# Can inequalities be added the way that equations can be added?

For example, if you have $$x > 6$$ and $$y > 4$$, then obviously $$x + y > 10$$. If you have $$x > 6$$ and $$y < 4$$, then the second inequality can be changed to $$-y > -4$$. Now adding $$x > 6$$ gives $$x-y > 2$$, which is obviously true. Now suppose you have $$x + y > 6$$ and $$x - y > 4$$. Adding the inequalities, the $$y$$ terms cancel and you end up with $$2x > 10$$ or $$x > 5$$. It is not intuitively obvious to me that this holds true, but I could not come up with a counter-example. I can see that you can't subtract inequalities, but is it always okay to add them?

• You do what you can logically justify. Mathematics is not about mysterious rules; everything you teacher told you is "ok" can be justified (hopefully...). $2x=x+y+x-y>6+x-y>6+4=10$, so, yes Sep 1, 2023 at 21:42
• If they go the same way, you can add them. If they go opposite ways and you add them, there will be a fight and the strongest direction will win. Sep 1, 2023 at 21:59
• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Sep 1, 2023 at 22:23
• – MJD
Sep 1, 2023 at 23:36

Consider any two inequalities $$a < b$$ and $$c < d$$, where each of $$a$$, $$b$$, $$c$$, and $$d$$ is real. Since $$a < b$$, it must be the case that (colloquially speaking) shifting each of them the same distance up or down the real line won’t disrupt things. In other words, $$a + c < b + c$$. (Note that this last bit wasn’t a proof, but merely an attempt to help you hone your intuition.)

But by the exact same kind of reasoning, it follows from $$c < d$$ that $$b + c < b + d$$. And now, all we need is the transitivity of less-than: Since

$$a + c < b + c < b + d,$$

we conclude that $$a + c < b + d$$.

• Looks like we were typing the same thing at the same time, down to the choice of variables! Great minds think alike. :-) Sep 1, 2023 at 22:03

I'll assume you know that you can add the same thing to both sides of an inequality, no? That is, if you know that $$x \lt y$$ then $$x +z \lt y+z$$ is valid.

Now, your question is: If I know that $$a \lt b$$ and $$c \lt d$$ does this imply that $$a+c \lt b+d \text{ ?}$$

It does, and here's how to prove it. Add $$c$$ to both sides of your first inequality to get $$a+c \lt b+c$$ and then add $$b$$ to both sides of your second inequality to get $$b+c \lt b+d$$ and then just use transitivity ($$r \lt s$$ and $$s \lt t$$ imply $$r \lt t$$) on those last two inequalities to get the inequality you wanted.

• Then it is perfectly okay to add inequalities. In particular, the original solution I gave in this question is perfectly reasonable, though maybe not intuitively obvious. math.stackexchange.com/questions/4729228/… I was also thinking that you could go from equations to inequalities by saying that x > y is equivalent to saying that x = y + p for some positive number p. Sep 1, 2023 at 23:43

Say you have two bags of cookies, $$a$$ and $$b$$. A friendly baker comes by and offers to trade with you: you will give the baker your bag $$a$$ and in return you will get a larger bag $$c$$ which contains more cookies. That is, $$a. You like cookies, so you agree.

Now the friendly baker offers the same deal for bag $$b$$: you will give up $$b$$ and in return you get a bigger bag $$d$$ that again contains more cookies. That is, $$c. You make that trade and the baker leaves.

Later you open the bags and count up your cookies. You have $$c+d$$ now, and you had $$a+b$$ before. Is it possible that you might not have more cookies than before you made the trades?

No, it's obviously impossible.

But that's what it would mean if $$a and $$b but not $$a+b too. So that is also impossble.