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If solving for an equation one way gives me an undefined answer, does that mean solving it in other ways too will give me an undefined answer? If no, then how consistent is math? (I'm not sure if I'm using the term "consistent" correctly, since I'm aware that it's also used in Gödel's incompleteness theorems which I have very little knowledge about) I also apologize if the way I word my question is confusing.

For example, solving for $a$ in $ab=0$ by the "traditional" way (the way it's taught in elementary and high school) will be like: $$ab=0$$ $$a=\frac 0 b$$

Assuming $b \neq 0$, $$a=\frac{0}{b}=0$$

Hence, $a=0$.

But if $a=b$, then the only number $b$ can be is $0$ or else by the zero product definition, the product of $a$ and $b$ can't be $0$. But this brings a problem because it'll show $b=0$ which contradicts what I assumed and will show that $a=\frac 0 b=\frac 0 0=0$. And $\frac 0 0$ is undefined. But if I try solving for it by taking the square roots of both sides: $$ab=0$$ $$aa=0$$ $$a^2=0$$ $$\sqrt{a^2}=|a|=\sqrt 0$$ $$a=±0=0$$

It gives me an answer (that's not undefined). This confuses me, because I always thought that solving an equation in one way will give me the same and consistent answer as solving the equation in any other way. Maybe I'm getting some concepts wrong from the start which needs to be corrected.

So if solving for an equation gives me an undefined answer, then does that mean I'll need to try to solve for it in another way to see if I can get a defined answer? If so, then how "far" or how many ways should I try until I get the correct answer?

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    $\begingroup$ What does “But if $a=b$” have to do with this? If $b \neq 0$, then $a=0$ is the only solution, while if $b=0$, then any $a$ satisfies the equation. That's all there is to it. $\endgroup$
    – Hans Lundmark
    Mar 15, 2022 at 15:05
  • $\begingroup$ @HansLundmark But I assumed that $b \neq 0$ for $a = \frac 0 b = 0$ so that I won't have $0/0$, doesn't that contradicts with what I assumed if I was given $ab=0$ and $a=b$? $\endgroup$
    – Mohammad muazzam ali
    Mar 15, 2022 at 15:09
  • $\begingroup$ @ryang Do you mean that the assumption $b \neq 0$ could be false at the end? If the assumption turns out to be false, then does that count as a contradiction? I mean, the reason I assumed $b \neq 0$ is so that I can avoid $\frac 0 0$ (which is undefined) and have $0$. $\endgroup$
    – Mohammad muazzam ali
    Mar 15, 2022 at 15:27
  • $\begingroup$ It seemed to me as if you were talking about only the equation $ab=0$, with $b$ given and $a$ sought, and then the additional assumption "if $a=b$" just came out of the blue. If instead you changed to talking about something different, namely the system of equations $a=b$ and $ab=0$ where $a$ and $b$ are both sought, then the only solution is $(a,b)=(0,0)$, since $ab=0$ is true if and only if $a=0$ or $b=0$ (and there's no need to divide by $b$ in order to draw that conclusion, it's a fundamental property of multiplication). $\endgroup$
    – Hans Lundmark
    Mar 15, 2022 at 17:18
  • $\begingroup$ @ryang but wouldn't that gives $a = \frac 0 b = \frac 0 0$? I thought division by zero is undefined and that any division or fraction properties won't work if that happens, and hence, I can't have $\frac 0 b = 0$. $\endgroup$
    – Mohammad muazzam ali
    Mar 15, 2022 at 23:21

1 Answer 1

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You're either forgetting that $b≠0$ is merely an assumption, or forgetting to consider the meaning of that word. When you also know that $a=b,$ then that assumption simply isn't viable. After all, an assumption is tentative and falsifiable, and just a premise.

While it is valid to introduce an assumption then derive conclusions from it, these conclusions are provisional and are firmed only when that assumption gets justified (then it is no longer merely an assumption). If it gets falsified, then we can conclude that its negation is true. If we can neither justify nor falsify that assumption, then we simply cannot derive any conclusion from our working, because our argument may be valid but unsound.

@ryang Do you mean that the assumption $b≠0$ could be false at the end? If the assumption turns out to be false, then does that count as a contradiction?

Indeed. The contradiction comes from the supposition $\color{red}{b≠0}$ clashing with the subsequent derivation $b=0;$ since the supposition is the only possible culprit, it must be actually false; in other words, it must be that $\color{green}{b=0}.$

@ryang but wouldn't that give $a=\dfrac0b=\dfrac00\;?$ I thought division by zero is undefined and that any division or fraction properties won't work if that happens, and hence, I can't have $\dfrac0b=0.$

Since we've determined that $\color{green}{b\text{ is indeed zero}},$ then naturally $ab=0\kern.6em\not\kern-.6em\implies a=\frac0b.$

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  • $\begingroup$ This has been an answer I've been searching for, but can you explain in a little more detail what you mean by the last sentence? (that $b$ is indeed zero and $ab=0\kern.6em\not\kern-.6em\implies a=\frac0b$. I don't really understand what does the crossed out arrow mean) $\endgroup$
    – Mohammad muazzam ali
    Mar 16, 2022 at 7:35
  • $\begingroup$ (i) You know that division by zero is undefined, and (ii) you have determined (from the indirect, contradiction proof above) that in the given exercise, $b$ is zero. Thus, you know that in the given exercise, dividing by $b$ is illegal. Thus, in the given exercise, $ab=0$ does not imply that $a=\frac0b.$ $\endgroup$
    – ryang
    Mar 16, 2022 at 7:45

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