# Reversible and irreversible operations in elementary algebra

I am in high school algebra, solving typical equations such as rational, irrational, quadratic, etc., and I have come across the idea of extraneous solutions. My textbook does not touch upon the idea of extraneous solutions and how they relate to reversible and irreversible operations, and I can't find much online. Could I get an explanation of what exactly reversible and irreversible operations mean?

I know that squaring is irreversble, as is multiplying both sides of an equation by $$0,$$ but why is multiplying both sides by $$x$$ (where $$x$$ is the variable in the equation) irreversible?

Is it also irreversible if I start with $$2x^2 = x,$$ and divide by $$x$$ to get $$2x = 1,$$ in which case $$0$$ is no longer a solution?

Why is it irreversible if I have $$\displaystyle \frac{x(2x + 1)}{x} = 0$$ and cancel out $$x$$ to get $$2x + 1 = 0$$?

Finally, why is it irreversible if I have $$2x + 1 = 0$$ and multiply both sides to get $$\displaystyle \frac{x(2x + 1)}{x} = 0$$?

These are the four cases I am most interested in. I want to understand them specifically, because I don't want my maths to be ambiguous.

• This question overlaps with yours : math.stackexchange.com/questions/1515492/…
– user186104
Nov 16, 2015 at 3:09
• The reason why $f(x)=\frac{x(2x+1)}{x}$ and $g(x)=2x+1$ are considered to be different is because $f(0)$ is not defined and technically has a hole there (you are not allowed to divide by zero) whereas $g(0)$ is perfectly well defined. You do have that $f(x)=g(x)$ almost everywhere but due to that slight difference, we still say $f\neq g$. Nov 16, 2015 at 3:14
• The "reversible/irreversible" language may be nonstandard (at least I've not heard it before). Hence the difficulty searching for it. Are these terms from your textbook? Nov 16, 2015 at 4:27

If you are given Equation B and the operation that resulted in Equation B (pretend this is all you know) and you can determine the exact equation in the previous line (call it Equation A), then this is a reversible step. For example, if I told you that I added 3 to both sides of Equation A to get x=5 (x=5 would be Equation B in my explanation), then you could reverse this operation by subtracting 3 from both sides to find that Equation A is x-3=2.

If you can't be sure what the line before is, then this is an irreversible step. For example, if I told you that I squared both sides of Equation A to get x^2=4, there's no way of knowing exactly what Equation A is. It could be either x=-2 or x=2. Or if I told you that I multiplied both sides of an equation by zero to get 0=0, then you have no idea what the previous line is because it could be anything. These steps are irreversible.

What you call operations on equations are logical inferences. If you "operate" on $$A=B$$ to get $$A'=B'$$ then what you have actually done is assert that $$A=B\implies A'=B'.$$ Assertions must be justifiable by the basic laws of arithmetic and the laws of logic. Instead of "operating", write in the "$$\implies$$"or "$$\iff$$ and ask "How do I know this inference is valid?

It might be that $$A'=B'$$ does $$not$$ imply $$A=B.$$ Reversibility means you have $$A=B\iff A'=B'.$$

E.g. $$x=2\implies x^2=2^2\iff x\in \{-2,2\}$$ is logical, but $$x^2=2^2$$ does not imply $$x=2.$$

E.g. $$\frac {x(2x+1)}{x}=0$$ implies $$x\ne 0$$ because if $$x=0$$ then the number $$0$$ cannot be equal the non-existent $$\frac {0(2 \cdot 0+1)}{0}.$$ Therefore $$\frac {x(2x+1)}{x}=0\implies [\,\frac {x(2x+1)}{x}=0\land x\ne 0 \,].$$ And you know from basic arithmetic that if $$x\ne 0$$ then $$\frac {xy}{x}=y.$$

For the other direction we have $$2x+1=0\implies x\ne 0$$ (Why?) so $$2x+1=0\implies [2x+1=0\land x\ne 0]\implies 0=\frac {x\cdot 0}{x}=\frac {x(2x+1)}{x}.$$

Could I get an explanation of what exactly reversible and irreversible operations mean?

Here and here are longer answers, but the short of it is that a valid operation on statement $$A(x)$$ to derive statement $$B(x)$$—here, valid just means that $$A(x)$$ universally implies $$B(x)$$—is said to be reversible (creates no extraneous solution) if $$A(x){\iff}B(x),$$ and irreversible (creates some extraneous solution) if $$A(x){\kern.8em\not\kern-.8em\iff} B(x),$$ that is, if there exists some $$x$$ for which $$B(x)$$ is true but $$A(x)$$ is false.

why is multiplying both sides by $$x$$ (where $$x$$ is the variable in the equation) irreversible?

Because $$f(x)=g(x)\kern.8em\not\kern-.8em\impliedby xf(x)=xg(x)$$ (put $$x=0, f(x)=x, g(x)=2x$$).

Is it also irreversible if I start with $$2x^2 = x,$$ and divide by $$x$$ to get $$2x = 1,$$ in which case $$0$$ is no longer a solution?

If the domain includes $$0,$$ then this operation is not even valid (so, discards some solution), since $$2x^2 = x\kern.6em\not\kern-.6em\implies 2x = 1$$ (put $$x=0$$), in which case the question of reversibility is moot.

If the domain excludes $$0,$$ this operation is actually reversible, since $$2x^2 = x\iff 2x = 1.$$

Why is it irreversible if I have $$\displaystyle \frac{x(2x + 1)}{x} = 0$$ and cancel out $$x$$ to get $$2x + 1 = 0$$?

If the implicit domain restriction $$x\ne0$$ is made explicit, then this operation is in fact reversible: $$\frac{x(2x + 1)}{x} = 0 \iff 2x + 1 = 0 \;\land\; x\ne0$$ that is, $$x\ne0\implies\left(\frac{x(2x + 1)}{x} = 0 \iff 2x + 1 = 0\right).$$

Finally, why is it irreversible if I have $$2x + 1 = 0$$ and multiply both sides to get $$\displaystyle \frac{x(2x + 1)}{x} = 0$$?

In practice, it is evident that $$x\ne0,$$ so $$\dfrac xx=1,$$ so this operation is reversible: $$2x + 1 = 0\iff\frac{x(2x + 1)}{x} = 0.$$