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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 $f^n-g^n=0$.

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    $\begingroup$ $\large {f(x) = x^n}\,$ is a single-valued function so $\,\large{a= b\,\Rightarrow\,f(a) = f(b)}\ $ $\endgroup$ – Bill Dubuque Nov 28 '16 at 18:07
  • $\begingroup$ Contrast that to: the root of $\,x = 1\,$ equals the root of $\,2x=2\,$ vs. the root of $\,x^2 = 1\,$ equals the root of $\,2x^2=2.\,$ The former is true because there "root of" is a single valued function, but the latter is not (one could choose either the root $\,x =1\,$ or $\,x = -1).\,$ We partly work around the latter problem by normalizing to a choice of positive roots (or principal branches). $\endgroup$ – Bill Dubuque Nov 28 '16 at 18:16
  • $\begingroup$ My comment wasn't actually as accurate as it could be. It would be more accurate to say that 'f(a)=f(b) --> a=b' if 'f' is bijective. $\endgroup$ – lordoftheshadows Nov 28 '16 at 18:18
  • $\begingroup$ @BillDubuque $x^2+x=5$; second order polynomial, 2 solutions. $(x^2+x)^2=5^2 \Rightarrow x^4+2x^3+x^2=25$; 4th order polynomial; 4 solutions; Is it so trivial?. Why exactly 2 of the last equation, out its 4 solutions are the 2 solutions of the former. Forget the examples; I just need a theorem if there is. $\endgroup$ – J. Manuel Nov 28 '16 at 18:29
  • $\begingroup$ The theorem is in the first comment. It's not clear precisely what is unclear about that to you. $\endgroup$ – Bill Dubuque Nov 28 '16 at 18:33
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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.

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  • $\begingroup$ The only exactly the same thing as I understand would be (not even so sure…) the trivially $\rm 5=5$ (or pick any number). $\rm (4+1)$ is not exactly the same thing as $\rm 5$. $\endgroup$ – J. Manuel Nov 28 '16 at 19:18
  • $\begingroup$ The expression 4+1 denotes exactly the same thing as 5. In fact, in many situations the latter is defined to be exactly the former. $\endgroup$ – Mariano Suárez-Álvarez Nov 28 '16 at 19:35
  • $\begingroup$ Arithmetic expressions denote numbers. There is exactly one number which is the successor of the successor of the successor of the successor of the successor of zero, and 5, 4+1, 2^2+1, 20-15, and many others are different ways of eriting that very same natural number. When you write 6-1=3+2 you are saying that the left hand side denotes a number, that the right hand side denotes a number and that the number denotes in each case is exactly the same one. $\endgroup$ – Mariano Suárez-Álvarez Nov 28 '16 at 19:39
  • $\begingroup$ I never wrote and I never meant that identities are trivialities: that would be a silly and, to boot, a false claim. Showing that two things are in fact the same things can be very non-trivial. $\endgroup$ – Mariano Suárez-Álvarez Nov 28 '16 at 23:02
  • $\begingroup$ You seem to be somewhat confused about what the meaning of equality is. Equality means exactly what I described: two things are equal when they are the same thing, and there is no more to it than that. $\endgroup$ – Mariano Suárez-Álvarez Nov 28 '16 at 23:02
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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.

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