I'm going to give this a go - but don't promise too much.
It Step 2 you've determined that 1 is an element of 'A'. 'A' being the subset of Z
++ = {n E Z
++ | n
3 + 2n = 3m for some 'm' E Z
++
In step 3, you assume that a positive integer 'k' is an element of 'A', i.e. k
3 + 2n is divisible by 3. Now you need to deermine that k + 1 E 'A', i.e. (k + 1)
3 + 2(k + 1) is divisible by 3.
Now they leave out a couple of steps in the study guide, but yes you need to multiple out and play a bit with association (said under correction?) to isolate the product as show in the book.
After initial multiplication:
= k
3 + 3k
2 + 5k + 3
= (k
3 + 2k) + (3k
2 + 3k + 3)
This first section is what we are looking for and is divisible by 3 by definition.
In the second section each member is divisible by 3, and hence the Sum is also divisble by 3.
You have proven that if k E 'A', then k + 1 E 'A', blah blah blah and that A is in fact equal to Z
++.
