Answer:
Let A be the set of all rational numbers p such that p2<2, and let B be the set of all rational numbers p such that p2>2. It can be shown that A has no largest element and that B has no smallest element by associating to each p>0 the number
q=p−p2−2p+2
and then proving that p∈A⇒q∈A and p∈B⇒q∈B.
Explanation:
We want to find a rational number q such that
2–√<q<p.
One idea is try to lower p, like
q=k−1kp<p
with a suitable rational k. If you write the condition 2–√<q, you get
2–√<k−1kp
and obtain
k≥pp−2–√.
Any positive rational number k above that bound gives you a good q. In order to get a rational number, you try to get rid of the 2–√ by rationalization
pp−2–√=p(p+2–√)p2−2
and 2>2–√, so
k=p(p+2)p2−2>p(p+2–√)p2−2.
Substituting it in q, you obtain
q=k−1kp=p−pk=p−p2−2p+2.