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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.

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    $\begingroup$ This question overlaps with yours : math.stackexchange.com/questions/1515492/… $\endgroup$
    – user186104
    Nov 16, 2015 at 3:09
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    $\begingroup$ 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 $f\neq g$. $\endgroup$
    – JMoravitz
    Nov 16, 2015 at 3:14
  • $\begingroup$ 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? $\endgroup$ Nov 16, 2015 at 4:27

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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.

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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 $$2x+1=0\implies [2x+1=0\land x\ne 0]\implies 0=\frac {x\cdot 0}{x}=\frac {x(2x+1)}{x}.$$

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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: $$2x + 1 = 0\iff\frac{x(2x + 1)}{x} = 0.$$

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