Question

Let N = (x) x is multiple of 2) and P = (x) x is a multiple of 6). Describe the intersection of N and P.

Answer 1

This sounds like sets...

The first few terms of N include 2, 4, 6, 8, 10, 12, etc...

and the first few terms of P include 6, 12, 18, 24, etc.

So N and P appear to intersect every time there is a multiple of 6...
in other words, 2 and 4 are in N but not in P, so they are are not in the intersection
6 is in both N and P
8 and 10 are in N but not P, so they are not in the intersection...
12 is in both N and P

So it appears the intersection of N and P is just the set of multiples of 6... in other words, the intersection is equal to P.

This happens whenever one set completely encompasses another smaller set.

For example, the intersection of the set of all letters with the set of just vowels is equal to the set of just vowels.. no consonants qualify, and all 5 vowels are in both the set of all letters AND the set of just vowels, so those 5 letters are in the intersection.

Answer 2

You lost me.. I'm confused.

Answer 3

Sorry :)

If N is multiples of 2, and P is multiples of 6, then when do N and P have the same numbers? N "intersects" with P whenever a number is both a multiple of 2 AND a multiple of 6. But since ALL multiples of 6 are also multiples of 2, then ALL numbers in P are also part of N, so the intersection is equal to the set P.

Essentially, you are saying that "all multiples of 6 are also multiples of 2".

Answer 4

& that's the answer? o: