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I am working on the following short question

Denote by $\boldsymbol{D} $ the set of all finite decimals. Show that sup$\left \{ a \in \boldsymbol{D} | a^2 \leq 3 \right \}=\sqrt{3}$

I think I've got it, and I understand the question. I am just not sure my answer is complete enough.

I use the inequality $a^2 \leq 3$.

$$- \sqrt{3} \leq a \leq \sqrt{3}$$

So any $M \geq \sqrt{3}$ is an upper bound. So $a \leq \sqrt{3} \leq M$, showing that sup$\left \{ a \in \boldsymbol{D} | a^2 \leq 3 \right \}=\sqrt{3}$. Does this suffice? It is for an assignment, so should I go into the definition of a supremum? Why isn't the supremum 3?

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Unfortunately, that's not enough. This merely shows that $\sqrt{3}$ is an upper bound. You want to show something like "if $b < \sqrt{3}$ then $b$ is not an upper bound." – Gyu Eun Lee Jan 17 '13 at 20:32
In other words, if $b<\sqrt{3}$ then there is an $a\in D$ such that $b<a$. – Thomas Andrews Jan 17 '13 at 20:33
you should clarify whether or not $\sqrt{3}$ is assumed to exist – cats Jan 17 '13 at 22:15
up vote 4 down vote accepted

I'm assuming that your question actually wants you to prove the existence of $\sqrt{3}$ as a limit of finite decimals, in which case none of the answers are correct, because they all work under the assumption that $\sqrt{3}$ exists.

I'll assume that $\mathbb{R}$ has the least upper-bound property (as well as the Archimedean property, ordering, etc.), which is usually what you're given in these types of problems. First, $3$ is obviously an upper bound for $S = \{a\in D\mid a^2 \le 3\},$ so $S$ has a least upper bound, say $s.$ Suppose that $s^2 > 3.$ Then let $\epsilon = \frac{1}{n} < \frac{s^2-3}{2\cdot s};$ the existence of $n$ is guaranteed by the Archimedean property. We then have $(s-\epsilon)^2 > s^2 - 2s\epsilon > s^2 - (s^2 - 3) = 3,$ so $(s-\epsilon)$ is an upper bound for $S$ as well, which is a contradiction since $s$ is the least upper bound of $S.$

Now, suppose $s^2 < 3.$ Let $\epsilon < \min\left(1, \frac{3-s^2}{2s+1}\right).$ It's not hard to show then that $(s+\epsilon)^2 < 3.$ The point now is that there exists a rational number $r = \frac{p}{q}$ in the interval $(s, s+\epsilon).$ This is an exercise that you should have already proven (it uses the Archimedean property), so I will just cite it here. Note that there exists $n$ such that $10^{-n} < r - s.$

Since we can perform (approximate) division (for a finite number of digits - you should verify division actually works in the reals as well, identifying an infinite string of 9's after the decimal point with 1), perform the division $p$ divided by $q$ up to the $n$-th digit after the decimal point; call this truncated decimal $r'$. What follows the $n$-th decimal is strictly less than $10^{-n}$ (another thing to prove). Hence, $s < r' \le r < s+\epsilon,$ from which it follows that $r' \in S.$ Then, $s$ is not an upper bound of $S$ since $r' \in S,$ contradiction.

By the Trichotomy Property of the reals, $s^2 = 3$ since $s^2 \not< 3$ and $s^2 \not> 3.$

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You have showed that $\sqrt{3}$ is an upper bound of the set $\{a\in \mathbf{D}\mid a^2\leq 3\}$, but not that it is the least upper bound. Suppose that $M$ is another upper bound of the set, i.e. $a\leq M$ for every $a\in\mathbf{D}$ with $a^2\leq 3$.

Assume for contradiction that $M<\sqrt{3}$. But there exists an $a\in \mathbf{D}$ in between $M$ and $\sqrt{3}$, i.e. $M<a<\sqrt{3}$, which shows that $M$ is not an upper bound, and hence the assumption that $M<\sqrt{3}$ is false. We conclude that $\sqrt{3}\leq M$ for any upper bound which exactly means that $\sqrt{3}$ is the supremum.

Note that actually $\sup\{a\in\mathbb{R}\mid a^2\leq 3\}=\sqrt{3}$.

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As others have said, the supremum isn't $3$ since we're looking at the set of all finite decimals such that $a^2\leq3$. Taking the square root of both sides says that $a\leq\sqrt{3}$. I would approach this in the following way:

Let $\sqrt{3}=1.a_1a_2\ldots$ and $A=\{a\in \mathbf{D}:a^2\leq3\}$. Let $(\sqrt{3})_n=1.a_1a_2\ldots a_n$. Note that for all $n$, $$(\sqrt{3})_n<\sqrt{3},$$ thus $(\sqrt{3})_n\in A$. Now, observe that $$\sqrt{3}-(\sqrt{3})_n<10^{-n+1}.$$ By the Archimedean property, for any $\varepsilon>0$, we can make the right hand side as small as we want, hence we can make $(\sqrt{3})_n$ as close to $\sqrt{3}$ as we want. This shows explicitly that the least upper bound has to be $\sqrt{3}$.

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The supremum isn't $3$, because the supremum is the smallest upper bound and I'm sure you could find another $b < 3$ for which $b^2 \leq 3$ holds.

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A contradiction argument works well here: if $S$ is the supremum and $S < \sqrt{3}$, then you can find a finite decimal $d$ that satisfies $S < d < \sqrt{3}$, which contradicts $S$ being an upper bound of the set you have there.

In general, to be a supremum you must be:

  • An upper bound, and furthermore,
  • the minimal upper bound

You were fine on the first point, but not on the second.

To answer your last question of why the supremum is not 3 - it's because there exists a lower upper bound ($\sqrt{3}, 2, 2.1$, etc.)

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