Question

How can I prove that `QQ` is incomplete by using the fact that `RR` is archimedean ?

For this proof I have to show that:
Let `S = {x in QQ | x^2 <= 2}`, then `EE` an upper bound in `QQ`.
supremum`(S) !in QQ`

Answer 1

See explanation...

Explanation:

We do not need to look at `RR` to prove that `QQ` is incomplete in the order theoretical sense.

Let `S = { x in QQ | x^2 <= 2 }`

First note that there is no rational number `x` such that `x^2 = 2`.

So we can write:

`S = { x in QQ | x^2 < 2 }`

Note that if `x >= 2` then `x^2 > 2` so `x !in S`

So `2 in QQ` is an upper bound of `S`

Suppose `a` is a positive rational number with `a^2 < 2`

Let:

`b = (4a)/(a^2+2)`

Then:

`b - a = (4a)/(a^2+2) - a = (4a-(a^3-2a))/(a^2+2) = (a(2-a^2))/(a^2+2) > 0`

and:

`2-b^2 = 2- ((4a)/(a^2+2))^2`

`color(white)(2-b^2) = 2 - (16a^2)/(a^4+4a^2+4)`

`color(white)(2-b^2) = (2a^4+8a^2+8 - 16a^2)/(a^4+4a^2+4)`

`color(white)(2-b^2) = (2(a^4-4a^2+4))/(a^4+4a^2+4)`

`color(white)(2-b^2) = (2(a^2-2)^2)/(a^4+4a^2+4) > 0`

So we have:

`a < b` and `b^2 < 2`

So `a` is not a supremum of `S`.

So `QQ` does not contain a supremum of `S`.

That is `QQ` is incomplete.