# Thread: Set problem

1. ## Set problem

If $\displaystyle A = \{0, {1\over2} , 1, {3\over2} , 2, {5\over2} , 3, {7\over2} , 4, {9\over2} , 5\}$, write down the following set;

$\displaystyle \lbrace2n + 1 \in A : n \in A\rbrace$

The solution is $\displaystyle \{1,2,3,4,5\}$ but I can't get my head round why, could someone please explain this to me?

Cheers.

2. Hi.

Originally Posted by Gaz
If $\displaystyle A = \{0, {1\over2} , 1, {3\over2} , 2, {5\over2} , 3, {7\over2} , 4, {9\over2} , 5\}$, write down the following set;

$\displaystyle \lbrace2n + 1 \in A : n \in A\rbrace$

The solution is $\displaystyle \{1,2,3,4,5\}$ but I can't get my head round why, could someone please explain this to me?

Cheers.
It is $\displaystyle n \in A$,
what is n? n = 0 or n = 1/2 or n = 1 or n = 3/2 or n = 2 or ... or n = 5

It should be $\displaystyle 2n+1 \in A$

Check it for all n

n= 0:

2*0+1 = 1, is $\displaystyle 1 \in A$? Yes it is! Therefor our solution contains 2*0+1 = 1

n = 1/2 :

2*1/2 + 1 = 2,
is 2 in A? Yes it is, therefor 2 is a solution, too
...
...
...
n = 5:

2*5 + 1 = 10+1 =11
Is 11 in A? No, its not! Therefore 11 is not a solution.

Do you think you find the other solutions by yourself?

Regards, Rapha

3. Right. I've got it now. I know what my problem was!!

Thanks!