thing as putting \(b\) into \(f,\) because \(a\) and \(b\) are identical.

From this point of view, solving an equation amounts to applying a sequence of functions in order to generate an equation whose solutions are easy to read off. What this sort of argument shows is that the new equations are logical consequences of the original equation.

So, for example, suppose you're given that

$$x + 1 = 2$$and you want to subtract one from both sides. Then just apply the function \(f(s) = s - 1,\) yielding

$$x = f(x + 1) = f(2) = 1.$$Indeed there are some subtleties here. First of all, just because this last equation is a consequence of the original equation doesn't mean the last implies the original. (That would amount to the very common mistake of thinking a conditional and its converse are logically equivalent.) In other words, for an arbitrary function \(f(a) = f(b)\) need not imply that \(a = b:\) the operation you perform on both sides might not be reversible. (It was in the example I just gave because the function I applied was linear, and all (non-constant) linear functions have inverses that don't require domain restrictions, which makes the transformation "reversible." Unfortunately in school almost all the examples we start out with are linear, so we have our intuition about equation-solving trained on a very special set of examples, which don't illustrate what can happen in general.)

The failure of \(f(a) = f(b)\) to imply \(a = b\) explains why certain operations, for example, squaring both sides, might generate "extraneous solutions." For example, if you apply the function \(f(s) = s^2\) to the equation \(x = 1,\) you deduce that \(x^2 = 1.\) You could then apply the function \(g(s) = \sqrt{s}\) to deduce that \(\lvert x \rvert = 1,\) and from this, deduce that either \(x = 1\) or \(x = -1.\) But \(x = -1\) doesn't satisfy the original equation, because the second step we took, squaring both sides, isn't reversible. Implication is not commutative.

Another subtlety is that applying a certain transformation to both sides may require you to make an assumption without even realizing it. In other words, some operations, such as dividing by \(x,\) tacitly carry certain restrictions. The function \(f(s) = s / x,\) for example, requires that \(x \ne 0;\) otherwise the value of the function doesn't make sense. So if you have \(x^2 = x,\) and you apply the function \(f(s) = s / x\) to both sides, you're tacitly assuming that \(x \ne 0.\) That's why in other cases you might lose solutions rather than generate extraneous ones.

Of course, not all equations have solutions. For example, applying \(f(s) = s - x\) to the equation

$$x = x + 1$$yields \(0 = 1.\) What this argument shows is that

$$(\exists x : x = x + 1) \Longrightarrow 0 = 1.$$By contraposition, we conclude that

$$\not\exists x : x = x + 1$$or in other words, there is no value of \(x\) that satisfies the equation \(x = x + 1,\) because assuming there is would give us a contradiction.

## Inequalities

This idea can also be extended to inequalities. If you apply a function \(f(x)\) to the statement \(a \lt b,\) you'd typically like to conclude something like \(f(a) \lt f(b),\) or \(f(a) \gt f(b).\) In other words, you want to know whether the process preserves the direction of the inequality, or reverses it.

But to draw such a conclusion, you generally need to know whether \(f(x)\) is

- increasing, i.e. \(a \lt b \Longrightarrow f(a) < f(b)\) or
- decreasing, i.e. \(a \lt b \Longrightarrow f(a) \gt f(b)\)

on the interval from \(a\) to \(b.\) For example, \(f(x) = x + 2\) is always increasing, and \(g(x) = -x\) is always decreasing, so applying \(f\) to \(a \lt b\) yields

$$a + 2 = f(a) \lt f(b) = b + 2$$but applying \(g\) yields

$$-a = g(a) \gt g(b) = -b$$This second fact is just what we mean when we say "multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality." It's simply a consequence of the fact that the function \(f(x) = -x\) is decreasing.

What about squaring both sides of an inequality? In that case we're dealing with the function \(h(x) = x^2,\) which is decreasing on \((-\infty,\,0),\) and increasing on \((0,\,\infty),\) so you have to be careful about "squaring both sides" of an inequality. If \(a \lt b,\) and \(b \lt 0,\) then \(h\) is decreasing on \((a,\,b),\) so

$$a^2 = h(a) \gt h(b) = b^2$$but if \(a \gt 0,\) the inequality would be flipped.

This article was mostly taken from here. Primarily, I have modified the original article because the author provided incorrect definitions for increasing and decreasing functions. I have also improved the typesetting for some of the math. Additionally, I chose to remove several rants made by the original author, as well as some offshoots which were unnecessary. For example, the author provided a definition for identities, which would already be known to the reader, and thus, unnecessary.