determining if a number is an element in a set?

Question

Let A = (X element R | {x,x^2,x^3....} intersecting Z (integers) not equal to the empty set

is 5 an element of A, is 1/2 an element of A?? thanks

i thought that 5 would be and 1/2 wouldnt because 5 is an integer and 1/2 isnt...

Answer 1

It depends on what x is. If x is between 0 and 1, then 5 (and basically any integer) is not in {x, x^2, ...} for sure (check this). If x is larger than 1, then 1/2 is not in {x, x^2, ...} for sure (check this). Since A is nonempty, we know from above that x is not between 0 and 1 (0= 1. If x=1, then 5 is not in there and neiter is 1/2. If x=2, then 5 and 1/2 are not in A. You need to investigate your problem futher (I am assuming that x is a fixed element in R).

If x is not a fixed element, then the question does not make sense. The first set would just be all of R. So you would be looking at the intersection of R with Z, which is just Z. So you are definitely looking at a set for some fixed x. This means that 5 is not always in your set A, you will only have some of the integers and not all of them in A. You need to be careful with the wording.

Answer 2

Yes, 5 belongs to A (as 5 belongs to Z) and 1/2 doesn't, as (0.5)^n < 1 for each n>0

Note: x=sqrt(2)=1.4142... belongs to A, as x^2=2. The same for each radical x=n^(p/q), as x^q=n^p is integer

Regarding impeachrob's reply: As the question is written, A is the set of all reals x such that a power of x (at least) belongs to Z.

Answer 3

You're right. But to formally prove it you need to show that 1/2 does not belong to A.

Since A = R intersection Z all you need to prove is that 1/2 does not belong to one of the component groups, so if 1/2 doesnt belong to Z or R it cant belong to R intersection Z.

hope this helps