Theorem stating that raising to $\rm n^{th}$ power both sides of an equation *is ok*.

Question

In a previous question about “squaring both sides of an equality” I made the following observation in my answer.

1. Algebra’s fundamental theorem states: Any $\rm n^{th}$ order polynomial has $\rm n$ roots.

2. Then, if one increase the order of a polynomial by $\rm m$ (let’s say $\rm 1^{st}$ to $\rm 2^{nd}$) the number of roots will increase from $\rm n$ to $\rm n+m$, Therefore raising the power makes things different.

3. But raising to the same power both sides of an equation makes the solution set of the $\rm (n+m)^{th}$ order polynomial to contain the solutions set of the $\rm n^{th}$ polynomial.

I don’t have any problem assuming the first 2 paragraphs. However, when I made the last one “But rising to the same power both sides…” unwillingly I made a leap of faith, since it seemed so obvious that it didn’t deserve more to say. But now, if any one asks me to quote that, I wouldn’t know any theorem stating that. It is really simple to show it when we have numbers, but not when we have functions (or equations); more importantly, I showed in my 2nd paragraph that it “makes things different…” then, why the resulting solution set must contain the solution set of the original problem.

I can actually prove it, therefore, posting a proof or not is not important. I just want to know if there is already any theorem proving that the resulting solutions set of raising both sides of an equation must contain the solutions set of the original problem, so I can quote it; or it is so trivial that it’s not necessary to quote that at all (though my proof don’t seems to be trivial to me).

PS1: Please follow the paragraph as numbered in this question. Not the order of the paragraphs as in the linked answer.

PS2: Even though I concentrate in polynomials, it should be noticed that it is polynomial at its higher level of abstraction. It could be any equation in the form $f(x)=g(x)$, since at this level of abstraction, raising the power in both sides, becomes the polynominal $f^n-g^n=0$.

Answer 1

If you have a polynomial with $n$ solutions $a_1, a_2, \ldots , a_n$ then you have a factorisation $$ (x-a_1)(x-a_2)\ldots(x-a_n) = 0 $$

Raising each side to the same power doesn't give you any more roots, just increases the multiplicity of each root. This works when you have $0$ on the right hand side because that is the factorisation that gives you the roots.

I think the question you linked is slightly different because it isn't a polynomial (unless you count the infinite Taylor series). In any case the first answer explained it nicely. https://math.stackexchange.com/q/2024377

If you have an equation $f(x) = g(x)$ then $$ f(x)^2 = g(x)^2 \implies f(x)^2 - g(x)^2 = 0\implies (f(x) - g(x))(f(x)+g(x))=0$$

so that you have the solution set of $f(x)=g(x)$ (as intended) but also you've obtained solutions for $f(x)=-g(x)$. As you said though, the original solution set is clearly contained in the new equation

edit: if you use other powers then you still have the $f(x)-g(x)$ term in the complete factorisation, same conclusion. I think it's just a natural consequence of your theorem (and would therefore explain it as such), not aware of any other theorem stating that as a result.

Answer 2

When you write $$A=B$$ you are saying that that is on the left hand side of the equal sign is exactly the same thing as what is on the right hand side.

How could it be possible the case, then, that if you do something to $A$ and do the same thing to $B$ you do not get the same result?

Equality means that the two things are equal. It is as simple as that.