Expert: Paul Klarreich - 4/22/2006

Question
Hallo Sir;
Thank you for your efforts. If B is a set,includes the elements
B={0,3,6,9,12,....etc.}, and
B1={x:x=9n-3,n natural number},
B2={x:x=9n-6,n natural number}, and
B3={x:x=9n-9,n natural number}, are subsets of B.
so that (B1nB2)nB3={ } &  (B1UB2)UB3=B.
How can I prove that (B1nB2)nB3={ } and (B1UB2)UB3=B ?
thanks you again, and I hope you answer my question.  

Answer
Hello, Ahmad,

Subject:  Provement
Question:  Hallo Sir;
Thank you for your efforts. If B is a set,includes the elements
B={0,3,6,9,12,....etc.}, and

B1={x:x=9n-3,n natural number},

B2={x:x=9n-6,n natural number}, and

B3={x:x=9n-9,n natural number}, are subsets of B.

so that (B1nB2)nB3={ } & (B1UB2)UB3=B.
How can I prove that (B1nB2)nB3={ } and (B1UB2)UB3=B ?

thanks you again, and I hope you answer my question.
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With sets, the notation is sometimes hard to send through the (crude) facility of the internet.  So I am going to assume your problem says:

Prove that (B1 * B2) * B3 ={ }
and that (B1 + B2) + B3 = B.

where I use '+' for the union and '*' for the intersection of sets.  And I also assume that 'natural number' means {1,2,3,4...}, not including zero, and I write N for the set of natural numbers.

1. You want to prove that your three sets B1, B2, B3 have no elements in common.  Now if you write out:

B1 = {6, 15, 24, .. }
B2 = {3, 12, 21, .. }
B3 = {0, 9, 18, ..  }

Can some number be a member of both B1 and B2?

If x is in B1, then  x = 9n1 - 3  for some n1.
If x is in B2, then  x = 9n2 - 6  for some n2.

Solve those equations:

9n1 - 3 = 9n2 - 6
3 = 9(n2 - n1)

n2 - n1 = 1/3

So if n1 is a natural number, then n2 is not, and vice versa.  So B1 * B2 is empty and the intersection of the empty set with anything is empty.  

2. You want to prove that your three sets B1, B2, B3 together add up to B.  To do this, find a way to write any element of B as one of the members of B1,b2, or B3.

B = {0,3,6,9,12,..} = {x : x = 3(k-1), where  k is in N}

[You need the 3(k-1) instead of 3k because you are including zero.]

Now any k in N can be written in the form  3n, 3n-1, 3n-2, where n is in N.

If k = 3n,   then x = 3(k-1) = 3(3n-1)   = 9n - 3, and x is in B1
If k = 3n-1, then x = 3(k-1) = 3(3n-1-1) = 9n - 6, and x is in B2
If k = 3n-2, then x = 3(k-1) = 3(3n-2-1) = 9n - 9, and x is in B3

I think that does it.