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?

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 $(a+b,c+d)=(a,b)+(c+d)$.
