Download Let S be the set of all positive rational numbers x such that x 2 < 3

Let S be the set of all positive rational numbers x such that x2 < 3. This
set is nonempty and bounded above. (Can you prove this clearly?) I claim
that it does not have a maximum. To prove this, let us recall that if r were
a maximum, we would have to have r ∈ S and x ≤ r for every x ∈ S. Let
us show that this is impossible. Precisely, we shall show that if r ∈ S, then
there exists x ∈ S with x > r.
Here is the proof. Since r ∈ S, r2 < 3. Thus 3 − r2 is a positive (rational)
number, and hence so is = 3 − r2 . We will show that there is a positive
number δ such that x := r + δ ∈ S. Then x > r is the number we want.
There are of course many possibilities for δ. For example, we can choose any
δ which is less than min{ 1+2r
, 1}. Here is why this works.
x2 =
=
=
<
(r + δ)2
r2 + 2rδ + δ 2
3 − + δ(2r + δ)
3 − + δ(2r + 1)
This last inequality is a consequence of the fact that δ ≤ 1. Since δ <
δ(2r + 1) < , so we get
x2 < 3 − + = 3,
,
1+2r
as required.
Exercise: Prove that the set of rational numbers x such that x2 > 3 has
no minimum.
Solution: Actually this is trivial as stated, since we can take x to be
a very negative number. I should have again said that the set of positive
rational numbers such that x2 > 3 has no minimum. This can be proved the
same way as the original problem, but it is a little easier. Suppose x > 0 and
x2 > 3. Let := x2 − 3, which is positive. I claim that there is a positive
number x0 < x such that x02 > 3. In other words, I claim there is a positive
number δ < x such that (x − δ)2 > 3. Take any positive δ which is less than
the minimum of x and /2x. Such a δ exists because x and /2x are both
positive. Then 2δx < , and
(x − δ)2 = x2 − 2δx + δ 2
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Since δ 2 > 0, we get
(x − δ)2 > x2 − 2δx
> 3 + − 2δx
> 0.
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