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## Homework Statement

Show that the set ##\{x \in \mathbf Q; x^2< 2 \}## has no least upper bound in ##\mathbf Q##; using that if ##r## were one then ##r^2=2##. Do this assuming that the real field haven't been constructed.

## Homework Equations

N/A

## The Attempt at a Solution

Attempt at proof:

##r\in Q## is an upper bound of the set if and only if ##r^2 \ge 2##.

Choose any ##r_1 \in \mathbf Q## that satisfies ##r_1^2 >2##. Then ##r_1## is an upper bound of the set.

Set ##r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}>2##.

Note that ##r_2 \in \mathbf Q## and ##r_2 < r_1##. ##r_2## is also an upper bound since ##r_2^2>2##.

This shows that for every upper bound ##r_1## with ##r_1^2 > 2## there's always possible to find an upper bound ##r_2## where ##r_2<r_1##. Hence the only possible least upper bound satisfy ##r^2=2## but there is no ##r\in \mathbf Q## that satisfy that. Hence the set have no least upper bound in ##\mathbf Q##.

Is the above correct? Anything I can do to improve it? I'm also wondering about the choice of ##r_2##. I remembered that from an earlier example in our book but I'm not sure I could've come up with a choice like that by myself without a lot of trial and error. How would I go about finding a choice like that in the first place?