Algebra: What allows us to do the same thing to both sides of an equation?

Solution 1:

Equations

Here's one way of looking at it: if $a=b$, then $f(a)=f(b)$, no matter the function $f(x)$.

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 $$x=f(x+1)=f(2)=1$$

Aren't we done? We've isolated $x$, so what more is there to do? The problem is that high school trains us to stop thinking once we get to this point. (Actually it trains us not to think about what we're doing at all and instead to rely on the process to do the thinking for us. But manipulating symbols, by itself, doesn't constitute a mathematical argument. Experts often omit the details, but that's because they know how to fill them in if they needed to; beginners should be taught how to fill in the logical details above and beyond the symbol manipulation.)

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." I put quotation marks around that phrase, because it's something of a misnomer: they aren't actually solutions to the (original) equation, precisely because they're extraneous. So, 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 $|x|=1$. At this stage you could analyze the problem into cases (depending on whether $x$ is positive or negative) using the definition of absolute value and deduce that either $x=1$ or $x=-1$. But this doesn't mean that either answer is a solution to the original equation. (Obviously $x=-1$ doesn't satisfy the original equation!) That's because the second step, of squaring both sides, isn't reversible. The chain of implication doesn't flow all the way backwards.

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 $f(s)=\frac{s}{x}$, for example, requires that $x\neq0$; otherwise the value of the function doesn't make sense. So if you have $$x^2=x$$ and you apply the function $f(s)=\frac{s}{x}$ to both sides, you're tacitly assuming that $x\neq0$. 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 $$(\exists x)(x=x+1)\implies0=1$$ By contraposition we conclude that $$\lnot(\exists x)(x=x+1)$$ or in other words that there is no value of $x$ that satisfies the equation $x=x+1$, because assuming there is such a value leads us into a contradiction. (As this example illustrates, if we're being totally rigorous we should really pay attention to quantifiers. But that's more than you're asking.)

And some equations are, in fact, true for all values of the variables. Such equations are called identities. A silly example is $$x=x$$ but a slightly more interesting example is $$x^2-1=(x+1)(x-1)$$ If you try to solve identities like this one, you'll produce a tautology like $0=0$.

Inequalities

For what it's worth, you can also extend this idea to the logic of inequalities. If you apply a function $f(x)$ to the statement $a<b$, you'd typically like to conclude something like $f(a)<f(b)$ or $f(a)>f(b)$. 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., $a<b\iff f(a)<f(b)$ or
• decreasing, i.e., $a<b\iff f(a)>f(b)$

on the interval from $a$ to $b$. So, for example, $f(x)=x+2$ is always increasing, and $g(x)=-x$ is always decreasing, so applying $f$ to $a<b$ yields $$a+2=f(a)<f(b)=b+2$$ but applying $g$ yields $$-a=g(a)>g(b)=-b$$ 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.

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

$$a^2=h(a)>h(b)=b^2$$

but if $a>0$ then the inequality is reversed.

Summary

The moral of this story is:

when you do something to both sides of an equation or inequality, think carefully about what function you're applying. In particular, for equations, think about whether

• It's invertible on the same domain you start out with.

• Its application requires you to make any assumptions.

For inequalities, you also want to think about whether

• the function is increasing or decreasing on the interval defined by the inequality you're starting from.

Solution 2:

What allows us to do the same thing to both sides of an equation?

It depends what thing is. If thing is a function, then it is OK because that's the definition of function. If thing is not a function, watch out.

Example:

$$ \frac{1}{2} = \frac{2}{4} $$

...take the numerator of both sides...