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.