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When solving equations like

$$\begin{align} 4x-4 &=\frac{(2x)^2}{x} \\ -4 &= \frac{4x^2}{x} -4x \\ -4 &= 4x -4x \\[0.2em] -4 &= 0\end{align}$$

using the equality-symbol feels like abuse of notation, since you'll end up with $-4=0$, which is not an equality. For instance I feel it would be better to write

$$\begin{align} 4x-4 &\:\Box\:\frac{(2x)^2}{x} \\ -4 &\:\Box\: \frac{4x^2}{x} -4x \\ -4 &\:\Box\: 4x -4x \\[0.4em] -4 &\:\Box\: 0 \\[0.3em] -4 &\neq 0\end{align}$$

So I was wondering if there's a symbol or any other notations being used when trying to solve such an equation where you don't know if there's an equality?

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    $\begingroup$ I'm sure a lot of people have wondered about this. I do to, and so did math.stackexchange.com/questions/885192/… and math.stackexchange.com/questions/109161/… $\endgroup$ – David K Sep 19 '14 at 14:00
  • $\begingroup$ Go ahead and use a box if you want. Maybe you are trying to decide among $<$, $>$, and $=$, for example. Make sure your steps are reversible, and go for it. $\endgroup$ – GEdgar Sep 19 '14 at 14:03
  • $\begingroup$ I found your questions when I searched around, but it bothered me that they were for inequalities with > and <, and not inequalities with = and ≠. Your ideas were nice though. $\endgroup$ – Frank Vel Sep 19 '14 at 14:07
  • $\begingroup$ I sometimes use a question mark, for instance when comparing two numbers: $\sqrt{2}-1\mathrel{?}1/2$; $\sqrt{2}\mathrel{?}3/2$; $2\sqrt{2}\mathrel{?}3$; $8\mathrel{?}9$ (going to a new line, usually, at the blackboard). So at the end we know that ? is $<$, because we used only transformation that don't change the direction of inequalities. Any symbol is good. $\endgroup$ – egreg Sep 19 '14 at 15:02
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I put a question mark above an equal sign, like this: $\stackrel{?}=$. (In MathJax, just type $\stackrel{?}=$.)

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    $\begingroup$ Or even $\overset{?}=$. $\endgroup$ – TheSimpliFire Feb 9 at 19:17
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It's perfectly fine to have equality signs.

When you solve equations, what you really do is say

Assume that the following is true

$$4x-4=\frac{(2x)^2}{x}$$

then

$$-4=0$$

is true.

Contradiction, the original assumption is false.

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  • $\begingroup$ Agreed, and I would add that it's useful to be clear in the surrounding text about whether any given equation is a statement of something known to be true or the starting point of a possible proof by contradiction. $\endgroup$ – David Z Sep 19 '14 at 20:15
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    $\begingroup$ This style only works if one knows from the outset that the equation is not going to have any solutions. But it seems to me that OP is in a context where one is trying to establish whether there are any solutions (and in the end it turns out: no). $\endgroup$ – Marc van Leeuwen Sep 21 '14 at 6:35
  • $\begingroup$ @MarcvanLeeuwen Actually if there is solutions, this does work, it's not a proof by contradiction, though. You just end up with: Assume that $x-2=5$ is true, then $x=7$ is true $\endgroup$ – Alice Ryhl Sep 21 '14 at 10:37
  • $\begingroup$ I feel these are two different questions: "Is there a solution to this equation?" and "What is the relation between these two expressions?" and I was hoping for an answer to the latter. In any case their answer might be the same "There is a solution ⇔ These expressions have an equality", but I was hoping for some new notations to avoid having to assume things to begin with. $\endgroup$ – Frank Vel Sep 21 '14 at 13:03
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I like to put the $\iff$ (if and only if) symbol at the beginning of every new line, like this: $$\begin{align} 4x-4 &=\frac{(2x)^2}{x} \\ \iff-4 &= \frac{4x^2}{x} -4x \\ \iff-4 &= 4x -4x \\[0.2em] \iff-4 &= 0\end{align}$$ So $4x-4=\frac{(2x)^{2}}{x}$ if and only if $-4=0$, which is true.

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    $\begingroup$ Maybe a different last word? ;-) $\endgroup$ – egreg Sep 19 '14 at 15:03
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    $\begingroup$ Sorry, that's a little ambiguous isn't it. I mean that the entire statement is true, obviously $-4 \ne 0$! $\endgroup$ – preferred_anon Sep 19 '14 at 15:05
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    $\begingroup$ in the last sentence, you mean $(2x)^2$ not $2x^2$ in the numerator of the fraction. $\endgroup$ – Joao Sep 20 '14 at 4:31
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    $\begingroup$ But your second $\iff$ is not justified, since in the direction $\Leftarrow$ you are dividing by $x$. (You may consider yourself saved here by the fact that for $x=0$ it reads "meaningless $\Leftarrow$ false", but you don't know that yet when you first write it.) The point is, the reasoning is really $\implies$ here, and arriving at an absurdity one concludes that the initial equation cannot be satisfied, which is what $\implies$ gives you. $\endgroup$ – Marc van Leeuwen Sep 21 '14 at 6:43
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    $\begingroup$ I should have put $(x \ne 0)$ in there somewhere, but in truth I was being lazy abd just copied the block of equations straight from the OP. Well spotted though! $\endgroup$ – preferred_anon Sep 21 '14 at 8:52

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