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It has been well-documented why certain functions applied to equations do not add extraneous solutions. In this post, the accepted response shows that if the applied function is injective, then it is guaranteed that the solution set is preserved.

However, the response does not explore my issue (at least not explicitly). Namely, if the applied function is not injective, is the solution set always expanded? If it is preserved sometimes, when and why?

I can answer my first question by providing the example below:

First, consider the following simple equation:

$$ x = 4 $$

$$ x^2 = 16 $$

$$ x = \pm4 $$

As we can see, only one of the resulting solutions satisfies the original equation; we have gained an extraneous solution.

Now, consider another equation:

$$ \sqrt5 = \sqrt{1 + x^2} $$

$$ 5 = 1 + x^2 $$

$$ 4 = x^2 $$

$$ x = \pm 2 $$

The resulting solutions satisfy the original equation, meaning that squaring both sides (applying a non-injective function) did not add another solution. Then, it is not the case that using a non-injective function will necessarily expand the solution set.

However, one question remains: When will applying a non-injective function expand the solution set, and when will it not? I may be misinterpreting the linked post, but this discussion is not highlighted.

LateGameLank
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    Because in the second case you are (or should be) using the true theorem $[B=\sqrt A\Leftrightarrow (B\ge 0\land A=B^2)]$ whereas in the first case you aren't using the true theorem $[A=B\Leftrightarrow ((A\le 0\land B\le 0\land A^2=B^2)\lor (A\ge 0\land B\ge 0\land A^2=B^2))]$ – Sassatelli Giulio Dec 28 '24 at 20:18
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    @SassatelliGiulio: your point is that the definition of $\sqrt{t}$ makes an arbitrary but convenient decision about the which of the two possible solutions to $z^2 = t$ to choise. You should write your comment up as an answer. – Rob Arthan Dec 28 '24 at 21:07
  • @RobArthan Oh, fat chance I'm doing that. – Sassatelli Giulio Dec 28 '24 at 22:37
  • @SassatelliGiulio: the MSE police (of which I am definitely not a member) are deeply against comments that answer questions. Hence my tip. – Rob Arthan Dec 29 '24 at 00:56
  • @RobArthan The way I see it, I'm not preventing anybody from answering, if anything I'm making their job easier. Of course nobody will, because by now even the most dim-witted individual with a bachelor's degree can recognize unwelcome content. – Sassatelli Giulio Dec 29 '24 at 01:24
  • Well, now we have an awkward situation. I thought this question was already answered elsewhere, but then you pointed out in an edit that while your question is quite similar to the other question, the answers there really don't cover your particular case. And yet the question was closed after your edit. I'm voting to reopen. – David K Dec 30 '24 at 18:56
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    The accepted answer of the linked question starts out by saying it's safe to apply a function that is injective over all real numbers. But later it give some other cases, such as functions that are defined only on some real numbers, and the sine function when you know the number on each side of the original equation is between $-\pi/2$ and $\pi/2.$ – David K Jan 01 '25 at 17:40
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    I've added an answer to the linked question that explains things differently than the accepted solution; I focus on how to make an equivalent statement (same solution set) for any equation. I used your equation with $\sqrt5$ as one of my examples. – David K Jan 04 '25 at 16:15

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The way algebra is taught most of the time with equations written below other equations means that the top equation implies the bottom one, but not always in reverse. For example, the first set of equations can be written more accurately as:

$x=4 \implies x^2=16 \iff x=\pm \sqrt{16} \iff x=\pm 4 $

Note how you cannot go back from $x^2=16$ to $x=4$.
That's because you basically lose the information $x>0$

The theorem you use to "square" both sides of the equation is this:
$a^n=b^n \iff a=b$ if $n$ is odd
$a^n=b^n \iff${ $a=b $ OR $ a=-b $} if $n$ is even.

Meaning that regardless of $a=b$ or $a=-b$ ($a$ and $b$ having the same or opposite sign) you can say that $a^{2k}=b^{2k}$, but to go back, you must know which one holds true.

So I guess one way to solve this is to make a note on the first line that $x>0$ and use that information at the end. Note how the equation you get at the end isn't wrong either, it says that x is either 4 or -4, which is a true statement, but less specific.

I'm not sure if there's a formal and general way of dealing with this sort of "accidental generalizations" but if there is I'd be glad to have it explained to me.

S. McDonald
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  • This response is similar to the discussion here. This is not my concern. Rather, I am interested in understanding when a non-injective function (such as squaring) does preserve the solution set. – LateGameLank Dec 30 '24 at 17:18