W, J, H? Abusive notation. Anyway, lol...
1) It tells you that the real numbers are composed of irrational and rationals. In other words, real numbers are composed of numbers that cannot be expressed as a fraction of integers (irrational ones) and those that can be expressed as a fraction of integers (rational ones). Thus:
2) Clearly the natural numbers are a subset of the integers. Any natural number in the set {1,2,3,...} is a part of the set of integers {...,-2,-1,0,1,2,3,...}. So their union is just the integers again. Thus:
3) The intersection of the irrationals and the integers is the set of all elements that BOTH irrational and integer. Can you find any such numbers? Can you find numbers that cannot be expressed as a ratio between two integers and yet are still integers?
4) The irrationals a proper subset of the reals, so if you are asking the intersection, it is the set of all numbers that both real and irrational. But by definition, all irrationals are real. So the set that you get for your answer is just the irrationals once again.
In general, here are a few tips for sets:
-If
is a subset of
(written as
), then their intersection is
. Example: the irrationals are a subset of the reals, so their intersection is just the irrationals.
-If
is a subset of
, then their union is
. Example: the irrationals are a subset of the reals, so their union is just the reals.
-If
and
are disjoint sets (meaning that they share no elements), then their intersection is the empty set. Example: the intersection between the negative integers and the positives is empty (no such number is both negative and positive at the same time)
-If
and
are disjoint sets, then their union is simply a new set that contains every element of both
and
. Example: the union of all the nonnegative integers and the negative integer is simply the set of integers because every integer is either negative or nonnegative.
Try to finish the rest yourself and show us your work. We can check it over for you.