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This is probably a little basic, but say we want to 'prove' that 'If $x + 1 = x(1 + a)$ then $ax = 1$.

Now, back in high school I'd have just gone for the line-by-line method, i.e.

\begin{align*} x + 1 &= x(1 + a)\\ x + 1 &= x + ax\\ 1 &= ax \end{align*}

But now that I'm starting to write simple proofs it seems like the examples I'm given make more use of phrasing like '[...] $x + 1 = x + ax$ which implies that $1 = ax$ [...]' instead of the above. I'm not sure if it's just the style of this professor to write it that way, but are these two things equivalent? Is the line-by-line layout above just the same as saying $$x + 1 = x(1 + a) \iff x + 1 = x + ax \iff 1 = ax?$$ or perhaps is it the softer $$x + 1 = x(1 + a) \implies x + 1 = x + ax \implies 1 = ax?$$

Basically, what do the line breaks in the line-by-line method say about each statement that follows another? I wonder if it's just generally ambiguous and so to be avoided in favor of the specific implication terminology?

  • 3
    They say nothing and such writings are not grammatically correct. Use the proper symbols, or write the relationship between the equations in words. – David Mitra Jan 21 '14 at 21:31
  • The author should be clear about it. Usually, one expects $\iff$s. – Pedro Jan 21 '14 at 21:36
  • I don't have enough reputation to vote up yet, but thank-you both. It was my inkling that they were ambiguous/meaningless and that I should use the symbols/proper statements. – user122881 Jan 22 '14 at 07:33
  • When I used this multiline notation on one of exams with clear intention of them being connected by $\iff$ I scored 0 points. Examiner said there is $\implies$ by default but he was trying to find "error" in everyone's test and I will not tell you what he was called. – Trismegistos Jan 22 '14 at 11:48
  • @Trismegistos, sorry to hear that mate. People who insist on silly conventions disgust me. – goblin GONE Sep 16 '14 at 10:35

2 Answers2

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Please, always add $\Leftrightarrow$ or $\Rightarrow$ (or even $\Leftarrow$) between such lines. Better still, add some narrative text. Nothing is harder to guess for the reader than whether you are making equivalence transforms or are deriving necessary conditions or are heuristically looking for sufficient conditions. Remember that you should test solutions you found to see if they fulfil the original equation? That is precisely because non-equivalences occur with usual transforms and may introduce spurious solutions (e.g. if you multiply both sides of an equation with $a-b$, that is not an equivalence transfrom if it may happen that $a=b$; the alternative would be to introduce cumbersome case distictions in all subsequent lines). And don't just consider someone giving marks for the solution your reader - you yourself should also be sure what you meant when you wrote that! Mentioning a reason why two lines are equivalent (or whatever) would also be helpful.

In your specific example, you want to prove "If $A$ then $Z$", so there is no need to have equivalences in each step (and hence no need to care if they are); $A\Rightarrow B\Rightarrow\ldots\Rightarrow Z$ suffices. In such a one-directional proof you may happily do things like "multiply both sides with $0$" or the like.

  • In your specific example, you wnat to prove "If A then Z", so there is no need to have equivalences in each step (and hence no need to care if they are); $A \implies B \implies \ldots \implies Z suffices. ### Of course, that makes sense. I was getting caught up in making the statements completely accurate when it doesn't really matter in some cases. Guess I should focus more on what I'm actually trying to prove sometimes. Thanks.

    – user122881 Jan 22 '14 at 07:35
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Usually, without precision (which is bad), I would expect $\implies$, and not $\iff$.

It's clear when solving an equation: let's solve

$$\sqrt{1-x}=x$$ $$1-x=x^2$$ $$x=\frac{1\pm\sqrt{5}}{2}$$

The last step is simply solving a trinomial, I don't show the details as they are not important here. What is important, is that you have some trouble with the root $$x=\frac{1-\sqrt{5}}{2}<0$$

Because then you have the square root of a negative number in your original equation.

Hence, usually, when you solve an equation, using implications, you must absolutely check your solutions are actually solutions of your equation: you know you have them all, but there may be wrong solutions too, like here.


Sometimes, it's even more subtle, as when solving a cubic equation.

Once you have your equation in the form $z^3+pz+q=0$, one method of solution consists in writing $z=u+v$, then

$$u^3+b^3+(3uv+p)(u+v)+q=0$$

And, you can solve this by writing

$$u^3+v^3=-q$$ $$uv=-p/3$$

Then a trinomial appears, and it's easy to conclude.

However, you don't know in advance that this way to proceed will give you any solution at all. It may happen that you fail to solve this system of two equations, even though the original equation has $3$ roots. And it may happen because you have added a condition, by separating an equality $A+B=0$ into $A=0$ and $B=0$. So the logical symbol here would be really some kind of $\Longleftarrow$.

  • Before you wrote this equation you have to put what is x. If it is real number than 1 - x >= 0 which means that x >= 1 and then ${1 - \sqrt 5 }\over 2$ can not be solution because it is smaller than 1. – Trismegistos Sep 16 '14 at 14:02
  • @Trismegistos Yes. The point is, there may be too many solutions at the end, if we use $\implies$ somewhere in the resolution. – Jean-Claude Arbaut Sep 16 '14 at 14:05