1.6 Consequences of the completeness axiom

We first show that $\mathbb{R}$ contains $\sqrt{2}$: that is, there exists some element $s\in\mathbb{R}$ such that $s^{2}=2$ (so we have filled in at least one hole in comparison with the rationals).

Consider the set $A:=\{a\in\mathbb{R}:a^{2}<2\}$; this is a variant of the set we encountered in Example 1.14. Since $A\subseteq\mathbb{R}$ is nonempty (for instance, $1\in A$) and bounded above (for instance, $2$ is an upper bound), the completeness axiom (1.15) says that $A$ has a least upper bound in $\mathbb{R}$. That is, the real number $s:=\sup A$ exists.

We claim that $s=\sqrt{2}$; that is, $s^{2}=2$. We prove the claim using a Goldilocks approach: we show $s^{2}\not>2$ (not too hot) and $s^{2}\not<2$ (not too cold), so we must have $s^{2}=2$ (just right).

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  Suppose $s^{2}>2$. We want to show that this leads to a contradiction: our strategy is to find an upper bound for $A$ that is smaller than $s$ (which is supposed to be the least upper bound). Our candidate is a number of the form $s-1/n$, and we want to choose a value for $n$ so that $(s-1/n)^{2}>2$, since that will mean that for all $a\in A$, $(s-1/n)^{2}>2>a^{2}$, and so $s-1/n>a$.

  Since $s^{2}-2>0$, we can choose $n\in\mathbb{N}$ with $0<1/n<(s^{2}-2)/2s$ (this choice is based on some rough work: see Exercise 1.45). Then we have

  (1.6) (1.6) $$(s-1/n)^{2}=s^{2}-2s/n+1/n^{2}\geq s^{2}-2s/n>2,$$

  where the second step (“$\geq$”) is due to the fact that $1/n^{2}\geq 0$, and the final step is due to our choice of $n$. From (1.6) and the definition of $A$, it follows that $s-1/n>a$ for all $a\in A$. Hence $s-1/n$ is an upper bound for $A$ and $s-1/n<s$. This contradicts the fact that $s$ is the least upper bound for $A$, and so $s^{2}\not>2$.

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  Now suppose $s^{2}<2$. Here our strategy for deriving a contradiction is to find some $a\in A$ which is larger than $s$, which is supposed to be an upper bound for $A$. Our candidate is $s+1/n$ for a suitable choice of $n$.

  Since $2-s^{2}>0$, we can choose $n\in\mathbb{N}$ with $0<1/n<(2-s^{2})/(2s+1)$. Thus,

  (1.7) (1.7) $$(s+1/n)^{2}=s^{2}+2s/n+1/n^{2}\leq s^{2}+(2s+1)/n<2,$$

  where the second step is similar to that in (1.6), this time using $1/n^{2}\leq 1/n$ for $n\geq 1$, and the final step is due to our choice of $n$. From (1.7) and the definition of $A$, it follows that $s+1/n\in A$ and clearly $s+1/n>s$. However, this contradicts the fact $s$ is an upper bound for $A$, and so $s^{2}\not<2$.

Thus we have shown that $s=\sup A=\sqrt{2}$.

Let $\mathbb{R}^{\times}:=\mathbb{R}\setminus\{0\}$ be the punctured real line. Intuitively, we shall think of $\mathbb{R}^{\times}$ as a version of the number line which has a hole. Because of the presence of this hole, the completeness axiom fails for $\mathbb{R}^{\times}$ in the following sense: there exists a nonempty set $A\subseteq\mathbb{R}^{\times}$ which is bounded above in $\mathbb{R}^{\times}$ for which there is no least upper bound in $\mathbb{R}^{\times}$. Indeed, we can take $(-\infty,0)\subset\mathbb{R}^{\times}$.

This informal example involves decimal expansions of real numbers which you saw in IMU.

Suppose we wish to approximate $x=\sqrt{2}$ by rationals. We could consider the rational arising from the decimal expansion of $\sqrt{2}$:

If we choose our precision parameter $\varepsilon$ to be given by $\varepsilon:=0.01$, then $1.41$ approximates $\sqrt{2}$ with the required precision: $|1.41-\sqrt{2}|=0.0042\dots<0.01$. However, if we choose $\varepsilon:=0.0001$, then we need to use a longer decimal such as $1.4142$ to approximate $\sqrt{2}$ with the specified precision: our original choice $1.41$ no longer works since $|1.41-\sqrt{2}|=0.0042\dots>0.0001$, but the longer decimal $1.4142$ does work since $|1.4142-\sqrt{2}|=0.000013\dots<0.0001$.

Lemma 1.49 tells us that, no matter how tiny the precision parameter $\varepsilon$ is,⁴⁴ 4 For instance, we could take $\varepsilon:=10^{-10^{100}}$ which, if you think about it, is a really tiny number (if you try to write it down in the form $0.0000\dots 01$, then there would be nowhere near enough space for the number in the observable universe, even if each $0$ was just the size of an atom). we can always find a rational which approximates $\sqrt{2}$ with the required degree of precision. Intuitively, by travelling far enough down the decimal expansion, we can get as close as we like to $\sqrt{2}$.