# Assignment 2 Q 2iii

Posted by Tracey
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 Assignment 2 Q 2iii May 01, 2007 09:31PM Registered: 14 years ago Posts: 3,249 Rating: 0
I've done this question as per the examples in the study guide, I'm now stuck with step 3. Must I just multiply through? I did try that but I wasn't sure what the result is actually supposed to be. Is the final answer supposed to be something like 3(...) ?
 Re: Assignment 2 Q 2iii May 12, 2007 03:06PM Registered: 13 years ago Posts: 1,501 Rating: 0
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++ | n3 + 2n = 3m for some 'm' E Z++

In step 3, you assume that a positive integer 'k' is an element of 'A', i.e. k3 + 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:
= k3 + 3k2 + 5k + 3
= (k3 + 2k) + (3k2 + 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++.
 Re: Assignment 2 Q 2iii May 12, 2007 08:02PM Registered: 14 years ago Posts: 3,249 Rating: 0
Thanks Andre, that will definitely help. I was working along the same lines but I don't think my result was quite the same as yours so that's probably where I went wrong.
 Re: Assignment 2 Q 2iii May 12, 2007 08:40PM Registered: 13 years ago Posts: 1,501 Rating: 0
No problem.
I've done question 2-4. Still need to do 1 & 5, but I'm really delaying question 1. I'm still not comfortable recursion and strings
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