Assignment 2 Q 2iii 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(...) ?
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Sun, 18 Aug 2019 05:48:36 +0200Phorum 5.2.18http://osprey.unisa.ac.za/phorum/read.php?132,48910,49796#msg-49796Re: Assignment 2 Q 2iii
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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]]>AndreBCOS201VSat, 12 May 2007 20:40:43 +0200http://osprey.unisa.ac.za/phorum/read.php?132,48910,49795#msg-49795Re: Assignment 2 Q 2iii
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TraceyCOS201VSat, 12 May 2007 20:02:45 +0200http://osprey.unisa.ac.za/phorum/read.php?132,48910,49781#msg-49781Re: Assignment 2 Q 2iii
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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^{++}.]]>AndreBCOS201VSat, 12 May 2007 15:06:39 +0200http://osprey.unisa.ac.za/phorum/read.php?132,48910,48910#msg-48910Assignment 2 Q 2iii
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TraceyCOS201VTue, 01 May 2007 21:31:12 +0200