modern algebra?

Question

Let Z denote the set of all intebers, and let

A={x|x=3p-2 for some p in element Z}
B={x|x=3q+1 for some q in element Z}.

Prove that A=B.

Answer 1

A={x|x=3p-2=3q+1 for some p in element Z} if q=p+1 (1)

B={x|x=3q+1=3p-2 for some q in element Z} if p=q-1 (2)

and clearly, if q belongs to Z, p+1 also belongs to Z. The same can be said about p and q-1.

From (1), A is a subset of B
From (2), B is a subset of A

Therefore A=B.

Answer 2

Let x ∈ A. Then x = 3p - 2 for some p ∈ Z.
But then x = 3(p-1+1) - 2
= 3(p-1) + 3 - 2
= 3(p-1) + 1
Since p ∈ Z, p-1 ∈ Z and hence x ∈ B. This proves A ⊆ B.

Similarly, if x ∈ B then x = 3q + 1 for some q ∈ Z. Then x = 3(q+1) - 2 and q+1 ∈ Z, so x ∈ A. So B ⊆ A.

Since A ⊆ B and B ⊆ A, we must have A = B.

Answer 3

Well... let a be an element of A
Then a = 3p-2 for some p in Z
Then a = 3(q+1) -2 where q=p-1 and clearly q is in Z
Then a = 3q +3-2 = 3q+1
and so a is in B

So any element of A is in B, so A is a subset of B

You can do the same thing in reverse to show that B is a subset of A.

If 2 sets are subsets of each other, they are equal, so A=B

Answer 4

use the Division algorithm and/or congruence classes to show that -2 = 1 in Z3 (this is obviously true since -2 congruent 1 mod 3)