Let's say there's some relationship I'm trying to figure out between two values (e.g. is A > B or is B > A), and I want to use math to prove the relationship. Is there a common convention for how to do something like this? Which inequality symbol would I use between the two values if I don't know which value is bigger than which?

Here's a very simplistic example:

Say I'm not sure which of these is greater:

$\text{A}=\frac{3}{x^2+1}$ $\text{B}=\frac{3}{x^2}$

And I want to start using math to figure this out. Initially I set up an equation such as:

$\frac{3}{x^2+1} ? \frac{3}{x^2}$

And then I solve it:

$3x^2 ? 3\left(x^2+1\right)$



At this point, it's obvious that the $?$ symbol should be $<$. So I can now go backwards and replace each $?$ with a $<$, ultimately getting the following:

$\frac{3}{x^2+1} < \frac{3}{x^2}$

I just arbitrarily chose the ? symbol for this example. Is there a standard symbol for doing something like this?

Additionally, if I were to multiply either side by a negative value, I would need to remember to flip the symbols at that step. Is there any easy way to "mark" such a step so that I don't forget to flip the sign?

This is especially problematic if you are multiplying by a term which may be negative. In this specific example, $x^2$ and $x^2+1$ are always positive, so it's not a problem. However, if one of these terms was $x^2-1$ it would be a lot trickier... You would need need some kind of note such as: If$x^2-1<0$flip the sign here.

  • $\begingroup$ Maybe either of $\gtreqless$ or $\lesseqgtr$? $\endgroup$ Feb 14 '12 at 3:44
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    $\begingroup$ Funny you should ask this: I was just working to determine an unknown inequality. In cases like this, I usually use $\oslash$, and if somewhere in the course of my computations I perform an operation that reverses the inequality, I draw the slash the other way. $\endgroup$
    – user2093
    Feb 14 '12 at 3:51
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    $\begingroup$ How about "Suppose $A\ge B$". It can lead you to contradiction or obvious truth, so you just write the conclusion. $\endgroup$ Feb 14 '12 at 13:57

The general way to do such a thing will be to start with $A-B$ and proceed to show it is positive or negative.

For instance, in the case you have given us,

$$\begin{align}A-B&= \dfrac{c}{x^2-1}-\dfrac{c}{x^2}\\&=c\cdot \left(\dfrac{x^2-x^2+1}{x^2(x^2-1)}\right)\\&=\dfrac{c}{x^2}\cdot \dfrac{1}{x^2-1}\end{align}$$

Now the resulting expression $A-B$ is positive when $c(x^2-1)\ge 0$ and negative otherwise.

(Note that $A$ in not defined when $x= \pm 1$, and this is exempted from the domain of the problem.)

During the second reading, I find that you have made a mistake in what you have shown us. It happens in the step where you cross multiply $x^2$ to one side and $x^2-1$ to the other.

While it is true that $x^2 \ge 0$, it is not always true that $x^2-1 \ge 0$. This in fact fails, when $|x| \le 1$.

Hope this is convincing that this method is more comfortable than inventing Exotic Symbols.

  • $\begingroup$ Yeah, I just saw that and corrected it. This way seems more complicated, but is less error prone when there is potential for multiplication by -1. $\endgroup$
    – Senseful
    Feb 14 '12 at 4:05
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    $\begingroup$ I said more complicated... $\endgroup$
    – Senseful
    Feb 14 '12 at 4:21
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    $\begingroup$ Ya I don't want to get into a debate, I removed the last part of that comment. Your method works, thanks for the answer. $\endgroup$
    – Senseful
    Feb 14 '12 at 4:27
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    $\begingroup$ I'm going to update the original question so that it's simpler to follow. I'm letting you know in case you want to update your answer. $\endgroup$
    – Senseful
    Feb 14 '12 at 4:32
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    $\begingroup$ @Kannappan: It doesn't seem polite to ask "Do you think this is complicated?" - perhaps I am just misunderstanding your tone, which electronic communication makes difficult to discern. $\endgroup$ Feb 14 '12 at 10:44

This is entirely concerned with your specific example, but is a good thing to remember when working with inequalities:

If $A>B>0$, then dividing a positive quantity $C$ by $A$ will produce something smaller than dividing $C$ by $B$.

So ${3\over x^2+1}<{3\over x^2}$.

Informally "dividing by something smaller gets you something bigger".


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