The following exercises focus on the completeness of \(\mathbb R\). Their central theme is the least upper bound axiom: every nonempty subset of \(\mathbb R\) that is bounded above has a supremum in \(\mathbb R\).
This principle is what sets \(\mathbb R\) sharply apart from \(\mathbb Q\), and it underlies several cornerstone results of analysis, such as the existence of suprema and infima, the nested intervals theorem, and the convergence of bounded monotone sequences.
Exercise 1 — level ★☆☆☆☆
Let
\[ A=(0,1). \]
Find the supremum, infimum, maximum, and minimum of \(A\), explaining the role played by the completeness of \(\mathbb R\).
Answer
We have
\[ \sup A=1,\qquad \inf A=0. \]
The set \(A\) has neither a maximum nor a minimum.
Solution
The set \(A=(0,1)\) is nonempty and bounded above. For instance, \(1\) is an upper bound of \(A\), since every \(x\in A\) satisfies
\[ x\lt 1. \]
By the completeness axiom of \(\mathbb R\), the set \(A\) has a supremum.
Let us show that
\[ \sup A=1. \]
The number \(1\) is an upper bound of \(A\). Moreover, if \(M\lt 1\), we can pick a number \(x\) with
\[ M\lt x\lt 1. \]
Choosing in addition \(x\gt 0\), we obtain \(x\in A\) and \(x\gt M\). Hence no number smaller than \(1\) can be an upper bound of \(A\). Thus \(1\) is the least of all upper bounds, that is,
\[ \sup A=1. \]
An analogous argument shows that \(0\) is a lower bound of \(A\) and that no number greater than \(0\) is a lower bound. Indeed, if \(m\gt 0\), we may take \(x\in(0,m)\); then \(x\in A\) but \(x\lt m\). Therefore
\[ \inf A=0. \]
The set has no maximum, because \(1\notin A\), and no minimum either, because \(0\notin A\).
This example highlights a key distinction: the supremum and infimum may well exist even when the maximum and minimum do not.
Exercise 2 — level ★☆☆☆☆
Let
\[ A=\left\{1-\frac1n:n\in\mathbb N,\ n\ge 1\right\}. \]
Find the supremum, infimum, maximum, and minimum of \(A\).
Answer
We have
\[ \sup A=1,\qquad \inf A=0. \]
The set has minimum \(0\) but no maximum.
Solution
The elements of \(A\) are
\[ 0,\frac12,\frac23,\frac34,\ldots \]
Indeed, for \(n=1\) we get
\[ 1-\frac11=0. \]
For every \(n\ge 1\) we have
\[ 1-\frac1n\lt 1, \]
so \(1\) is an upper bound of \(A\).
We now show that it is the least upper bound. If \(M\lt 1\), then \(1-M\gt 0\). By the Archimedean property of the real numbers, there exists \(n\in\mathbb N\) with
\[ \frac1n\lt 1-M. \]
This inequality gives
\[ M\lt 1-\frac1n. \]
Hence \(M\) is not an upper bound of \(A\), and therefore
\[ \sup A=1. \]
Moreover, the first element of the set is \(0\), and all the others are greater than or equal to \(0\). Hence
\[ \inf A=0, \]
and \(0\) is in fact the minimum of \(A\).
The set has no maximum, since its supremum is \(1\) but \(1\notin A\).
Exercise 3 — level ★★☆☆☆
Let
\[ A=\{x\in\mathbb R:x^2\lt 2\}. \]
Find \(\sup A\) and explain why the completeness of \(\mathbb R\) is essential here.
Answer
We have
\[ \sup A=\sqrt2. \]
Solution
The set
\[ A=\{x\in\mathbb R:x^2\lt 2\} \]
consists of all real numbers whose square is less than \(2\). It contains both positive and negative numbers, and it is nonempty since, for example, \(1\in A\).
Furthermore, \(A\) is bounded above. Indeed, if \(x\ge 2\), then
\[ x^2\ge 4\gt 2, \]
so \(x\notin A\). Thus \(2\) is an upper bound of \(A\).
By the completeness of \(\mathbb R\), since \(A\) is nonempty and bounded above, the supremum
\[ \alpha=\sup A \]
exists.
Intuitively, \(A\) is made up of the numbers
\[ -\sqrt2\lt x\lt \sqrt2, \]
so its supremum is \(\sqrt2\).
The number \(\sqrt2\) is an upper bound of \(A\). Indeed, let \(x\in A\), so that \(x^2\lt 2\). If \(x\ge 0\), then \(x^2\lt 2\) yields \(x\lt\sqrt2\); if instead \(x\lt 0\), then certainly \(x\lt\sqrt2\). In either case \(x\le\sqrt2\).
Moreover, no number less than \(\sqrt2\) can be an upper bound. Indeed, if \(M\lt\sqrt2\), we may choose a real number \(x\) with
\[ \max\{M,0\}\lt x\lt\sqrt2. \]
Then \(x\gt 0\) and \(x\lt\sqrt2\), so
\[ x^2\lt 2. \]
Hence \(x\in A\) and, since \(x\gt M\), the number \(M\) fails to be an upper bound of \(A\).
We therefore conclude that
\[ \sup A=\sqrt2. \]
Completeness is essential because it guarantees that the supremum exists as a real number. Within the rationals, an analogous set would have a “gap” exactly at \(\sqrt2\), which is not rational.
Exercise 4 — level ★★☆☆☆
Let
\[ B=\{q\in\mathbb Q:q^2\lt 2\}. \]
Regarding \(B\) as a subset of \(\mathbb Q\), explain why it has no supremum in \(\mathbb Q\).
Answer
The set \(B\) has no supremum in \(\mathbb Q\).
Solution
The set
\[ B=\{q\in\mathbb Q:q^2\lt 2\} \]
is nonempty, since \(1\in B\), and it is bounded above in \(\mathbb Q\), since \(2\) is an upper bound.
Suppose, for contradiction, that \(B\) has a supremum in \(\mathbb Q\). Let
\[ s=\sup_{\mathbb Q} B. \]
Since \(1\in B\), we have \(s\ge 1\), so in particular \(s\gt 0\).
We now show that neither \(s^2\lt 2\) nor \(s^2\gt 2\) is possible.
First suppose that \(s^2\lt 2\). Choose a rational number \(h\gt 0\) so small that
\[ h\lt 1 \qquad \text{and} \qquad h\lt \frac{2-s^2}{2s+1}. \]
Then \(h^2\lt h\), and therefore
\[ (s+h)^2=s^2+2sh+h^2\lt s^2+2sh+h=s^2+h(2s+1)\lt 2. \]
Since \(s+h\in\mathbb Q\), this gives \(s+h\in B\), with \(s+h\gt s\). This contradicts the fact that \(s\) is an upper bound of \(B\).
Now suppose that \(s^2\gt 2\). Choose a rational number \(h\gt 0\) so small that \(h\lt s\) and
\[ h\lt \frac{s^2-2}{2s}. \]
Then
\[ (s-h)^2=s^2-2sh+h^2\gt s^2-2sh\gt 2. \]
We claim that \(s-h\) is still an upper bound of \(B\). Indeed, if \(q\in B\) and \(q\ge s-h\), then, since \(s-h\gt 0\), we would have
\[ q^2\ge (s-h)^2\gt 2, \]
contradicting \(q\in B\). Hence every \(q\in B\) satisfies \(q\lt s-h\), so \(s-h\) is an upper bound of \(B\).
But \(s-h\lt s\), contradicting the fact that \(s\) is the least upper bound.
Therefore neither \(s^2\lt 2\) nor \(s^2\gt 2\) is possible. We would have to have
\[ s^2=2. \]
This is impossible for \(s\in\mathbb Q\), because no rational number has square equal to \(2\).
Thus \(B\) is bounded above in \(\mathbb Q\), but it has no supremum in \(\mathbb Q\). This is one of the clearest ways in which \(\mathbb Q\) fails to be complete.
Exercise 5 — level ★★☆☆☆
Using the least upper bound axiom, prove that every nonempty subset of \(\mathbb R\) that is bounded below has an infimum.
Answer
Every nonempty subset of \(\mathbb R\) that is bounded below has an infimum.
Solution
Let \(A\subseteq\mathbb R\) be nonempty and bounded below.
Consider the reflected set
\[ -A=\{-x:x\in A\}. \]
Since \(A\) is nonempty, so is \(-A\).
Since \(A\) is bounded below, there is some \(m\in\mathbb R\) with
\[ m\le x \]
for every \(x\in A\). Multiplying by \(-1\) reverses the inequality:
\[ -x\le -m \]
for every \(x\in A\). Hence \(-m\) is an upper bound of \(-A\).
By the least upper bound axiom, the set \(-A\) — being nonempty and bounded above — has a supremum. Set
\[ \alpha=\sup(-A). \]
We claim that
\[ \inf A=-\alpha. \]
Since \(\alpha\) is an upper bound of \(-A\), for every \(x\in A\) we have
\[ -x\le \alpha. \]
Multiplying by \(-1\) gives
\[ x\ge -\alpha, \]
so \(-\alpha\) is a lower bound of \(A\).
Moreover, since \(\alpha\) is the least upper bound of \(-A\), the number \(-\alpha\) is the greatest lower bound of \(A\). Therefore
\[ \inf A=-\sup(-A). \]
Exercise 6 — level ★★☆☆☆
Prove that if \(A\subseteq\mathbb R\) is nonempty and bounded above, then for every \(\varepsilon\gt 0\) there exists \(a\in A\) such that
\[ \sup A-\varepsilon\lt a\le \sup A. \]
Answer
For every \(\varepsilon\gt 0\) there are elements of \(A\) arbitrarily close to the supremum from below.
Solution
Let
\[ \alpha=\sup A. \]
Since \(\alpha\) is an upper bound of \(A\), every \(a\in A\) satisfies
\[ a\le \alpha. \]
We must show that at least one element \(a\in A\) satisfies
\[ \alpha-\varepsilon\lt a. \]
Suppose, for contradiction, that this fails. Then every \(a\in A\) would satisfy
\[ a\le \alpha-\varepsilon, \]
which would make \(\alpha-\varepsilon\) an upper bound of \(A\).
But
\[ \alpha-\varepsilon\lt \alpha, \]
contradicting the fact that \(\alpha\) is the least upper bound.
Hence there must exist \(a\in A\) with
\[ \alpha-\varepsilon\lt a. \]
Since \(\alpha\) is also an upper bound, we have \(a\le \alpha\) as well, so
\[ \alpha-\varepsilon\lt a\le \alpha. \]
Exercise 7 — level ★★★☆☆
Using the completeness of \(\mathbb R\), prove that every increasing sequence that is bounded above converges.
Answer
Every increasing sequence of real numbers that is bounded above converges to its supremum.
Solution
Let \((a_n)\) be an increasing sequence that is bounded above. Consider the set of its values:
\[ A=\{a_n:n\in\mathbb N\}. \]
The set \(A\) is nonempty and bounded above, since the sequence is bounded above.
By the completeness of \(\mathbb R\), the supremum
\[ \alpha=\sup A \]
exists. We show that
\[ a_n\to \alpha. \]
Let \(\varepsilon\gt 0\). By the characterization of the supremum, there is an element \(a_N\in A\) with
\[ \alpha-\varepsilon\lt a_N\le \alpha. \]
Since the sequence is increasing, for every \(n\ge N\) we have
\[ a_N\le a_n. \]
Moreover, as \(\alpha\) is an upper bound of \(A\), we have
\[ a_n\le \alpha. \]
Hence, for every \(n\ge N\),
\[ \alpha-\varepsilon\lt a_n\le \alpha, \]
which implies
\[ 0\le \alpha-a_n\lt \varepsilon. \]
Therefore
\[ |a_n-\alpha|\lt \varepsilon \]
for all \(n\ge N\), and so \(a_n\to\alpha\).
Exercise 8 — level ★★★☆☆
Using the completeness of \(\mathbb R\), prove that every decreasing sequence that is bounded below converges.
Answer
Every decreasing sequence that is bounded below converges to its infimum.
Solution
Let \((a_n)\) be a decreasing sequence that is bounded below. Consider the set
\[ A=\{a_n:n\in\mathbb N\}. \]
The set \(A\) is nonempty and bounded below.
By Exercise 5, which follows from the least upper bound axiom, \(A\) has an infimum. Set
\[ \alpha=\inf A. \]
We want to show that
\[ a_n\to\alpha. \]
Let \(\varepsilon\gt 0\). By the characterization of the infimum, there is an element \(a_N\in A\) with
\[ \alpha\le a_N\lt \alpha+\varepsilon. \]
Since the sequence is decreasing, for every \(n\ge N\) we have
\[ a_n\le a_N. \]
Moreover, since \(\alpha\) is a lower bound of \(A\), we have
\[ \alpha\le a_n \]
for every \(n\). Hence, for every \(n\ge N\),
\[ \alpha\le a_n\lt \alpha+\varepsilon. \]
Therefore
\[ |a_n-\alpha|\lt \varepsilon \]
for all \(n\ge N\), and so \(a_n\to\alpha\).
Exercise 9 — level ★★★☆☆
Prove that the sequence
\[ a_n=1-\frac1n \]
converges, using the monotone convergence theorem.
Answer
The sequence converges, and
\[ \lim_{n\to+\infty}\left(1-\frac1n\right)=1. \]
Solution
Consider
\[ a_n=1-\frac1n. \]
We first show that \((a_n)\) is increasing. Computing:
\[ a_{n+1}-a_n= \left(1-\frac1{n+1}\right)-\left(1-\frac1n\right) = \frac1n-\frac1{n+1}. \]
Since
\[ \frac1n-\frac1{n+1}=\frac1{n(n+1)}\gt 0, \]
the sequence is increasing.
Moreover, for every \(n\ge 1\),
\[ a_n=1-\frac1n\le 1. \]
Thus \((a_n)\) is increasing and bounded above.
By the monotone convergence theorem, which follows from the completeness of \(\mathbb R\), the sequence converges.
Since the set of its values has supremum equal to \(1\), the limit is
\[ \lim_{n\to+\infty}a_n=1. \]
Exercise 10 — level ★★★☆☆
Let \([a_n,b_n]\) be nonempty closed bounded intervals such that
\[ [a_1,b_1]\supseteq [a_2,b_2]\supseteq [a_3,b_3]\supseteq\cdots. \]
Using the completeness of \(\mathbb R\), prove that
\[ \bigcap_{n=1}^{+\infty}[a_n,b_n]\ne\varnothing. \]
Answer
A nested (decreasing) sequence of nonempty closed bounded intervals has nonempty intersection.
Solution
Since the intervals are nested, each left endpoint \(a_n\) is less than or equal to every right endpoint \(b_m\). Indeed, for any indices \(m\) and \(n\), the nesting always gives
\[ a_n\le b_m. \]
Consider the set of left endpoints:
\[ A=\{a_n:n\in\mathbb N\}. \]
This set is nonempty, and it is bounded above, since every \(b_m\) is an upper bound of \(A\).
By the completeness of \(\mathbb R\), the supremum
\[ \alpha=\sup A \]
exists.
Since \(\alpha\) is an upper bound of \(A\), we have
\[ a_n\le \alpha \]
for every \(n\).
On the other hand, each \(b_n\) is an upper bound of \(A\). As \(\alpha\) is the least upper bound, we must have
\[ \alpha\le b_n \]
for every \(n\).
We have thus shown that, for every \(n\),
\[ a_n\le \alpha\le b_n. \]
This means that
\[ \alpha\in[a_n,b_n] \]
for every \(n\). Hence
\[ \alpha\in\bigcap_{n=1}^{+\infty}[a_n,b_n]. \]
Therefore the intersection is nonempty.
Exercise 11 — level ★★★☆☆
Show that the conclusion of the previous exercise can fail in \(\mathbb Q\).
Answer
In \(\mathbb Q\), a nested sequence of closed rational intervals can have empty intersection.
Solution
Consider rational intervals that approximate \(\sqrt2\) without containing it as a rational number.
For instance, choose two rational sequences \((a_n)\) and \((b_n)\) such that
\[ a_n\lt \sqrt2\lt b_n \]
and
\[ b_n-a_n\to 0. \]
Consider the sets
\[ I_n=[a_n,b_n]\cap\mathbb Q. \]
We can arrange for them to be nested:
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots. \]
In \(\mathbb R\), the intersection of the real intervals \([a_n,b_n]\) would be the singleton
\[ \{\sqrt2\}. \]
However,
\[ \sqrt2\notin\mathbb Q. \]
So, working inside \(\mathbb Q\), no rational number is common to all the intervals:
\[ \bigcap_{n=1}^{+\infty}I_n=\varnothing. \]
This shows that the nested intervals theorem relies on the completeness of \(\mathbb R\) and fails, in general, in \(\mathbb Q\).
Exercise 12 — level ★★★☆☆
Let \(A\subseteq\mathbb R\) be nonempty and bounded above. Prove that there exists a sequence \((a_n)\) of elements of \(A\) such that
\[ a_n\to \sup A. \]
Answer
Any supremum can be approximated by a sequence of elements of the set.
Solution
Set
\[ \alpha=\sup A. \]
For each \(n\in\mathbb N\), apply the characterization of the supremum with
\[ \varepsilon=\frac1n. \]
This yields an element \(a_n\in A\) such that
\[ \alpha-\frac1n\lt a_n\le \alpha. \]
We thus obtain a sequence \((a_n)\) of elements of \(A\).
From the double inequality
\[ \alpha-\frac1n\lt a_n\le \alpha \]
we get
\[ 0\le \alpha-a_n\lt \frac1n. \]
Since
\[ \frac1n\to 0, \]
the squeeze theorem gives
\[ \alpha-a_n\to 0. \]
Hence
\[ a_n\to\alpha=\sup A. \]
Exercise 13 — level ★★★★☆
Prove that if \(A,B\subseteq\mathbb R\) are nonempty, \(A\) is bounded above, and
\[ a\le b \]
for every \(a\in A\) and every \(b\in B\), then
\[ \sup A\le \inf B. \]
Answer
If every element of \(A\) lies to the left of every element of \(B\), then \(\sup A\le\inf B\).
Solution
By hypothesis,
\[ a\le b \]
for every \(a\in A\) and every \(b\in B\).
Fix any \(b\in B\). Then \(b\) is an upper bound of \(A\), since every element of \(A\) is less than or equal to \(b\).
Since \(\sup A\) is the least upper bound of \(A\), we have
\[ \sup A\le b \]
for every \(b\in B\).
Hence \(\sup A\) is a lower bound of \(B\).
Since \(\inf B\) is the greatest lower bound of \(B\) and \(\sup A\) is a lower bound of \(B\), it follows that
\[ \sup A\le \inf B. \]
Exercise 14 — level ★★★★☆
Let \(A,B\subseteq\mathbb R\) be nonempty and bounded above. Prove that
\[ \sup(A+B)=\sup A+\sup B, \]
where
\[ A+B=\{a+b:a\in A,\ b\in B\}. \]
Answer
The supremum of the sumset is the sum of the suprema.
Solution
Set
\[ \alpha=\sup A,\qquad \beta=\sup B. \]
For every \(a\in A\) we have \(a\le\alpha\), and for every \(b\in B\) we have \(b\le\beta\). Hence
\[ a+b\le \alpha+\beta. \]
Thus \(\alpha+\beta\) is an upper bound of \(A+B\), so
\[ \sup(A+B)\le \alpha+\beta. \]
We now prove the reverse inequality. Let \(\varepsilon\gt 0\). By the characterization of the supremum, there exist \(a_\varepsilon\in A\) and \(b_\varepsilon\in B\) such that
\[ \alpha-\frac{\varepsilon}{2}\lt a_\varepsilon\le\alpha \]
and
\[ \beta-\frac{\varepsilon}{2}\lt b_\varepsilon\le\beta. \]
Adding these term by term gives
\[ \alpha+\beta-\varepsilon \lt a_\varepsilon+b_\varepsilon \le \alpha+\beta. \]
Since \(a_\varepsilon+b_\varepsilon\in A+B\), no number less than \(\alpha+\beta\) can be an upper bound of \(A+B\).
Therefore
\[ \sup(A+B)=\alpha+\beta=\sup A+\sup B. \]
Exercise 15 — level ★★★★☆
Let \(A\subseteq\mathbb R\) be nonempty and bounded above, and let \(\lambda\gt 0\). Prove that
\[ \sup(\lambda A)=\lambda\sup A, \]
where
\[ \lambda A=\{\lambda a:a\in A\}. \]
Answer
For \(\lambda\gt 0\), the supremum behaves well under multiplication:
\[ \sup(\lambda A)=\lambda\sup A. \]
Solution
Set
\[ \alpha=\sup A. \]
For every \(a\in A\) we have
\[ a\le\alpha. \]
Since \(\lambda\gt 0\), multiplying by \(\lambda\) preserves the direction of the inequality:
\[ \lambda a\le\lambda\alpha. \]
Hence \(\lambda\alpha\) is an upper bound of \(\lambda A\).
We must show that it is the least one. Let \(\varepsilon\gt 0\). By the characterization of the supremum, there exists \(a_\varepsilon\in A\) such that
\[ \alpha-\frac{\varepsilon}{\lambda}\lt a_\varepsilon\le\alpha. \]
Multiplying by \(\lambda\) gives
\[ \lambda\alpha-\varepsilon \lt \lambda a_\varepsilon \le \lambda\alpha. \]
Since \(\lambda a_\varepsilon\in\lambda A\), every number less than \(\lambda\alpha\) fails to be an upper bound of \(\lambda A\).
Therefore
\[ \sup(\lambda A)=\lambda\alpha=\lambda\sup A. \]
Exercise 16 — level ★★★★☆
Prove that the Archimedean property follows from the completeness of \(\mathbb R\): for every \(x\in\mathbb R\) there exists \(n\in\mathbb N\) such that
\[ n\gt x. \]
Answer
The set \(\mathbb N\) is not bounded above in \(\mathbb R\).
Solution
Suppose, for contradiction, that the property is false. Then there would be a real number \(x\) with
\[ n\le x \]
for every \(n\in\mathbb N\). In other words, \(\mathbb N\) would be bounded above.
Since \(\mathbb N\) is nonempty, the completeness of \(\mathbb R\) would give a supremum
\[ \alpha=\sup\mathbb N. \]
As \(\alpha\) is the least upper bound, the number \(\alpha-1\) cannot be an upper bound of \(\mathbb N\).
Hence there exists \(n\in\mathbb N\) such that
\[ n\gt \alpha-1. \]
Adding \(1\) gives
\[ n+1\gt \alpha. \]
But \(n+1\in\mathbb N\), contradicting the fact that \(\alpha\) is an upper bound of \(\mathbb N\).
Therefore \(\mathbb N\) is not bounded above, and so for every \(x\in\mathbb R\) there exists \(n\in\mathbb N\) such that
\[ n\gt x. \]
Exercise 17 — level ★★★★☆
Using the Archimedean property, prove that for every \(\varepsilon\gt 0\) there exists \(n\in\mathbb N\) such that
\[ \frac1n\lt\varepsilon. \]
Answer
The sequence \(\displaystyle \frac1n\) converges to \(0\).
Solution
Let \(\varepsilon\gt 0\). Then
\[ \frac1\varepsilon\gt 0. \]
By the Archimedean property, there exists \(n\in\mathbb N\) such that
\[ n\gt \frac1\varepsilon. \]
Since both sides are positive, taking reciprocals reverses the inequality and yields
\[ \frac1n\lt \varepsilon. \]
This proves that, for every \(\varepsilon\gt 0\), there exists \(n\) with \(\frac1n\lt\varepsilon\).
Moreover, if \(m\ge n\), then
\[ 0\lt \frac1m\le \frac1n\lt\varepsilon. \]
Hence
\[ \frac1n\to 0. \]
Exercise 18 — level ★★★★★
Prove that between any two real numbers \(a\lt b\) there is at least one rational number.
Answer
The rational numbers are dense in \(\mathbb R\).
Solution
Let \(a,b\in\mathbb R\) with
\[ a\lt b. \]
Then
\[ b-a\gt 0. \]
By the Archimedean property, there exists \(n\in\mathbb N\) such that
\[ n(b-a)\gt 1. \]
Equivalently,
\[ nb-na\gt 1. \]
The gap between the real numbers \(na\) and \(nb\) is therefore greater than \(1\). Hence there is an integer \(m\in\mathbb Z\) such that
\[ na\lt m\lt nb. \]
Dividing by \(n\gt 0\) gives
\[ a\lt \frac mn\lt b. \]
The number
\[ q=\frac mn \]
is rational and lies in the interval \((a,b)\).
Thus between any two distinct real numbers there is always at least one rational number.
Exercise 19 — level ★★★★★
Prove that between any two real numbers \(a\lt b\) there is at least one irrational number.
Answer
The irrational numbers are dense in \(\mathbb R\).
Solution
Let \(a,b\in\mathbb R\) with
\[ a\lt b. \]
We seek an irrational number lying between \(a\) and \(b\).
Consider the numbers
\[ a-\sqrt2 \quad \text{and} \quad b-\sqrt2. \]
Since
\[ a-\sqrt2\lt b-\sqrt2, \]
by the density of the rationals in \(\mathbb R\) there is a rational \(q\in\mathbb Q\) such that
\[ a-\sqrt2\lt q\lt b-\sqrt2. \]
Adding \(\sqrt2\) to all three terms gives
\[ a\lt q+\sqrt2\lt b. \]
The number \(q+\sqrt2\) is irrational. Indeed, were it rational, then
\[ \sqrt2=(q+\sqrt2)-q \]
would be a difference of two rationals and hence rational — a contradiction.
Therefore there is an irrational number lying between \(a\) and \(b\).
Exercise 20 — level ★★★★★
Let \((a_n)\) be a Cauchy sequence in \(\mathbb R\). Explain why the completeness of \(\mathbb R\) implies that \((a_n)\) converges.
Answer
Every Cauchy sequence of real numbers converges to a real number.
Solution
A sequence \((a_n)\) is a Cauchy sequence if, for every \(\varepsilon\gt 0\), there exists \(N\in\mathbb N\) such that, for all \(m,n\ge N\),
\[ |a_n-a_m|\lt\varepsilon. \]
We first observe that every Cauchy sequence is bounded. Taking \(\varepsilon=1\), there exists \(N\in\mathbb N\) such that, for every \(n\ge N\),
\[ |a_n-a_N|\lt 1. \]
Hence, for every \(n\ge N\),
\[ a_n\in(a_N-1,a_N+1). \]
The finitely many terms \(a_1,\ldots,a_{N-1}\) are also bounded, so the whole sequence \((a_n)\) is bounded.
By the Bolzano-Weierstrass theorem, which follows from the completeness of \(\mathbb R\), the bounded sequence \((a_n)\) has a convergent subsequence. Thus there exist a real number \(L\) and a subsequence \((a_{n_k})\) such that
\[ a_{n_k}\to L. \]
We now prove that the whole sequence converges to \(L\). Let \(\varepsilon\gt 0\). Since \((a_n)\) is Cauchy, there exists \(N_1\in\mathbb N\) such that, for all \(m,n\ge N_1\),
\[ |a_n-a_m|\lt \frac{\varepsilon}{2}. \]
Since \(a_{n_k}\to L\), there exists \(k\) such that \(n_k\ge N_1\) and
\[ |a_{n_k}-L|\lt \frac{\varepsilon}{2}. \]
Then, for every \(n\ge N_1\), we have
\[ |a_n-L|\le |a_n-a_{n_k}|+|a_{n_k}-L|\lt \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \]
Therefore
\[ a_n\to L. \]
Thus every Cauchy sequence of real numbers converges to a real number. This is one of the fundamental equivalent forms of the completeness of \(\mathbb R\).