Maths help

[Luckeh!!!!]

Yes it is my homework

Yes I should be doing it myself

Yes I've looked at my notes.

and No, I can't do it which is why i'm here.

question pic

Anyone have a clue?

(cookie to the first person who gets it right)

[Intrepid00]

If you asked me back in High School I could have helped you. Since I am not in School anymore I forgot how to do the other 85% of the Math I learned cause I will never need it in real life and if I do I can find it in a book.

[Ebany]

Do your own work you munter.

MrL

[Luckeh!!!!]

Originally Posted by Ebany

Do your own work you munter.

MrL

no cookie for you then :tongue:

[alastair18]

Well this isn't entirely rigorousbut hopefully my hand waving will help you get you an answer.

The first part is rather straight forward. Just follow the example and repeat it 3 times.
S = {1} therefore N = 2
S = {1,2} therefore N = 6
S = {1,2,3} therefore N = 14

The second part asks for a proof and so hopefully if I tell you the way I thought about it you can do something similar.
r can be every integer between 0 and 2^(n+1) - 1.
You are asked to prove that there is one N for each r.
N is made of the sum of the terms in S.

If you examine the terms in S they are just all the powers of 2. Now if you consider adding any powers of 2 you can get any integer. Its just like the binary number system. For example if you want 13 you will want 2^0 + 2^2 + 2^3 so S = {0,2,3}

So the maximum value of r is the sum of all the terms. r = 2^0+2^1+2^2+....+2^n which can be easily shown to be 2^(n+1) -1
Once you have got this far you know that for each r you have 1 subset and r can be anything from 0 to 2^(n+1) - 1 thus you have 2^(n+1) subsets.

Hope this helps

[Luckeh!!!!]

Originally Posted by alastair18

Well this isn't entirely rigorousbut hopefully my hand waving will help you get you an answer.

The first part is rather straight forward. Just follow the example and repeat it 3 times.
S = {1} therefore N = 2
S = {1,2} therefore N = 6
S = {1,2,3} therefore N = 14

The second part asks for a proof and so hopefully if I tell you the way I thought about it you can do something similar.
r can be every integer between 0 and 2^(n+1) - 1.
You are asked to prove that there is one N for each r.
N is made of the sum of the terms in S.

If you examine the terms in S they are just all the powers of 2. Now if you consider adding any powers of 2 you can get any integer. Its just like the binary number system. For example if you want 13 you will want 2^0 + 2^2 + 2^3 so S = {0,2,3}

So the maximum value of r is the sum of all the terms. r = 2^0+2^1+2^2+....+2^n which can be easily shown to be 2^(n+1) -1
Once you have got this far you know that for each r you have 1 subset and r can be anything from 0 to 2^(n+1) - 1 thus you have 2^(n+1) subsets.

Hope this helps

Yes that does help alot actually, thnx very much

alastair18_cookie_counter++

[SYMM]

Doing it formally by induction would probably be better (base-case of showing N=1, then different inductive steps for odd/even numbers?), but if this means nothing to you, proove it as Alistair said...

[DaWoodster]

Lie, lies, lies I tell you!, the answer is 42.
What was the question?

[Cyp]

Originally Posted by DaWoodster

Lie, lies, lies I tell you!, the answer is 42.
What was the question?

{1,3,5}

[Bloomers III]

And They* Say Maths Is Pointless

*the Daily Mail