Real Analysis

Rational Numbers

The Set of Rational Numbers \(\mathbb{Q}\)

Construction of \(\mathbb{Q}\)

Assume \(\mathbb{Z}\), the integers, their arithmetic and order.

An equivalent class is a set whose members satisfy an equivalent relation. For example, \((1, 3) \sim (2, 6)\) are equivalent ordered pairs from the class “\(\frac{1}{3}\)”.

Informally, \(\mathbb{Q}\), the rational numbers, is the set of all such equivalent classes. Formally, \(\mathbb{Q} = \left \{\frac{m}{n} : m, n \in \mathbb{Z}, n \neq 0 \right \}\), where \(\frac{m}{n}\) is an equivalent class of ordered pairs \((m, n)\) with relation \((p, q) \sim (m, n)\) if \(pn = qm\) and \(q, n \neq 0\).

Check that \(\sim\) is an equivalent relation:

1. Reflexivity: Since \(mn = nm\) for all \(m, n \in \mathbb{Z}\) and \(n \neq 0\), \((m, n) \sim (m, n)\) is true.
2. Symmetry: Assume \((p, q) \sim (m, n)\) where \(p, q, m, n \in \mathbb{Z}\) and \(q, n \neq 0\). We have \(pn = qm\), which is the same as \(mq = np\), so \((m, n) \sim (p, q)\). Thus, \((p, q) \sim (m, n) \implies (m, n) \sim (p, q)\).
3. Transitivity: Assume \((a, b) \sim (c, d)\) and \((c, d) \sim (e, f)\) where \(a, b, c, d, e, f \in \mathbb{Z}\) and \(b, d, f \neq 0\). We have \(ad = bc\) since \((a, b) \sim (c, d)\) and \(cf = de\) since \((c, d) \sim (e, f)\). We also have \[\begin{align} & ad = bc \\ \implies & adf = bcf & \text{(since } f \neq 0 \text{)} \\ \implies & adf = bde & \text{(since } cf = de \text{)} \\ \implies & af = be & \text{(by cancellation law on } \mathbb{Z} \text{)} \\ \implies & (a, b) \sim (e, f) \end{align}\] Thus, \((a, b) \sim (c, d)\) and \((c, d) \sim (e, f) \implies (a, b) \sim (e, f)\).

Properties of \(\mathbb{Q}\)

Addition

The chosen definition of addition should not depend on the representative chosen, i.e. it should be well-defined. The following definition is not well-defined. \[ \frac{a}{b} + \frac{c}{d} = \frac{a + c}{b + d} \] Check: \(\frac{1}{2} + \frac{2}{3} = \frac{3}{5}\) but \(\frac{2}{4} + \frac{2}{3} = \frac{4}{7}\).

A well-defined but useless definition is as follows. \[ \frac{a}{b} + \frac{c}{d} = \frac{0}{1} \]

The correct definition is as follows. \[ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \] To show that the definition is well-defined, we must show that if \((a, b) \sim (a', b')\) and \((c, d) \sim (c', d')\) then \((ad + bc, bd) \sim (a'd' + b'c', b'd')\).

Proof. Assume \((a, b) \sim (a', b')\) and \((c, d) \sim (c', d')\). So, \(ab' = ba'\) and \(cd' = dc'\). Since \(ab' = ba'\), \(ab'dd' = ba'dd'\), which is the same as \(adb'd' = bda'd'\). Since \(cd' = dc'\), \(cd'bb' = dc'bb'\), which is the same as \(bcb'd' = bdb'c'\). Hence, \(adb'd' + bcb'd' = bda'd' + bdb'c'\), which means \((ad + bc)b'd' = bd(a'd' + b'c')\). So, \((ad + bc, bd) \sim (a'd' + b'c', b'd')\). Thus, if \((a, b) \sim (a', b')\) and \((c, d) \sim (c', d')\) then \((ad + bc, bd) \sim (a'd' + b'c', b'd')\). QED.

Multiplication

Definition of multiplication on \(\mathbb{Q}\): \[ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]

Proof of well-definedness. Assume \((a, b) \sim (a', b')\) and \((c, d) \sim (c', d')\). So, \(ab' = ba'\) and \(cd' = dc'\). Hence, \(ab'cd' = ba'dc'\), which is the same as \(acb'd' = bda'c'\). So, \((ac, bd) \sim (a'c', b'd')\). Thus, if \((a, b) \sim (a', b')\) and \((c, d) \sim (c', d')\) then \((ac, bd) \sim (a'c', b'd')\). QED.

\(\mathbb{Q}\) is an Extension of \(\mathbb{Z}\)

\(\left\{\frac{n}{1} : n \in \mathbb{Z} \right\}\) behaves the same as \(\mathbb{Z}\). The correspondence is \(\frac{n}{1} \longleftrightarrow n\).

Proof. Assume \(m, n \in \mathbb{Z}\). Note that \(\frac{n}{1} + \frac{m}{1} = \frac{n \cdot 1 + m \cdot 1}{1 \cdot 1} = \frac{n + m}{1}\) and \(\frac{n}{1} \cdot \frac{m}{1} = \frac{n \cdot m}{1 \cdot 1} = \frac{nm}{1}\). QED.

Order on \(\mathbb{Q}\)

An order on a set \(S\) is a relation, denoted \(\lt\), satisfying

1. (law of trichotomy) if \(x, y \in S\), exactly one of these is true: \(x < y\), \(x = y\), \(y < x\), and
2. (transitivity) if \(x, y, z \in S\), \(x < y\), and \(y < z\), then \(x < z\).

A set \(S\) is called an ordered set if there is an order on it.

For example, in \(\mathbb{Z}\), \(m < n\) if \(n - m\) is positive. Another example is, in \(\mathbb{Z} \times \mathbb{Z}\), \((a, b) < (c, d)\) if \(a < c\) or \(a = c\) and \(b < d\) (dictionary order).

\(\frac{m}{n}\) is positive if \(mn > 0\).

Proof of well-definedness. Assume \((a, b) \sim (a', b')\) and \(\frac{a}{b}\) is positive. Since \(\frac{a}{b}\) is positive, \(ab > 0\). Since \((a, b) \sim (a', b')\), \(ab' = ba'\), so \(a'b'b = ab'b'\), which implies \(a'b'b > a\) since \(b'b' > 0\) for all \(b' \neq 0\). Because \(a'b'b > a\), \(a'b'bb > ab\) since \(b \neq 0\). This implies \(a'b'bb > 0\) and thus \(a'b' > 0\) because \(bb > 0\) for all \(b \neq 0\). Hence, \(\frac{a'}{b'}\) is positive. Therefore, if \((a, b) \sim (a', b')\) and \(\frac{a}{b}\) is positive then \(\frac{a'}{b'}\) is positive. QED.

Order on \(\mathbb{Q}\): \(\frac{m}{n} < \frac{m'}{n'}\) if \(\frac{m'}{n'} - \frac{m}{n}\) is positive (subtraction is defined by addition on the negation of the second operand).

Proof of well-definedness. Assume \(\frac{a}{b} < \frac{c}{d}\), \((a, b) \sim (a', b')\), and \((c, d) \sim (c', d')\). Since \(\frac{a}{b} < \frac{c}{d}\), \(\frac{c}{d} - \frac{a}{b}\) is positive, which means \((cb - ad)bd > 0\), so either \(cb - ad > 0\) and \(bd > 0\) or \(cb - ad < 0\) and \(bd < 0\). Since \((a, b) \sim (a', b')\), \(ab' = ba'\). Since \((c, d) \sim (c', d')\), \(cd' = dc'\). There are four cases to consider.

Case 1: \(cb - ad > 0\), \(bd > 0\), and \(b'd' > 0\). We have \[\begin{align} & cb - ad > 0 \\ \implies & cb > ad \\ \implies & cbb'd' > adb'd' & \text{(since } b'd' > 0 \text{)} \\ \implies & c'bb'd > a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' > a'd' & \text{(since } bd > 0 \text{)} \\ \implies & c'b' - a'd' > 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' > 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}\]

Case 2: \(cb - ad > 0\), \(bd > 0\), and \(b'd' < 0\). We have \[\begin{align} & cb - ad > 0 \\ \implies & cb > ad \\ \implies & cbb'd' < adb'd' & \text{(since } b'd' < 0 \text{)} \\ \implies & c'bb'd < a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' < a'd' & \text{(since } bd > 0 \text{)} \\ \implies & c'b' - a'd' < 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' < 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}\]

Case 3: \(cb - ad < 0\), \(bd < 0\), and \(b'd' > 0\). We have \[\begin{align} & cb - ad < 0 \\ \implies & cb < ad \\ \implies & cbb'd' < adb'd' & \text{(since } b'd' > 0 \text{)} \\ \implies & c'bb'd < a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' > a'd' & \text{(since } bd < 0 \text{)} \\ \implies & c'b' - a'd' > 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' > 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}\]

Case 4: \(cb - ad < 0\), \(bd < 0\), and \(b'd' < 0\). We have \[\begin{align} & cb - ad < 0 \\ \implies & cb < ad \\ \implies & cbb'd' > adb'd' & \text{(since } b'd' < 0 \text{)} \\ \implies & c'bb'd > a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' < a'd' & \text{(since } bd < 0 \text{)} \\ \implies & c'b' - a'd' < 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' < 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}\]

Therefore, if \(\frac{a}{b} < \frac{c}{d}\), \((a, b) \sim (a', b')\), and \((c, d) \sim (c', d')\) then \(\frac{a'}{b'} < \frac{c'}{d'}\). QED.

Write \(x > y\) for \(x < y\). Write \(x \leq y\) for \(x < y\) or \(x = y\).

\(x^2 = 2\) has No Solution in \(\mathbb{Q}\)

Proof (by constradiction). Assume that \(x^2 = 2\) has a solution in \(\mathbb{Q}\), i.e. let \(x = \frac{p}{q}\), where \(p, q \in \mathbb{Z}\), \(q \neq 0\), and \(p\) and \(q\) are in lowest terms, i.e. they have no common factors.

So, \(\left(\frac{p}{q}\right)^2 = 2\), hence \(p^2 = 2q^2\). Then \(p^2\) is even (divisible by 2). Then \(p\) is even (because if \(p\) were odd, \(p^2\) would be odd).

So, \(p = 2m\) for some \(m \in \mathbb{Z}\), hence \(p^2 = 4m^2\) and \(4m^2 = 2q^2\). Then \(q^2 = 2m^2\).

Then \(q^2\) is even, hence \(q\) is even. This contradicts the fact that \(p\) and \(q\) are in lowest terms.

Therefore, \(x^2 = 2\) must have no solution in \(\mathbb{Q}\). QED.

\(\mathbb{Q}\) is a Field

A field is a set \(F\) with two operations \(+\), and \(\times\), satisfying the following axioms:

• (A1). \(F\) is closed under \(+\).
• (A2). \(+\) is commutative.
• (A3). \(+\) is associative.
• (A4). \(F\) has an is additive identity. Call it \(0\).
• (A5). Every element has an additive inverse.
• (M1). \(F\) is closed under \(\times\).
• (M2). \(\times\) is commutative.
• (M3). \(\times\) is associative.
• (M4). \(F\) has an is multiplicative identity. Call it \(1\).
• (M5). Every element except \(0\) has a multiplicative inverse.
• (D). \(\times\) distributes over \(+\).

In \(\mathbb{Q}\), \(0\) element is \(\frac{0}{1}\). \(1\) element is \(\frac{1}{1}\).

\(\mathbb{Z}\) is not a field because it does not satisfy (M5).

\(\mathbb{Q}\) is an ordered field, which is a field with an order such that the order is preserved by the field operations, i.e.

1. \(y < z \implies x + y < x + z\), and
2. \(y < z, x > 0 \implies xy < xz\).