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Math and science::Analysis::Tao::05. The real numbers

# $$x^2 = 2$$. There exists a positive real whose square is 2. Proposition.

This proposition's proof is enabled by the existance of the least upper bound for sets of reals.

### Proof

Let $$E$$ be the set $$\{y \in R: y \ge 0 \text{ and } y^2 < 2 \}$$. $$E$$ has an upper bound of 2, as if $$y > 2$$ then $$y^2 > 4 > 2$$. As $$E$$ is a non-empty set of reals and has an upper bound, it must have a least upper bound. Let $$x = \sup E$$ be the least upper bound of $$E$$.

Exactly one of $$x^2 < 2, x^2 = 2 \text{ or } x^2 > 2$$ must be true. We proceed to show that $$x^2 < 2$$ and $$x^2 > 2$$ lead to contradictions and must be false.

#### $$x^2 < 2$$ must be false.

Assume $$x^2 < 2$$ is true. (This means that $$x \in E$$, but we must go further to impose that $$x$$ is not an upper bound of $$E$$). Consider $$(x + \varepsilon)^2$$, where $$0 < \varepsilon < 1$$ is a real:

\begin{align}(x + \varepsilon)^2 &= x^2 + 2x\varepsilon + \varepsilon^2 \\ &\le x^2 + 4\varepsilon + \varepsilon^2 && \text{as } x \le 2 \\ &\le x^2 + 5\varepsilon && \text{as } \varepsilon < 1 \\\end{align}

We can choose an $$\varepsilon$$ such that $$x^2 + 5\varepsilon < 2$$. From above, $$(x + \varepsilon)^2 \le x^2 + 5\varepsilon$$, so it must be the case that $$(x + \varepsilon)^2 < 2$$ is also true. This implies that $$x + \varepsilon \in E$$. As $$x + \varepsilon > x$$, $$x$$ cannot be an upper bound of $$E$$, which is a contradiction. Thus, $$x^2 < 2$$ is false.

#### $$x^2 > 2$$ must be false.

A similar argument to above. Assume $$x^2 > 2$$. Consider $$(x - \varepsilon)^2$$, where $$0 < \varepsilon < 1$$ is a real:

\begin{align}(x - \varepsilon)^2 &= x^2 - 2x\varepsilon + \varepsilon^2 \\ &> x^2 - 4\varepsilon + \varepsilon^2 && \text{as } x < 2 \\ &> x^2 - 4\varepsilon && \text{as } \varepsilon > 0 \\\end{align}

Since $$x^2 > 2$$, we can choose an $$\varepsilon$$ such that $$x^2 - 4\varepsilon > 2$$ (why? This reasoning tripped me up when I first attempted the proof). In this case, $$(x-\varepsilon)^2 > 2$$ is also true.

We want $$x-\varepsilon$$ to be an upper bound, and to show this, it must be true that $$x - \varepsilon \gt y$$ for all $$y \in E$$. We show this by contradiction: if $$x-\varepsilon < y$$ for some $$y \in E$$, then $$(x-\varepsilon)^2 < y^2 < 2$$ , a contradiction. Thus $$x - \varepsilon$$ is an upper bound for E.

As $$x - \varepsilon$$ is an upper bound for E and is less than $$x$$, it cannot be true that $$x$$ is a least upper bound. Thus, $$x^2 > 2$$ mus be false.

As both $$x^2 < 2$$ and $$x^2 > 2$$ are false, it must be the case that $$x^2 = 2$$ is true.