Posted by slow_eddy

Announcements | Last Post | |
---|---|---|

myUnisa availability 21 to 24 March 2019 | 03/17/2019 02:24PM | |

SoC Curricula | 09/30/2017 01:08PM | |

Demarcation or scoping of examinations and assessment | 02/13/2017 07:59AM | |

School of Computing Short Learning Programmes | 11/24/2014 08:37AM | |

Unisa contact information | 07/28/2011 01:28PM |

(Really I should put my skeleton of 4(b) up as well - even though that's off to the casualty ward).

Let the expression "blah blah is odd" be P(n).

Base Case.

1^{2} + 5(1) + 1 = 7, which is odd. So P(1) is true.

Induction Hypothesis.

Imagine for a moment that for k >= 1 P(k) just happens to be true.

In other words k^{2} + 5k + 1 is an odd number.

So by definition it has the general form 2m+1, where m is an integer.

So, equivalently, k^{2} + 5k = 2m (just some even number)

I'm pretty sure the "even number angle" is superfluous, but something in me just likes it, so there.

Now we examine the nature of P'(k + 1)

This is the assertion that "blah blah is even" or "da dee da is odd" (that this is true). So plug in k+1 and see what it does.

(k+1)^{2} + 5(k+1) ¿ is even ?

= k^{2} + 2k + 1 + 5k + 5

= k^{2} + 5k + 2k +6

Now we notice that k^{2} + 5k = 2m, *by the induction hypothesis*. So we continue ...

= 2m + 2k + 6 .... which is starting to look very much like an even number. We confirm that:

= 2(m+k+3)

So it follows from the induction hypothesis that P(k+1) is indeed TRUE.

So you could plug in a 1 in the place of k. Now you have P(k=1) is TRUE (base case), so then P(k+1=1+1=2) is TRUE. (Because we've just proven that this follows given the truth of the previous one). And so from P(2) we go to P(3). 3 to 4. 4,5. 5,6. .... infinity, if you please.

The short way of saying it is that "by the principle of mathematical induction blah blah blah".

Sometimes it's good to remind oneself of the mechanics of how it works at the very end.

OK. So now I offer you a skeleton of a 4b whose wheels fell off for me.

Let the expression "blah blah is odd" be P(n).

Base Case.

1

Induction Hypothesis.

Imagine for a moment that for k >= 1 P(k) just happens to be true.

In other words k

So by definition it has the general form 2m+1, where m is an integer.

So, equivalently, k

I'm pretty sure the "even number angle" is superfluous, but something in me just likes it, so there.

Now we examine the nature of P'(k + 1)

This is the assertion that "blah blah is even" or "da dee da is odd" (that this is true). So plug in k+1 and see what it does.

(k+1)

= k

= k

Now we notice that k

= 2m + 2k + 6 .... which is starting to look very much like an even number. We confirm that:

= 2(m+k+3)

So it follows from the induction hypothesis that P(k+1) is indeed TRUE.

So you could plug in a 1 in the place of k. Now you have P(k=1) is TRUE (base case), so then P(k+1=1+1=2) is TRUE. (Because we've just proven that this follows given the truth of the previous one). And so from P(2) we go to P(3). 3 to 4. 4,5. 5,6. .... infinity, if you please.

The short way of saying it is that "by the principle of mathematical induction blah blah blah".

Sometimes it's good to remind oneself of the mechanics of how it works at the very end.

OK. So now I offer you a skeleton of a 4b whose wheels fell off for me.

I tried to do the short version that they seem to be using in tut102:

Base Case: Where n=1, we have 1^{2}+5(1)+1 = 2(3) + 1

Inductive Step:

Assume the formula holds for n, then since n^{2}+5n+1 is odd, there exist by the definition of odd, an integer m such that n^{2}+5n+1 = 2m+1

So,

(n+1)^{2} + 5(n+1) + 1

= n^{2} + 2n + 1 + 5n + 5 + 1

=(n^{2}+5n+1) + 2n + 5 + 1

=(2m + 1) + 2n + 6

=2(m+n+3) + 1

Thus, the formula holds for n+1 if it holds for n. By the principle of Mathematical Induction it means that the formula holds for all n >= 1

Base Case: Where n=1, we have 1

Inductive Step:

Assume the formula holds for n, then since n

So,

(n+1)

= n

=(n

=(2m + 1) + 2n + 6

=2(m+n+3) + 1

Thus, the formula holds for n+1 if it holds for n. By the principle of Mathematical Induction it means that the formula holds for all n >= 1

@hexium, your algebra's dead right according to my reckoning, but you need to watch how you express the "n+1" step.

In particular the word "So" is potentially misleading. If you mean, "So now we do n+1..." (which is I think all you really want to say there), then of course it would be fine.

However, "So" looks much too much like "So then if that's so then this is so". Which is not correct. P(n+1) doesn't follow*until* the algebra shows that this is so. You don't start with "So"; you *end* with "So". I would use a different expression. Find another short word that says "We're now going to look at the expression with n+1 substituted in, and see what it entails."

Also (but I don't think this is a major point), I reckon it's better to use the "k notation". k is some arbitrary number "in the middle of nowhere, if you like", where n is part of the sequence. I won't labour the point, because it's probably unimportant - or even "Egal".

In particular the word "So" is potentially misleading. If you mean, "So now we do n+1..." (which is I think all you really want to say there), then of course it would be fine.

However, "So" looks much too much like "So then if that's so then this is so". Which is not correct. P(n+1) doesn't follow

Also (but I don't think this is a major point), I reckon it's better to use the "k notation". k is some arbitrary number "in the middle of nowhere, if you like", where n is part of the sequence. I won't labour the point, because it's probably unimportant - or even "Egal".

That's what I don't like about these subjects... in 201 they show you one way of doing it... then in 361 another...(using n all the way in stead of using k for instance) and they give oh so few examples that it feels like we are supposed to "get it" through osmosis or something...

I will keep the "so" in mind and rather wrte "Now"

I will keep the "so" in mind and rather wrte "Now"

Yes, I think "now" will do me fine, too. Unless I get all enthusiastic about induction in mid-exam...

I wonder if they've ever asked for course of values induction. Probably safer to look that up again once the conundrum of the diamond is abandoned or solved. (It's a real headbanger that one, isn't it?)

I wonder if they've ever asked for course of values induction. Probably safer to look that up again once the conundrum of the diamond is abandoned or solved. (It's a real headbanger that one, isn't it?)

Well one can't reason directly about diamond in an arbitrary accessible world, and the ~

in the front of ~ box ~ would seem to effectively block any attempt to get some "working string" in a dotted box...

Yes, it's a bugger. Probably the thing to do is to stop attempting to solve, and just try to do all possible manipulations on that particular structure?

Probably a better way would be to assign natural meanings to every part of every statement, and then try to reason from one to the other in words. Sometimes one can then work back to symbols. (And at least what one's doing seems to make some kind of real sense at the time).

in the front of ~ box ~ would seem to effectively block any attempt to get some "working string" in a dotted box...

Yes, it's a bugger. Probably the thing to do is to stop attempting to solve, and just try to do all possible manipulations on that particular structure?

Probably a better way would be to assign natural meanings to every part of every statement, and then try to reason from one to the other in words. Sometimes one can then work back to symbols. (And at least what one's doing seems to make some kind of real sense at the time).

Sorry, only registered users may post in this forum.