I found the following on Wikipedia, on the page for Inequalities:

If $a<b$ and $c<d$ then $a+c < b+d$.

It references Intermediate Algebra, but I don't see this specific property there.

Is there a name for this particular property? Is there a proof for this property from the other properties?


This holds in any ordered field (or more generally, partially ordered group); the only property we need to take advantage of is translation invariance and transitivity. That is, the properties that $$a<b\Leftrightarrow a+c<b+c$$ $$a<b \text{ and } b<c\Rightarrow a<c$$

Starting with $$a<b$$ we can, using translation invariance, add a constant to both sides: $$a+c<b+c$$ We can, using translation invariance, we can also establish, from $c<d$ translated by $b$ get that $$b+c<b+d$$ and using transitivity gives $$a+c<b+d$$

  • $\begingroup$ Very elegant, thank you! $\endgroup$ – Gustav Bertram Nov 29 '14 at 16:36
  • $\begingroup$ Why does it have to be a field? Wouldn't you only need an ordered group since there is only one operation? $\endgroup$ – D_R Nov 29 '14 at 16:44
  • $\begingroup$ @D_R That's a good point and I edited to include it; it even holds in any partially ordered group, I suppose - the only difference in the proof of that would be that left-addition and right-addition might not be the same, so you need an additional property (i.e. $a<b\Leftrightarrow c+a<c+b$). $\endgroup$ – Milo Brandt Nov 29 '14 at 16:51
  • $\begingroup$ Partial ordering has only less than or equal to comparisons? $\endgroup$ – D_R Nov 29 '14 at 16:55
  • $\begingroup$ @D_R Partial ordering has that $a\leq b$ and $b\leq a$ could both be false - that is, $a$ and $b$ could be incomparable (whereas total ordering asserts that at least one of those holds). The difference between $\leq$ and $<$ is purely one of notation, since you can easily define one given the other. $\endgroup$ – Milo Brandt Nov 29 '14 at 16:58

I don't know about the name.

But proof: yes. If $p$ and $q$ are positive numbers, then $p + q$ is a positive number; this follows from a commonly used postulate of the real number system. We can take $a < b$ to mean $b - a$ is positive.

Now let $p = b - a$ and $q = d - c$. Both are positive hence $p + q = (b + d) - (a + c)$ is as well. Therefore

$$a < b \ \text{ and } \ c < d \ \ \Longrightarrow \ \ a + c < b + d$$


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