I would like to know the logic behind adding two intervals $$[a, b]$$ $$[c, d]$$ together and for the sake of simplicity let's say that the numbers in the interval are just integers and not real numbers. I know that the idea is to just add the lowest and highest of the two $$[a+c, b+d]$$ and this is the result of the addition.

My question is whether this is just defined to be like this mathematically or is there a real world example that can shed some light on why it is calculated so?

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    $\begingroup$ This is simply a consequence of adding linear inequalities. $\endgroup$
    – Deepak
    Mar 13, 2021 at 10:30
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    $\begingroup$ Suppose we know that a < x < b, and c < x < d. Then (a + c) < x < (b + d). Thus, ([a, b] + [c, d]) = [(a + c), (b + d)]. $\endgroup$ Mar 13, 2021 at 10:31
  • $\begingroup$ @Doug, that shows $[a,b]+[c,d]$ is contained in $[a+c,b+d]$, but you need something more to show they're equal. $\endgroup$ Mar 13, 2021 at 10:52
  • $\begingroup$ @GerryMyerson You're right. Oops. $\endgroup$ Mar 13, 2021 at 11:03
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    $\begingroup$ @DougSpoonwood - Did you mean "$a \le x \le b$ and $c \le y \le d$. Then $(a+c) \le x + y \le (b+d)$"? Without $x + y$, what you wrote has far more wrong than Gerry Myerson pointed out. (And of course $[\;,\;]$ mean $\le$, not $<$.) $\endgroup$ Mar 13, 2021 at 17:58

2 Answers 2


Here's a proof that if $r$ is in $[a+c,b+d]$ then $r=x+y$ for some $x$ in $[a,b]$ and some $y$ in $[c,d]$:

First, note that $a+c\le r\le a+d$ or $a+d\le r\le b+d$.

In the former case, there must be $y$ in $[c,d]$ such that $r=a+y$. Let $x=a$.

In the latter case, there must be $x$ in $[a,b]$ such that $r=x+d$. Let $y=d$.

Note: I'm not sure that this engages with OP's concerns. And I suspect that what I've written here has been written before on this site, probably multiple times. But it does seem to speak to some of the comments on the question.

  • $\begingroup$ Sure you don't have some typos here? For example $a+c \le r \le a + d$? How can this be in all cases? $\endgroup$ Mar 22, 2021 at 16:01
  • $\begingroup$ I didn't say it was true in all cases. I wrote $a+c\le r\le a+d$ OR $a+d\le r\le b+d$. $\endgroup$ Mar 23, 2021 at 9:01

Since none of the comments (nor the single answer) provide a "real world example" as the question wanted, I wanted to provide a simple one.

Let's say you are going to buy some apples from two stores. Store X will give you a minimum of $a$ apples, and a maximum of $b$ apples, thus its range is $[a, b]$. Meanwhile, store Y will give you a minimum of $c$ apples and a maximum of $d$ apples, thus its range is $[c, d]$.

In the worst case scenario, both stores will give you their minimum amount of apples. That would be $a$ apples from store X and $c$ apples from store Y. And in the best case scenario, they give you $b$ and $d$ apples, respectively. Thus, the new range that represents the sum of the number of apples that you can get from the two stores would be $[a+c, b+d]$.


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