A bounded set of real numbers is given. Your task is to find the
supremum. (See below for definitions and
theoretical background.) Enter a decimal number or fraction into the
form field and press the `Enter`

key. If your choice is not
the supremum, a hint is printed, giving either (i) a *larger
element of the set* if the choice is too small, or
(ii) a *smaller upper bound of the set* if the choice is
is too large.

A bonus is awarded for deducing the supremum in a small number of attempts. More challenging examples have a higher bonus value. The current bonus and difficulty level are shown. Reloading the page resets your score.

Determine the supremum of the set

Answers may be entered as decimals or fractions, e.g., $1.25$ or $5/4$.

$\sup S = $

A non-empty set $S$ of real numbers is *bounded above* if
there exists a real number $b$ such that $x \leq b$ for
every $x$ in $S$. Geometrically, a set $S$ is bounded
above if some real number $b$ lies to the right of $S$ on
the number line.

If $b$ is an upper bound of $S$, and if $b \lt b'$, then $b'$ is also an upper bound of $S$ by transitivity of inequality: $x \leq b \lt b'$ for every $x$ in $S$. Geometrically, if $b$ lies to the right of $S$ and $b'$ lies to the right of $b$, then $b'$ lies to the right of $S$. In particular, upper bounds are never unique.

A real number $\beta$ is a *least upper bound*
or *supremum* of $S$ if (i) $\beta$ is an upper
bound of $S$, and (ii) we have $\beta \leq b$ for every
upper bound $b$ of $S$.

At most one real number is the supremum of $S$: If
$\beta$ and $\beta'$ satisfy conditions (i) and (ii), then
$\beta \leq \beta'$ because $\beta'$ is an upper bound and
$\beta$ satisfies (ii). Reversing roles, $\beta' \leq \beta$.
Geometrically, the supremum of $S$ is the *leftmost point lying on
or to the right of $S$*.

The *completeness property* of the real number system says that
if $S$ is a non-empty set of real numbers and $S$ is bounded
above, then $S$ has a (unique) supremum.

Conceptually, the supremum of a bounded set $S$ plays the role of the largest element, even if $S$ has no largest element. For example, the set $S = (-\infty, 0)$ of negative real numbers has no largest element, but does have a supremum, namely $0$.

To see how completeness might fail to hold, consider the set $S$
of negative real numbers in the “universe”
of *non-zero* real numbers. No negative number $x$ is an
upper bound of $S$, because $x \lt x/2 \lt 0$. That is,
$x/2 \in S$, so $x$ does not lie entirely to the right
of $S$. Conversely, every positive real
number $b$ *is* an upper bound of $S$, since
$x \lt 0 \lt b$ for all $x$ in $S$. However,
$0 \lt b/2 \lt b$, so $b$ is not the *smallest* upper
bound of $S$. That is, no non-zero real number is a supremum
of $S$.

For a more “natural” example of non-completeness, consider
the “universe” of rational numbers and the bounded set
$S = \{x : x^{2} \lt 2\}$. There is no *rational*
number $\beta$ that (i) lies on or to the right of $S$
and (ii) is the leftmost number with
property (i). Intuitively, the rational number system has a
“gap” at $\sqrt{2}$, just as the set of non-zero real
numbers has a gap at $0$. More precisely, an upper bound
of $S$ must be greater than $0$, since $1 \in S$. It
therefore suffices to prove that no positive rational number is a
supremum of $S$.

If $b = p/q$ is a positive rational number, then either $2 \lt b^{2}$
or $b^{2} \lt 2$, since $2$ has no rational square root. Further,
$b$ is an upper bound of $S$ if and only if $x \lt b$ for
all rational $x$ with $x^{2} \lt 2$, if and only if
$2 \lt b^{2}$. One direction is easy: If $b$ is *not* an
upper bound of $S$, there exists an $x$ in $S$ such
that $0 \lt b \lt x$. Squaring gives $0 \lt b^{2} \lt x^{2} \lt 2$.
Contrapositively, if $2 \lt b^{2}$ (and $b$ is a positive
rational), then $b$ is an upper bound of $S$. The inverse
(if $b^{2} \lt 2$, then $b$ is not an upper bound of $S$) is
shown below.

If $2 \lt b^{2}$, then \[ b' = \frac{1}{2}\left(b + \frac{2}{b}\right) = \frac{b^{2} + 2}{2b} = \frac{1}{2}\left(\frac{p}{q} + \frac{2q}{p}\right) = \frac{p^{2} + 2q^{2}}{2pq} \] is also positive and rational. Algebra gives \begin{gather*} b' - b = \frac{1}{2}\left(b + \frac{2}{b}\right) - b = \frac{1}{2}\left(-b + \frac{2}{b}\right) = \frac{2 - b^{2}}{2b} \lt 0, \\ (b')^{2} - 2 = \left(\frac{b^{2} + 2}{2b}\right)^{2} - 2 = \frac{b^{4} + 4b^{2} + 4}{4b^{2}} - \frac{8b^{2}}{4b^{2}} = \frac{b^{4} - 4b^{2} + 4}{4b^{2}} = \left(\frac{b^{2} - 2}{2b}\right)^{2} \gt 0. \end{gather*} That is, $b' \lt b$ (by the first line) and $2 \lt (b')^{2}$ (by the second).

If instead $b^{2} \lt 2$, then $b_{0} = 2/b$ is a positive rational number whose square is larger than $2$. The argument of the preceding paragraph constructs a positive rational number $b_{0}'$ smaller than $b_{0}$ whose square is larger than $2$. Setting $b' = 2/b_{0}'$, we have a positive rational number $b'$ satisfying $b \lt b'$ and $(b')^{2} \lt 2$. In words, a positive rational number $b$ satisfying $b^{2} \lt 2$ is not an upper bound of $S$.

Combining these observations, if $b$ is a positive rational
number and $b^{2} \lt 2$, then $b$ is not an upper bound
of $S$, while if $2 \lt b^{2}$, then some smaller positive
rational is also an upper bound of $S$; that is, $b$ is not
the *smallest* upper bound of $S$. It follows that
$S = \{x : x^{2} \lt 2\}$ has no rational supremum.