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?
 A: 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
$$(a,b)+(c,d)=\{a\}+(0,b-a)+\{c\} + (0,d-c).$$
A: In terms of sets and set notation:
$(a,b)$ = all the points of R that are between a and b exclusively =$\{x\in \mathbb R| a < x < y\}$
If $A$ and $B$ are sets, than we say $A + B =\{x+y|x \in A; y \in B \}$.
So the statement $(a,b)+(c,d) = (a+b,c+d)$ means that $(a,b) + (c,d)=\{x+y|a <x <b;c <y<d\}$ is the same as $(a+b, c+d) = \{z|a+b <z <c+d\}$
Proof:
1) if $z=x+y \in (a,b) + (c,d)$, that is, $a <x <b;c <y <d$ then $a+c <x+y <b+d$ so $(a,b)+(c,d) \subset (a+c,b+d) $.  That was easy.
2) if $z \not \in (a+b,c+d)$
a) if $z \le a + c$.  Let $x$ be any number such that $a < x < b$.  Then if  $z =x+y$ for some $y$ it follows that $y = z-x < a+c-a =c$ so $z \not \in (a,b) +(c,d)$.
b) Likewise if $z \ge b + d$.  Let $y$ be any number such that $c < y < d$.  Then if  $z =x+y$ for some $x$ it follows that $x = z-y > b+d-d =b$ so $z \not \in (a,b) +(c,d)$.
So $z \not \in (a+b,c+d)$ implies $z \not \in  (a,b)+(c,d)$.
So $(a,b)+(c,d) \subset (a+b,c+d)$
1 and 2 together mean $(a+b,c+d)=(a,b)+(c+d)$.
A: 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
$$[a,b] + [c,d] := [a+c,b+d]$$
