Show that $(a,b)+(c,d) = (a+c,b+d)$ Let $a&lt;b$ and $c&lt;d$ be real numbers. Show that $(a,b)+(c,d) = (a+c,b+d)$.


I don't understand the question. Since $(a,b)$ and $(c,d)$ are intervals, what does it mean to add them?
Hint: The set inclusion $(a,b) + (c,d)\subset (a+c, b+d)$ is easy to prove. The reverse inclusion is trickier. For the reverse inclusion, first prove the simpler case $a=c=0$, then apply this result to the general case, using the identity
$$(a,b)+(c,d)=\{a\}+(0,b-a)+\{c\} + (0,d-c).$$
Let function $f : [a,b] \times [c,d] \to \mathbb R$ be defined by

$$f (x,y) := x + y$$

We can find the minimum and maximum of $f$ by solving two linear programs, namely,

$$\begin{array}{ll} \text{minimize} &amp; x+y\\ \text{subject to} &amp; a \leq x \leq b\\ &amp; c \leq y \leq d\end{array}$$

and

$$\begin{array}{ll} \text{maximize} &amp; x+y\\ \text{subject to} &amp; a \leq x \leq b\\ &amp; c \leq y \leq d\end{array}$$

Since $\nabla f (x,y) = 1_2$, we conclude that:


the minimum of $f$ is $a+c$, which is attained at $(x,y) = (a,c)$.
the maximum of $f$ is $b+d$, which is attained at $(x,y) = (b,d)$.


Thus, interval addition is defined as follows

$$[a,b] + [c,d] := [a+c,b+d]$$