22

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?

Frank Vel
  • 5,507
  • 6
    I'm sure a lot of people have wondered about this. I do to, and so did http://math.stackexchange.com/questions/885192/solve-for-undetermined-inequality-symbol and http://math.stackexchange.com/questions/109161/symbol-notation-strategy-for-figuring-out-an-unknown-inequality – 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 Answers3

23

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.

Alice Ryhl
  • 8,001
  • 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
  • 3
    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 06:35
  • @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
21

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

JRN
  • 6,691
9

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.

preferred_anon
  • 17,765
  • 4
    Maybe a different last word? ;-) – egreg Sep 19 '14 at 15:03
  • 4
    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
  • 2
    in the last sentence, you mean $(2x)^2$ not $2x^2$ in the numerator of the fraction. – Joao Sep 20 '14 at 04:31
  • 2
    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 06:43
  • 2
    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 08:52