In class, we discussed the following proposition:
Proposition 1. The set
A = {x ∈ Q : x2 < 2}
has no largest element.
In order to prove this, we took an arbitrary positive x ∈ A and tried to find
y ∈ A such that x < y. Somewhat mysteriously, we chose
y =x−
x2 − 2
.
x+2
and argued that y ∈ A and y > x. The following is a sketch of how one might
come to find such a y.
The basic idea is that we’d like to find a positive z ∈ Q such that
y = x − (x2 − 2)z
lies in A. [Obviously if z is positive then y > x.] To this end, we observe that
y 2 − 2 = x2 − 2 − 2xz(x2 − 2) + (x2 − 2)2 z 2 = (x2 − 2)[1 − 2xz + (x2 − 2)z 2 ].
In order for y 2 − 2 to be negative, we thus need to find z so that f (z) =
1 − 2xz + (x2 − 2)z 2 is positive. Using the quadratic formula, we find that
z=
1
√
x± 2
makes f (z) = 0. Since x2 − 2 < 0, f (z) is a concave down parabola, and it
follows that any choice of z so that
0<z<
1
√
x+ 2
works. We want z to be rational, so we simply choose
z=
1
,
x+2
although any z of the form (x + n)−1 for n ≥ 2 is valid.
1