# Symbol for unknown relation?

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?

• 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/… – David K Sep 19 '14 at 14:00
• 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. – GEdgar Sep 19 '14 at 14:03
• 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. – Frank Vel Sep 19 '14 at 14:07
• 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. – egreg Sep 19 '14 at 15:02

## 3 Answers

I put a question mark above an equal sign, like this: $\stackrel{?}=$. (In MathJax, just type $\stackrel{?}=$.)

• Or even $\overset{?}=$. – TheSimpliFire Feb 9 '19 at 19:17

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.

• 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. – David Z Sep 19 '14 at 20:15
• 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). – Marc van Leeuwen Sep 21 '14 at 6:35
• @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 – Alice Ryhl Sep 21 '14 at 10:37
• 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. – Frank Vel Sep 21 '14 at 13:03

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.

• Maybe a different last word? ;-) – egreg Sep 19 '14 at 15:03
• Sorry, that's a little ambiguous isn't it. I mean that the entire statement is true, obviously $-4 \ne 0$! – preferred_anon Sep 19 '14 at 15:05
• in the last sentence, you mean $(2x)^2$ not $2x^2$ in the numerator of the fraction. – Joao Sep 20 '14 at 4:31
• 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. – Marc van Leeuwen Sep 21 '14 at 6:43
• 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! – preferred_anon Sep 21 '14 at 8:52