Show that $x_0^2 = 2$ is impossible by showing $x_0^2 = 2$, then $x_0$ is not rational

Question

The set S of rational numbers x with $x^2 \lt 2$ has rational upper bounds but no least rational upper bound. Suppose that S has a least rational upper bound and call it $x_0$.

Show that $x_0^2 = 2$ is impossible by showing $x_0^2 = 2$, then $x_0$ is not rational.

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Is this as simple as showing that $x_0 = \sqrt 2$ and obviously not rational? Or is that not robust enough?

Answer 1

Your logic is basically sound, but not quite watertight. When you say "showing that $x_0=\sqrt 2\;$", you are assuming the existence of the number $\sqrt 2$, which is not available when you are working with the rational numbers. You should instead say "showing that $x_0^2=2\;$".

Then, as you say, once you have shown that $x_0^2$ has to equal $2$, you can use the traditional ancient Greek argument to show that this leads to a contradiction.

And to show that $x_0^2=2$, suppose that $x_0^2>2$ and then exhibit a rational number $q$ such that $2<q^2<x_0^2$. (This requires a bit of thought.)

Answer 2

It sounds as if you have been set a very dreary logic-chopping problem that is actually incorrect. If you suppose that the set $S = \{x :\Bbb{Q} \mid x^2 < 2\}$ has a least upper bound $x_0$ that is rational, then you have made a false supposition and from it you can conclude any proposition you like, including the proposition that $x_0^2 = 2$.