Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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." –  neuguy 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 –  lyj Jan 17 '13 at 22:15

5 Answers 5

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 $s^2 \not> 3.$

share|improve this answer

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

share|improve this answer

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 $\sqrt{3}$.

share|improve this answer

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.

share|improve this answer

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.