Use the fact that the remainder of every integer when divided by 3 is 0, 1 or 2.
In general you prove that " ", for A and B sets, by proving " " and " ".
And you prove " " by starting "if " and using the properties of X and Y to conclude "then "
Here, you have and B= Z.
To show that is easy. is then
i)
Since Z is "closed under multiplication" x= 3q, for any q in Z, is in Z
or ii)
Since Z is "closed under multiplication" and "closed under addition", x= 3q+ 1, for any q in Z, is in Z.
or iii)
Again, since Z is "closed under multiplication" and "closed under addition", x= 3q+ 2, for any q in Z, is in Z.
Therefore A is a subset of Z.
To prove the other way is only slightly harder. If , then dividing x by 3 gives remainder 0, 1, or 2 since the remainder has to be a non-negative integer less than 3: x/3 "has quotient q with remainder r" so that x/3= q+ r/3 and x= 3q+ r.
i) If the remainder is 0 then x= 3q where q is the quotient so
ii) If the remainder is 1 then x= 3q+ 1 where q is the quotient so
iii) If the remainder is 2 then x= 3q+ 2 where q is the quotient so
In any case, so
Together those give .