# Show Set is Bounded

1. May 5, 2007

### sinClair

Hi everyone. I'm a math student still learning to do
proofs. Here is a problem I encountered that seems easy but
has me stuck.

1. The problem statement, all variables and given/known
data

Let a be a positive rational number. Let A = {x e Q (that
is, e is an element of the rationals) | x^2 < a}. Show that
A is bounded in Q. Find the least upper bound in R of this
set.

2. Relevant equations
None.

3. The attempt at a solution
So I want to show that there exists an M such that x < or =
to M for all x in A.
So for all x in A, x^2<a.
=> x < (a/x) if x>0 or x > (a/x) if x < 0
So it seems like I find an upper bound for x if x is
positive and a lower bound for x if x is negative but what
havn't acounted for the other cases.

Last edited: May 5, 2007
2. May 5, 2007

### quasar987

It does not suffice to find an individual bound for all the elements of A. You must find ONE bound that fits ALL the elements of A.

I suggest you treat the two cases a< or =1 and a>1 and find a bound for A is both cases.

3. May 5, 2007

### sinClair

So the set seems to be bounded by -1,1 if a<1 and sqrt(a) and -sqrt(a) if a>1. But I got this through taking square roots, which arn't there when x is an element of the rationals.

4. May 6, 2007

### quasar987

why not just use a if a>1 ?

5. May 6, 2007

### HallsofIvy

Staff Emeritus
Notice that you are also asked to find the least upper bound in R. That should be obvious.