# Can the results of an problem invalidate the process and therefore themselves?

For example, suppose that to solve an equation, one must divide both sides by x, and it is later discovered that x = 0 (and zero was a permissible value in the original equation). What would this mean for the equation and/or results?

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Let us take a simple example, $x(x^2+2)=x$. This has the obvious solution $x=0$. Now look for non-zero solutions. In the search for such a solution, division by $x$ is permissible. So such a solution must be a solution of $x^2+2=1$, which in the reals is impossible. There is no issue: we have found all the solutions. – André Nicolas May 8 '13 at 14:39
I'd think there would be other, similar cases. – Lee Sleek May 8 '13 at 14:42
Sure, there are plenty of such cases. The logic is always the same. When we divide by $x$, we are confining attention to the non-zero solutions. – André Nicolas May 8 '13 at 14:50

I think it's worth me giving a similar answer to the above (which is very good, and which I upvoted!) just to restate it with a different, more abstract emphasis...

In any attempt to make a logical deduction ("the set of solutions is exactly 4, -7.3 and 0"), you are forbidden from making any step which just maybe could be invalid. Therefore, your situation would not arise because the division is forbidden unless it is known to be legal.

Of course, when you're solving the problem you don't know whether 0 is possible or not, and so you need some way to check it out either way. We do this by considering both possibilities separately, in cases. This means adding an assumption, say "$x\neq 0$", and then continuing. You find results or contradictions, and you can deduce what the answer is. How?

Logically, you're now on firm ground, with a pair of statements like $$x\neq 0\implies x=6,-3.7,0$$ and $$x=0\implies x=7,-6$$

Then you look at what you've got. In the first case, we discover $x=6,-3.7$ are valid solutions, but $x=0$ isn't, because this doesn't really arise in the first case. In the second case, we see that we cannot have $x=0$ and hence there are no other solutions.

Then the set of solutions is $\{6,-3.7\}$. Done!

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I wouldn't say anything that includes notation such as "$x=6,-3.7,0$" is logically on firm ground. – Henning Makholm Feb 14 '15 at 4:07
That's a well-established shorthand which just means $x \in \{6, -3.7, 0\}$, and is hardly a real problem here. I judged it likely to be closer to the language of equation solving the poster would be familiar with. – Sharkos Feb 16 '15 at 13:33

Let's go back to the point where you made the step of dividing by $x$. Split your solution into two cases at this point: Case 1, $x\neq 0$. Then it is valid to divide by $x$. However, as you said, once you continue to do algebra you find that $x=0$, which contradicts your assumption. Thus, the case $x\neq 0$ is impossible. Case 2, $x=0$. Plug in $x=0$ and verify that it works. If it also gave a contradiction, then you would conclude there is no solution.

I remember this happening when I took multivariable calculus, doing problems on Lagrange multipliers. I think splitting into cases makes things a bit more clear.

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