(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Couple of Proofs (Regular Induction / Well Ordering)

Hi there everyone, I've been having a bit of trouble of solving these questions, so any help would be greatly appreciated:

1. The problem statement, all variables and given/known data

1: Prove, via regular induction, that it is possible to draw a line-segment of length exactly sqrt(n) using just a ruler and a compass for all integers n >= 1.

(Note that certain numbers like sqrt(2) and sqrt(3), etc. have infinitely long decimal representations, so they cannot be simply be measured off on a ruler.)

2. Relevant equations

HINT: You may find the following geometric fact useful: If C is any circle and t = abc is any triangle inscribed in C such that ac is a diameter and b is on the circumference, thn <abc = 90 degrees.

3. The attempt at a solution

So I know that through regular induction that the basis step should be n = 1; so the sqrt(1) is simply 1, which can be measured on a ruler. So for the Inductive step we assume that P(k) is true, and we need to show that P(k+1) is true, i.e, sqrt(k+1) can be measured drawn with a ruler and compass.

To tell you the truth, I really have no idea on how to tackle this problem. The only thing that I know is that if we do inscribe a triangle within a circle, then the only thing that I can feel certain about stating is that the diameter must be k, and I'm guessing that through the Pythagorean theorem, I have to show that the hypotenuse must be sqrt(k+1). But does that show that it can be drawn? And how would I go about showing that?

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1. The problem statement, all variables and given/known data

2: Prove the following directly using the well-ordering principle ( do not use induction): Every integer n >= 1 can be written as the sum of distinct powers of 2. Be sure to state clear how you use well-ordering in your proof.

2. Relevant equations

3. The attempt at a solution

To be honest, I don't even know how to start this problem. I met with my T.A and I know that I have to prove this by a contradiction using well ordering, and that I can rewrite k and k - 1 +1, but that's it really. Any help is appreciated.

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I have one more question, it's just making sure that my cases are correct.

Just to make sure, when proving the P(k+1) case, I can split up the k + 1 pile as r and (k+1)-r right? so the product of those two is kr + r - r^2. and then when I split up those two piles, I know that each of those piles are less than k, and due to the Inductive Hypothesis (which I left out here) I can write each of those splits as r(r-1)/2 and [(k+1)-r][([k+1]-r)+1]/2 and then I simply add these products together to get my k+1 counter result right (which should be (n+1)n/2 I think)? Assume that you are given a pile of n >=1 pebbles. You are to divide this

into n piles, each containing one pebble. You do this by arbitrarily splitting the pile into

two piles (not necessarily of equal size) and then successively splitting each of the resulting

piles into smaller piles until you end up with piles of size one. If at some stage you split a

pile into two piles of size r and s, then you add the product rs to a counter (which is set to

zero before this process begins). Prove, via strong induction, that no matter how you split

the piles, at the end the counter will always read n(n − 1)/2.

Anyways, thanks in advance!

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# Homework Help: Couple of Proofs (Regular Induction / Well Ordering)

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