# Meaning of an Addition of Inequalities

I'd like to apologize in advance if I use the wrong terms for some things; I'm a newbie and I haven't been studying math in English.

Say I have these inequalities: \begin{align*} x &\geq 5\\ x &\geq 7 \end{align*}

Their intersection would be the range in which both inequalities are true, so $$x \geq 7$$. Their union would be the range in which either inequality is true, so $$x \geq 5$$.

My question is, following this same sense: what would be the relationship between those two inequalities and the result of their addition?

That is, solving this as a "system of inequalities", by adding the two respective sides of the geq, would result in: \begin{align*} x+x &\geq 5+7 \\ x &\geq 6 \end{align*}

And, while it all seems incredibly trivial, I realized that I can't seem to explain to myself what this even means. That is, $$x \geq 6$$ isn't the union of the two original inequalities, nor is it their intersection; but what is it? When I "solve" a system of inequalities in this way, what exactly is it that I even do? What does this result set mean, in relation to the original inequalities that I "solved", and why does it qualify as a "solution" at all?

I do understand why if $$a \gt b$$ and $$c \gt d$$ then $$a + c \gt b + d$$; it makes perfect sense that the sum of two greater numbers would be greater than the sum of two smaller numbers. But other than that, I can't put into words the relationship of the first two and the result.

Some background information, to maybe help clarify what I ask: I had constructed the following four inequalities and wanted to algebrically show that they are true only if $$a$$ and $$b$$ are real numbers that are larger than $$1$$. \begin{align*} a b &\gt 0\\ a + b &\gt 0\\ ab &\gt a\\ ab &\gt b \end{align*}

Logically, I can definitely explain to myself why that is true (unless it isn't and I'm being silly, in which case, please do feel free to tell me so): the first two inequalities prove that $$a$$ & $$b$$ are positive, and the latter two prove that they're larger than 1. But when I tried to "solve" this as a system of inequalities -- i.e, add those inequalities to each other -- I realized that the final result is that $$ab \gt 0$$, and I couldn't exactly explain to myself what this "solution" even has to do with the inequalities it's composed of. I did, at that point, understand that an addition like that was not equivalent to an intersection, but that's about all I did understand.

• Welcome; please look at our MathJax guide for formatting. Commented Dec 5, 2019 at 20:58
• Thank you :) A quick google told me to use the Google charts API, which failed since I don't have image permissions or something along those lines. Will definitely look into your link!
– Shay
Commented Dec 5, 2019 at 21:03

First, the inequality you get by adding inequalities is implied by the inequalities you start with, but is not equivalent to them. That is, as you note, if $$a\gt b$$ and $$c\gt d$$, then $$a+c\gt b+d$$. But if $$a+c\gt b+d$$, you cannot then conclude that $$a\gt b$$ and $$c\gt d$$, or any other such combination. If it is the case that $$x\geq 5$$, and it is also the case that $$x\geq 7$$, then it is certainly the case that $$2x\geq 12$$, and hence that $$x\geq 6$$. This is not the best possible thing you can say about $$x$$, but it is nonetheless a true thing you can say about $$x$$ based on your assumptions.

That is, in general, adding inequalities will yield an inequality that is weaker than the conjunction of your original inequalities: any solution to the conjunction of the original inequalities will also be a solution to their sum, but not conversely.

Note that this is also true for equalities! If $$x=3$$ and $$y=5$$, then $$x+y=8$$; but from $$x+y=8$$ you cannot deduce that $$x=3$$ and $$y=5$$.

When you add inequalities (just as when you add equalities), you are simply obtaining a condition entailed by those inequalities. It's not as neat as taking the conjunction (intersection) or disjunction (union), because you do not have a bidirectional implication. Note that if $$S_1$$ is the solution to the first inequality, and $$S_2$$ is the solution to the second inequality, then

• $$x\in S_1\cup S_2$$ if and only if $$x\in S_1$$ or $$x\in S_2$$;
• $$x\in S_1\cap S_2$$ if and only if $$x\in S_1$$ and $$x\in S_2$$,

so you get information flowing in both directions. You don't get that with the sum (or product) of inequalities.

Second, to your system: you can't necessarily solve a system of inequalities by adding or subtracting, just like you can't always solve a system of equalities by adding and subtracting. It just may be that you can find a clever set of operations that will lead you to a conclusion that is narrow enough to identify $$a$$ and $$b$$ (just as with equalities), but then again it may not. However, that is fine: if from your assumptions you end up with the conclusion that you must have specific values of $$a$$ and $$b$$, then that means that if there is a solution, it must be the values of $$a$$ and $$b$$ that you found. But if not all steps you took are reversible, you must check that final determination in your original system. This, again, also occurs with equalities, as manipulation of equalities may lead to spurious solutions: from $$x=1$$, squaring you get $$x^2=1$$, which gives solutions $$x=1$$ and $$x=-1$$... but $$x=-1$$ does not solve the original equation. That's because the step of squaring is not "reversible" (it's an implication, not an "if and only if").

From $$ab\gt 0$$, you know that $$a$$ and $$b$$ have the same sign. If they are both negative, then $$ab\gt a$$ and $$ab\gt b$$ both hold, but then $$a+b\lt 0$$ does not. Thus, we conclude that $$a$$ and $$b$$ are both positive.

Then \begin{align*} ab\gt b &\iff ab-b\gt 0\\ &\iff b(a-1)\gt 0.\\ ab\gt a &\iff ab-a\gt 0\\ &\iff a(b-1)\gt 0. \end{align*} Since $$b\gt 0$$, then $$b(a-1)\gt 0$$ implies $$a\gt 1$$; and similalry, $$a(b-1)\gt 0$$ implies $$b\gt 1$$. So you end up with $$a\gt 1$$ and $$b\gt 1$$. Then you can verify that these all satisfy the original equations, so that's your solution.

• Just a small point: I did not claim that ab > 0 implies that a and b are positive, but that the combination of ab > 0 with a + b > 0 (the first two inequalities in that system) does :-) That should cover it, right? Otherwise, though, and more importantly: fantastic, expansive answer and I feel much better now; you've definitely cleared things up for me. Thanks a lot :)
– Shay
Commented Dec 5, 2019 at 21:44
• @ShyGuy: Quite, you did not say the first one did. I'll edit to remove that assertion. Commented Dec 5, 2019 at 21:46

The left side of the inequalities you have are both equivalent. The right side is not. That is, you are getting

2 left sides $$\geq$$ sum of the right sides

Which is equivalent to

left side $$\geq$$ average of the right side

This is just happenstance since the left sides of the inequalities are the same. If you think of the inequalities as sets of numbers, then the union of the 2 sets is all the elements they have in common, which is just $$[7, \infty)$$