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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?

Michael Munta
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    This is simply a consequence of adding linear inequalities. – Deepak Mar 13 '21 at 10:30
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    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)]. – Doug Spoonwood Mar 13 '21 at 10:31
  • @Doug, that shows $[a,b]+[c,d]$ is contained in $[a+c,b+d]$, but you need something more to show they're equal. – Gerry Myerson Mar 13 '21 at 10:52
  • @GerryMyerson You're right. Oops. – Doug Spoonwood Mar 13 '21 at 11:03
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    @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 $<$.) – Paul Sinclair Mar 13 '21 at 17:58
  • @Paul I was also very confused by Doug's comment. – Michael Munta Mar 13 '21 at 18:35
  • They were just expanding on Deepak's comment. When $A, B$ are sets of numbers, $A+B$ most commonly means the set $A + B = {a + b \mid a \in A, b \in B}$. If $A, B$ are intervals, the fact that if $u \le v, x \le y$, then $u + x \le x + y$, a basic result about inequalities and addition, implies the double-sided inequality I gave, which gives $[a,b] + [c,d] \subseteq [a+c, b+d]$. On the other hand, suppose $z \in [a+c, b+d]$. Choose some $x \in [a,b]$ and consider $y = z - x$. – Paul Sinclair Mar 13 '21 at 18:50
  • @PaulSinclair you're right. That's another error in there. – Doug Spoonwood Mar 13 '21 at 19:37
  • @Paul, if you choose the wrong $x$ in $[a,b]$ then $z-x$ won't be in $[c,d]$. – Gerry Myerson Mar 13 '21 at 22:22
  • "The most common use is in software, to keep track of rounding errors in calculations and of uncertainties in the knowledge of the exact values of physical and technical parameters. The latter often arise from measurement errors and tolerances for components or due to limits on computational accuracy. Interval arithmetic also helps find guaranteed solutions to equations (such as differential equations) and optimization problems. " https://en.wikipedia.org/wiki/Interval_arithmetic – Doug Spoonwood Mar 14 '21 at 16:51
  • Have your questions been answered now, Michael? or is there still something you'd like to have clarified? – Gerry Myerson Mar 15 '21 at 11:59
  • Earth to Michael: come in, please. – Gerry Myerson Mar 16 '21 at 22:39
  • Sorry I don't have time till weekend to study this. On the first glance I think everything is covered for now. – Michael Munta Mar 17 '21 at 10:14
  • OK. By the way, if you want to be sure I see a comment addressed to me, you have to put @Gerry in it, so I get notified. – Gerry Myerson Mar 18 '21 at 00:10

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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.

Gerry Myerson
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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 $[a+c, b+d]$.

Vadrif Draco
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