some math questions

[cipherpol]

Gonna need help understanding some of these proof questions

If there is a statement P(n) about natural numbers n, out of the 8 truth tables ([TTT], [TTF], [TFT], etc.) what values can the truth table [P(k), P(k+1), P(k)⇒P(k+1)] take?

Can someone give an example about one of the possibilities? Not sure how to go about writing this.

Thanks

EDIT: done

[Mao]

If I'm understanding this correctly, P(n) is an arbitrary function of n that returns T or F. k+1 and k are both arbitrary points on the domain of P. That is, I can rewrite the truth table as [P(m), P(n), P(m) implies P(n)] where m, n are both natural numbers.

Then the possible combinations are:
[TTT], [TFF], [FTT], [FFT]

[cipherpol]

Ah, I should have mentioned this was to do with induction. I remember my lecturer going on about how a false statement can imply anything, so I'm wondering whether it would be [FTT] instead of [FTF] in your post.

[Mao]

Ah, I should have mentioned this was to do with induction. I remember my lecturer going on about how a false statement can imply anything, so I'm wondering whether it would be [FTT] instead of [FTF] in your post.

Quite right, I wasn't thinking at the time. You are absolutely right.

(the source of my knowledge of truth tables goes as far as minecraft allows)

[cipherpol]

Find the 80th number in the bijection of the positive rational numbers with the natural numbers

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, 1/7, 3/5, ...

lol, is there a better way than drawing out the table to find the 80th number?

[/0]

The easiest way is probably to just draw out the table and in the manner of the sieve of eratosthenes, cross out rational numbers with . I mean, it's a practical way to reach 80, but might not be so useful for higher numbers.

Another way which might work... in the bijection, note that you only have rational numbers such that . So you could count all the pairs of numbers with until you reach the 80th.

A useful tool you might want to use is Euler's totient function. Let be the number of integers less than which are relatively prime to . Then

where are prime factors of . Proof can be found on wikipedia.

Actually, the diagonal counting method in your question makes it really annoying. If we used a counting method like in my picture, it would be much easier to find the 80th number :/

[cipherpol]

Thanks /0 =]

Not sure if we've even been through the material in the next question... :S

Identify the interesting points for the function f(z)=z+1/z.

[/0]

Let , then

We can view this as a parametric equation:



After playing around with the graph in Mathematica, I realised that as you vary and keep fixed, a point will do a loop-the-loop around one of the 'interesting points'.

If you do this for different , you find that the size of the loop changes. It is reasonable to assume that for a particular , the loop is so small that it becomes a cusp.

Then to find this cusp, we must simultaneously solve , .

Here's the notebook. If you don't have mathematica you can download a free mathematica player.

[Mao]

+1 for mathematica.