Show Set is Bounded: Prove A is Bounded in Q

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Homework Help Overview

The problem involves proving that a set A, defined by a positive rational number a and the condition x^2 < a, is bounded within the rational numbers Q. The task also includes finding the least upper bound of this set in the real numbers R.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find a single bound for all elements of A, considering different cases based on the value of a. There is also exploration of the implications of using square roots and the nature of rational numbers.

Discussion Status

The discussion is ongoing, with participants providing insights into the requirements for bounding the set A. Some suggestions have been made regarding specific bounds based on the value of a, but no consensus has been reached on the approach to take.

Contextual Notes

Participants note the importance of considering both positive and negative cases for bounding the set, as well as the requirement to find the least upper bound in R, which adds complexity to the problem.

sinClair
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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.

Homework Equations


None.

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.

Thanks for your help.
 
Last edited:
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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.
 
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.

Hm, I guess I have to think about this some more.
 
why not just use a if a>1 ?
 
Notice that you are also asked to find the least upper bound in R. That should be obvious.
 

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