Show that $(a,b)+(c,d) = (a+c,b+d)$ 
Let $a<b$ and $c<d$ 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?

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,
$$\begin{array}{ll} \text{minimize} & x+y\\ \text{subject to} & a \leq x \leq b\\ & c \leq y \leq d\end{array}$$
and
$$\begin{array}{ll} \text{maximize} & x+y\\ \text{subject to} & a \leq x \leq b\\ & c \leq y \leq d\end{array}$$
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
$$[a,b] + [c,d] := [a+c,b+d]$$
