# Home work with set builder notation

1. Sep 21, 2008

### kkp

Ok, I am needing help turning (2, 5, 10, 17) into set builder notation. I know to get these you add odd numbers 3, 5, 7 but I can't wrap my mind around putting this into notation.

2. Sep 21, 2008

### symbolipoint

In other words, you are looking for a formula. Check for a changing difference between consecutive terms.

3. Sep 22, 2008

### Tac-Tics

It's pretty pointless to use set builder notation for such a small set. (Homework always seems that way, doesn't it?)

Keep in mind that set builder notation is of the form

{expression | for <variable(s)> in {a bigger set} such that <condition>}

Here, you're working with integers, so the "bigger set" is going to be Z or Z+ or something.

The tricky part is figuring out a useful condition. For example, if your set was {2, 3, 5, 7, 11}, you could have said: {x | x in Z+ where x is prime and x <= 11}.

4. Sep 22, 2008

### CRGreathouse

This question is silly. Here are some equally silly answers.

$$\{n | n\in\{2, 5, 10, 17\}\}$$
$$\{n | (n-2)(n-5)(n-10)(n-17)=0\}$$
$$\{n^2+1 | 1\le n\le4\}$$

The polynomial in the second answer can be rewritten as n^4 - 34n^3 + 369n^2 - 1460n + 1700, if you prefer.

Last edited: Sep 22, 2008
5. Sep 24, 2008

### symbolipoint

How exactly to convert this into set builder notation, not sure; but I did some checking on the sequence of numbers.

The first term is obviously just 2.
After that, the next terms conform to 2 plus the sumation as index goes from 2 to i of three plus two times the expression (n-2);

In other words, I'm saying from the second term onward, the term is
2 + summation from 2 to i of (3 + 2(n-2)).

Some variation from that pattern might be possible (not sure) after n=4, since we might not be sure if only four terms as originally given were enough to build the pattern.