Can you help me subtract intervals? I was reading my abstract math textbook and they subtracted
$[3, 6] - [4, 8) = [3, 4)$. I was wondering if someone could write out how they got to $[3, 4$). I looked at wikipedia and it said I should go 
$[a, b] - [c, d] = [a-d, b-c]$. When I did this, I got $[-5, 2)$. I would be thankful for an explanation-- the book doesn't explain so it's probably really obvious-- but I don't know.
 A: You are looking at two different definitions of $A-B$:
Set difference: $A - B = \{ x\in A \, \mid \, x \notin B \}$ which in this case gives $$[3,6]−[4,8) = [3,4)$$
Interval arithmetic: $A - B = \{ x-y \in \mathbb{R} \, \mid \, x\in A,  \,y \in B \}$ which in this case gives $$[3,6]−[4,8) = (-5,2]$$
A: $$[3,6]\setminus[4,8)=([3,4)\cup[4,6])\setminus\left([4,6]\cup(6,8)\right)$$
$$=[3,4)\setminus(6,8)$$
$$=[3,4)$$
A: Let $I_1$ and $I_2$ be intervals in the real line. Then $I_1 - I_2 = I_1 - (I_1\cap I_2)$.
If $I_1 = [a,b]$ and $I_2 = [c,d]$ (note that $a < b$ and $c < d$) then there are some cases to consider:
If $d < a$ or $b < c$ then $I_1 - I_2 = I_1$ since $I_1\cap I_2 = \emptyset$.
If $a < c < b < d$ then $I_1 - I_2 = [a,c)$ since $I_1\cap I_2 = [c,b]$.
If $c < a < d < b$ then $I_1 - I_2 = (d,b]$ since $I_1\cap I_2 = [a,d]$.
If $a < c < d < b$ then $I_1 - I_2 = [a,c)\cup (d,b]$ since $I_1\cap I_2 = I_2$.
If $c < a < b < d$ then $I_1 - I_2 = \emptyset$ since $I_1\cap I_2 = I_1$.
A: This is the same like sets. Not subtracting intervals.
So we want to find the values that exists in $[3,4]$ but don't exists in $[4,8]$. It should be very clear how the author got $[3,4)$ 
notice that the answer is open at $4$ so $4$ itself is not included since $ 4\in [4,8]$
