I started university earlier this month, doing Mathematics and Economics. In one of the modules, Foundations, we're doing sets. I understand them, it's just the subtle nuances that I'm not entirely sure about. I have a few problems here I've done, though they're probably completely wrong...

1. If A = {1,2,3,...,10}, list the members of the following sets:
iii: {3x + 1 ∈ Z : x ∈ A} = {4,7,10,13,...,31}
iv: {n ∈ A : 3n + 1 ∈ Z} = {4,7,10}

3: Rewrite the following listed sets by specifying a defining property of the elements.
ii: B = {1,1/2,1/4,...} = {1^-2(n-1) ∈ B : n ∈ N}
iv: D = {2,5,10,17,...} = ? (ie. haven't gotten around to trying it yet)

Any help would be much appreciated. If you could maybe work through them for me, that would be great. Even if it isn't all of them.

I started university earlier this month, doing Mathematics and Economics. In one of the modules, Foundations, we're doing sets. I understand them, it's just the subtle nuances that I'm not entirely sure about. I have a few problems here I've done, though they're probably completely wrong...

1. If A = {1,2,3,...,10}, list the members of the following sets:
iii: {3x + 1 ∈ Z : x ∈ A} = {4,7,10,13,...,31}
iv: {n ∈ A : 3n + 1 ∈ Z} = {4,7,10}

3: Rewrite the following listed sets by specifying a defining property of the elements.
ii: B = {1,1/2,1/4,...} = {1^-2(n-1) ∈ B : n ∈ N}
iv: D = {2,5,10,17,...} = ? (ie. haven't gotten around to trying it yet)

Any help would be much appreciated. If you could maybe work through them for me, that would be great. Even if it isn't all of them.

First of all, I'll gladly help and don't mean to nag but honestly the only way to get math is to spend lots of time stuck on something until you figure it out.

For 1 iv, it's asking you for the n's in A so that 3n + 1 is an integer, but anything in A, when you multiply by 3 and add 1 will become an integer, so the set is actually A itself.

For 3ii, you want 1 divided by powers of two, so it's {1/(2^n), n ∈ Z, n>0}.
For 3iv, every term is gotten from the one before by adding 3, then 5, then 7, so on. So the nth term will be 2 + (0+3 + 5 + 7 + .... + 2n - 1) = 1+ 1 + 3 + 5 + ... + 2n - 1.

As it turns out, the sum of the first k odd numbers gives k^2 (example, the sum of the first 5 odd numbers is 1+3+5+7+9 = 25 which is 5^2). This is just an arithmetic sum so you may try to prove it, its not too hard. So back to our set, the nth term will be 1 + n^2. So your set is {n^2 + 1, n ∈ N}

let me say that i found this set stuff not actually important. its just formalities

or 2,5,10,17, the rule is

1 + 1 = n+1
2 + 2 +1 = 2n+1
3 +3 +3 +1 = 3n+1
4+ 4+4+4+1 = 4n+1




n^2 + 1

though by experimenation,

2n + (n-1)^2
also works

you can do this by trial and error, what i did was find some pattern. it takes a bit of mental stretching, but it is doable. to be honest, i actually got the second equation first, but after expanding , it equals the first equation. i would just draw a table

n-->number
1-->2
2-->5
3-->10
4-->17
so somehow the 1 turns into a 2. now by logic, you would probably think you either multiplied it by 2 or added 1
somehow, the 2 turns into a 5. you could either double it, then add 1, or square it, then add 1

personally, i would just start with the first few terms then work out a bunch of rules, and seeing which ones will work