Assignment 2 Semester 1

Hi, I'm hoping someone could lend some assistance. I seem to be having a weee bit of misunderstanding..

Q2c) I cant seem to prove it -> I end up with trying to prove the following equation (everything following the ^ is the exponent, everything following the / is under the line)

1/2^0 + 1/2^1 + ... + 1/2^k + 1/2^k+1 = 2 - 1/2^k+1

here we can simplify the first part up to 1/2^k to something based on the induction principle

ie. 1/2^0 + 1/2^1 + ... + 1/2^k = 2- 1/2^k (as stated by the induction principle)

that means that we have to prove (2- 1/2^k) + 1/2^k+1 = 2 - 1/2^k+1

either I am misunderstanding this, or making as dumb mistake, or I just cant do the algebra . any advice would be greatly appreciated.

Danie van Eeden
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I don't know exactly what your problem is, but if you've done the first two parts you should be quite well on your way. Now all you must do is :
1. Show that the expression satisfies the base case.
2. Take the expression, assume it's true for some arbitrary value (call that k), and then by just manipulating the expression (eg. ADDING THE SAME THING TO BOTH SIDES), show that "P(k+1)" is also true. In other words, keep working on it till you see "hey that's one of the forms of P(k+1)"...

OK, so there's your algebra in the terms of your earlier courses. Now what you do is, having convinced yourself by familiar ways that the statement is indeed true, is put that discovery inside the broader inductive principle framework the SG reveals. (ie. Induction in terms of Sets, basically).

It is a bit tricky (or I found it so), so don't get despondent if it goes slowly. It took me hours to get it figured (I hope, anyway). You want to make use of the fact that 2ⁿ⁺¹ = 2n(2), for one thing. (I actually don't remember in detail why that helped, this just rings a bell).

That's most of it. Just get LHS (or RHS if you really must) into the form it would be if it were P(k+1) instead of P(k).Then take RHS (or vice versa if you've chosen to go the other way), and rearrange it. To give yourself heart, put all that down on your answer sheet and imagine you'll get most of your marks for this part, anyway. Now take RHS and start to express it in different ways, until it's the RHS of of P(k+1).

If LHS of P(k+1) = RHS of P(k+1) then P(k+1) is true, right?

Hehe,

think I found it by luck.. i thinks. My algebra suddenly gave me an answer that seemed to be the same as the other side of the = sign.

Gotto double check again, but I might just have stumbled onto it.

thx for the reply

Danie van Eeden
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Good to hear it worked out. Nice feeling when the light shines in.

I think the big trick once that's done and checked out is to put it into the set-theoretical frame. I'm still far from comfortable with that, although I begin to see a "wider world of induction" for those who take the time to pace it through a few more times.

that always been the key to most i guess - to "take the time"
thx again

Danie van Eeden
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