4.3 Rational Numbers

Since the only integers that have a multiplicative inverse are \(1\) and \(-1\), it would be beneficial to expand our number system to include symbols that can operate as multiplicative inverses. In this process we allow the ordered pairs in \(\mathbb{Z}\times (\mathbb{Z}\setminus \{0\})\) to be written as a fraction, \[\frac{a}{b}\cong (a,b).\] But we also want the set to be closed under addition and multiplication. This results in a set that contains all possible rational expressions of integers whose denominator is non-zero. Therefore we create the set \[\widehat{\mathbb{Q}}= \left\{ \frac{m}{n}\middle \vert m,n \in \mathbb{Z}, n\neq 0\right\} \cong \mathbb{Z}\times (\mathbb{Z}\setminus \{0\}).\]

Note: in \(\widehat{\mathbb{Q}}\), \(\frac{2}{4}\) and \(\frac{1}{2}\) are distict elements. We will work on getting their equivalence as we create the rational numbers using a partition of \(\widehat{\mathbb{Q}}\) generated by an equivalence relation.

Similar to our process of constructing the integers from the natural numbers, we define a relation of two rational expressions by \[\frac{a}{b} \sim \frac{c}{d} \Leftrightarrow ad=bc.\]

We will now verify that this relation of rational expressions represents an equivalence relation.

Reflexive

One can see that this relation is reflexive, since if \(\frac{a}{b}\), with \(b\neq 0\), is a rational expression of integers that \(ab=ab\) and so \(\frac{a}{b}\) is equivalent to itself.

Symmetric

Since equality of integers is a symmetric relationship we have that for any integers \(a,b,c,d\) with \(ad=bc\) that \(bc=ad\) and so this equivalence of rational expressions is symmetric.

Transitive

The more challenging property of equivalence relations to prove is the transitive property. To prove transitivity, we assume that we have three rational expressions, \[\frac{a}{b}, \: \frac{c}{d}, \: \mbox{ and } \: \frac{m}{n},\] with \(b\), \(d\), and \(n\) all non-zero, with \[\frac{a}{b}\sim \frac{c}{d} \quad \mbox{ and } \quad \frac{c}{d} \sim \frac{m}{n}.\] From the definition of equivalence of rational expressions we have that \(ad=bc\) and \(cn=md\). From the properties of multiplication of integers we have that \[(ad)\cdot (cn) = (bc) \cdot (md),\] or equivalently, \[ancd = bmcd.\] This can then be rewritten as \[ (an-bm)\cdot (cd) =0.\] Since the integers have the property that the product of two non-zero integers is also non-zero, we know that either \(an-bm=0\) or \(cd=0\). Since \(d\) is assumed to be non-zero, \(cd=0\) would imply that \(c=0\). This would then imply that both \(ad\) and \(md\) are zero, since \(ad=bc\) and \(cn=md\). Since \(d\neq 0\) we would then have that \(a=0\) and \(m=0\), and consequently that \(an=bm=0\). Therefore, \(an-bm=0\) independently of \(cd=0\). Therefore, we have that \(an=bm\) and that \(\frac{a}{b}\sim \frac{m}{n}\), implying that equivalence of rational expressions is transitive.

Since this relation on rational expressions is an equivalence relation we should explore what the equivalence classes of these rational expressions look like.

We will first look at the rational expressions that are equivalent to \(\frac{0}{1}\). We see that

\[\left[\frac{0}{1}\right] = \left\{ \frac{a}{b} \: \middle \vert \: b\neq 0, \: 0\cdot b=1\cdot a\right\} = \left\{\frac{0}{b} \: \middle \vert \: b \neq 0 \right\}\] Similarly, \[\left[ \frac{1}{1}\right] = \left\{ \frac{a}{b} \: \middle \vert \: b\neq 0, \: 1\cdot b = 1 \cdot a\right\} = \left\{ \frac{a}{a} \: \middle \vert \: a \neq 0\right\}.\]

We can follow this process and see that we can embed the integers into this set of equivalence classes using the correspondence of \[ n\in \mathbb{Z} \quad \leftrightarrow \quad \left[\frac{n}{1} \right] .\]

4.3.1 Operations

Now that we understand the set of equivalence classes of rational expressions, we need to define our operations of addition and multiplication on this set and verify that these definitions are well-defined.

Definition 4.1 Let \(\left[\frac{a}{b}\right]\) and \(\left[\frac{c}{d}\right]\) be two equivalence classes of rational expressions with \(a,b,c,d\in \mathbb{Z}\) and \(b,d\neq 0\). We define addition and multiplication as \[\left[\frac{a}{b}\right] + \left[\frac{c}{d}\right] := \left[\frac{ad+bc}{bd}\right] \quad \mbox{ and } \quad \left[\frac{a}{b}\right] \times \left[\frac{c}{d}\right] := \left[ \frac{ac}{bd} \right].\]

Since \(b\) and \(d\) are non-zero integers, the no non-zero zero divisors property tells us that \(bd\neq 0\) and so the operations of addition and multiplication are closed. We can make sure that they operate appropriately on the equivalence classes by making sure that different representations for the original equivalence classes of rational expression generate a sum and product that are elements of the same equivalence class. The process is similar to the same process that was done with the integers. Since we did the proof for addition being well-defined on the integers, we will do the proof that multiplication is well-defined for the rational numbers.

Let \(\frac{a}{b}\sim \frac{a'}{b'}\) and \(\frac{c}{d} \sim \frac{c'}{d'}\). Using the definition of the equivalence classes, we have that \[a\cdot b' = b \cdot a' \quad \mbox{ and } \quad c \cdot d' = d \cdot c'.\]

Therefore, \[\begin{align*} (bd)\cdot (a'c') &= (ba') \cdot (dc') \quad \mbox{ by the commutative property of multiplication} \\ &= (ab') \cdot (dc') \quad \mbox{ since } ab' = ba' \\ &= (ab') \cdot (cd') \quad \mbox{ since } cd'=dc' \\ &= (ac) \cdot (b'd') \quad \mbox{ by the commutative property of multiplication} \end{align*}\] and so we have that \(\frac{ac}{bd} \sim \frac{a'c'}{b'd'}\).

So we have constructed a new number system that we call the rational numbers, \(\mathbb{Q}\). In order to simplify our notation, we will no longer write each rational number as an equivalence class. We will instead just keep in mind the equivalence class structure and use any of the possible elements of the equivalence class to represent the rational number. We will also often write rational expressions of the form \(\frac{n}{1}\) as the corresponding integer \(n\). We will also write \[\frac{-a}{b}, \quad \frac{a}{-b}, \mbox{ and }\quad -\frac{a}{b} \] as equivalent expressions.

Related Content Standards

  • (7.NS.2) Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.
    1. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as \((-1)(-1) = 1\) and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
    2. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If \(p\) and \(q\) are integers, then \(-(p/q) = (-p)/q = p/(-q)\). Interpret quotients of rational numbers by describing real world contexts.
    3. Apply properties of operations as strategies to multiply and divide rational numbers.

4.3.2 Order

We can define an order on \(\mathbb{Q}\) by first defining the relationship between a rational expression and \(0\), built on the order on the integers. We say that for a rational expression \(\frac{a}{b}\), with \(b\neq 0\) that \[\left(\frac{a}{b} < 0 \Leftrightarrow ab<0\right), \quad \left(\frac{a}{b}=0 \Leftrightarrow ab=0\right), \mbox{ and } \quad \left(\frac{a}{b} >0 \Leftrightarrow ab >0\right).\] From the no non-zero divisors property of the integers, we know that one, and only one, of the three statements can be true for a single rational expression. If \(\frac{c}{d}\sim \frac{a}{b}\), we use that \(ad=bc\) and \[(ad)^2 = (ad)(ad) = (ad)(bc) = (ab)(cd)\] to see that \(ab\) and \(cd\) have the same relationship to \(0\). So this ordering with respect to \(0\) is well-defined.

We can then extend the ordering to the entire set of rational numbers by defining \[\left(\frac{a}{b} < \frac{c}{d} \Leftrightarrow \left(\frac{a}{b} + \frac{-c}{d}\right) <0 \right), \quad \left(\frac{a}{b} = \frac{c}{d} \Leftrightarrow \left(\frac{a}{b} + \frac{-c}{d}\right) = 0 \right),\] \[\mbox{ and } \quad \left(\frac{a}{b} > \frac{c}{d} \Leftrightarrow \left(\frac{a}{b} + \frac{-c}{d}\right) > 0 \right).\]

If we have rational numbers \(\frac{a}{b}\), \(\frac{m}{n}\), and \(\frac{p}{q}\) with \[\frac{a}{b} < \frac{m}{n}, \quad \mbox{ and } \quad \frac{m}{n} < \frac{p}{q},\] then \[\frac{m}{n}+ \frac{-a}{b} >0, \quad \frac{p}{q}+ \frac{-m}{n} >0\] and so \[\frac{p}{q} + \frac{-a}{b} = \frac{p}{q}+ \frac{-m}{n} + \frac{m}{n}+ \frac{-a}{b}>0\] and this relation satisfies the properties of an order and we can extend the properties of inequalities from the integers.

Theorem 4.5 Let \(r,s,t\in \mathbb{Q}\). Then \[(1) \: (r=s) \Leftrightarrow (r+t=s+t), \quad (2) \: (r<s) \Leftrightarrow (r+t<s+t),\] \[(3) \: (r=s) \Leftrightarrow (r\times t=s\times t), \mbox{ with } t\neq 0 \quad \mbox{ and } \quad (4) \: (r<s) \Leftrightarrow (r\times t<s\times t), \mbox{ with } t> 0.\]

Proof. In order to get the idea of the proof we will prove parts (1) and (4) and leave the proofs of the other two parts to the reader.

Let \(r=\frac{a}{b}\), \(s=\frac{c}{d}\), and \(t=\frac{m}{n}\) be rational numbers with \(a,b,c,d,m,n\in \mathbb{Z}\) and \(b,d,n\neq 0\).

Part (1). Then \[(r=s) \Leftrightarrow (ad=bc) \Leftrightarrow \left( (ad)n^2 = (bc) n^2 \right) \Leftrightarrow \left( (an+bm)(dn)=(cn+dm)(bn) \right) \] since \(d\) and \(n\) are non-zero. We then see that \[\left( (an+bm)(dn)=(cn+dm)(bn) \right) \Leftrightarrow \left(\frac{an+bm}{bn} = \frac{cn+dm}{dn}\right) \Leftrightarrow (r+t=s+t)\] by the definition of equality of rational numbers and properties of the operations on the integers.

Part (4). We can assume without any loss of generality that \(b,d,n >0\) by changing the signs in the numerators, if needed. Then we know that \[(r<s) \Leftrightarrow ((s-r)>0 ) \Leftrightarrow \left( \frac{cb-ad}{bd} >0 \right) \Leftrightarrow \left( (cb-ad)(bd) >0\right) \Leftrightarrow \left( (cb-ad) >0\right)\] since \(bd>0\).

Similarly, \[\left( r\times t < s\times t\right) \Leftrightarrow \left( \frac{am}{bn} < \frac{cm}{dn} \right) \Leftrightarrow \left( \frac{bcmn - admn}{bdn^2} >0\right). \] Since \(b,d,n >0\), we know that this is equivalent to \[\left( (bcmn-admn)>0 \right) \Leftrightarrow \left( (cb-ad)(mn) > 0 \right) \Leftrightarrow \left( (cb-ad) >0 \right)\] since \(t>0\) and \(n>0\) implies that \(m>0\) and \(mn>0\).

Therefore we have that \[(r<s) \Leftrightarrow \left( r\times t < s\times t\right).\]

Note: The phrase “without loss of generality” is commonly used in mathematics when the narrowing of an argument to a particular case does not affect the overall argument. In this instance, we can assume that \(b>0\) because if we had the case of a rational number with \(b<0\) (\(\frac{3}{-7}\)), we could rewrite it as \(\frac{-3}{7}\).

Related Content Standards

  • (7.NS.1) Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
    1. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged.
    2. Understand \(p + q\) as the number located a distance \(|q|\) from \(p\), in the positive or negative direction depending on whether \(q\) is positive or negative. Show that a number and its opposite have a sum of \(0\) (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.
    3. Understand subtraction of rational numbers as adding the additive inverse, \(p- q = p + (-q)\). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
    4. Apply properties of operations as strategies to add and subtract rational numbers.

4.3.3 Algebraic Properties

Since the rational numbers were constructed from the integers, they maintain many of the corresponding properties of the integers. So if \(\frac{a}{b}\), \(\frac{m}{n}\), and \(\frac{p}{q}\) are elements of \(\mathbb{Q}\), we have the following properties.

Table 4.3: Properties of Rational Numbers
Property
\(\frac{a}{b}+\frac{m}{n} \in \mathbb{Q}\) \(\frac{a}{b}\cdot \frac{m}{n} \in \mathbb{Q}\) Closure
\(\frac{a}{b}+\left(\frac{m}{n}+\frac{p}{q}\right) =\left(\frac{a}{b}+\frac{m}{n}\right)+\frac{p}{q}\) \(\frac{a}{b}\cdot \left(\frac{m}{n}\cdot \frac{p}{q}\right) = \left(\frac{a}{b} \cdot \frac{m}{n}\right) \cdot \frac{p}{q}\) Associative Property
\(\frac{a}{b}+\frac{m}{n}=\frac{m}{n}+\frac{a}{b}\) \(\frac{a}{b}\cdot \frac{m}{n} = \frac{m}{n}\cdot \frac{a}{b}\) Commutative Property
\(\frac{a}{b}+0=\frac{a}{b}\) \(\frac{a}{b} \cdot 1 = \frac{a}{b}\) Identities
\(\frac{a}{b} \cdot \left(\frac{m}{n}+\frac{p}{q}\right) = \left(\frac{a}{b}\cdot \frac{m}{n}\right) + \left(\frac{a}{b} \cdot \frac{p}{q}\right)\) \(\left(\frac{a}{b}+\frac{m}{n}\right) \cdot \frac{p}{q} = \left(\frac{a}{b}\cdot \frac{p}{q}\right) + \left(\frac{m}{n}\cdot \frac{p}{q}\right)\) Distributive Property
\(\frac{a}{b} + \left(-\frac{a}{b}\right) =0\) If \(a,b\neq 0\), \(\frac{a}{b} \cdot \left(\frac{b}{a}\right) =1\) Inverses
If \(\frac{a}{b},\frac{m}{n}>0\), then \(\frac{a}{b}\cdot \frac{m}{n}>0\). If \(\frac{a}{b},\frac{m}{n}<0\), then \(\frac{a}{b}\cdot \frac{m}{n}>0\).
\(\frac{a}{b} \cdot \frac{m}{n}=0\) if, and only if, \(\frac{a}{b}=0\), \(\frac{m}{n}=0\), or both No zero divisors

4.3.4 Properties of Exponents

Using our definition of exponents from Section 4.1, we will expand the number system for the base from the integers to the rational numbers.

If \(\displaystyle{\frac{p}{q}}\) (with \(q\neq 0\)) is a rational number, for any natural number \(n\) we define \[\left(\frac{p}{q}\right)^n := \frac{p^n}{q^n}. \] Such a definition makes sense because we know from Section 4.1 that \(p^n\) and \(q^n\) are both well-defined integers, that \(q^n\neq 0\), and the ratio of such integers in a rational number.

We can then extend the number system for the exponents from the natural numbers to the integers for \(p,q\neq 0\) by defining \[\left(\frac{p}{q}\right)^{-n} = \left(\frac{q}{p}\right)^n\] for all \(n\in \mathbb{N}\). In the case that \(p=0\), this definition would not be well-defined and so we define \(0^n=1\) for all \(n\in \mathbb{Z}\).

Related Content Standards

  • (8.EE.1) Know and apply the properties of integer exponents to generate equivalent numerical expressions.

With the extended definition of exponents we still have for all rational numbers, \(a\), that \(a^0=1\) and \(a^1=a\). We also have from our definition that for \(a\neq 0\) and \(n\in \mathbb{Z}\), \[a^{-n}=\frac{1}{a^n}=\left(\frac{1}{a}\right)^n.\]

For \(m,n \in \mathbb{N}\) and integers \(p\) and \(q\), with \(q\neq 0\), \[\left(\frac{p}{q}\right)^m \cdot \left(\frac{p}{q}\right)^n = \left(\frac{p^m}{q^m} \right) \cdot \left(\frac{p^n}{q^n}\right) = \frac{p^m \cdot p^n}{q^m \cdot q^n} = \frac{p^{(m+n)}}{q^{(m+n)}} = \left(\frac{p}{q}\right)^{(m+n)}.\] This can then be extended to \(m,n\in \mathbb{Z}\) with

  • [Case 1:] If \(p=0\), then \(\frac{p}{q}=0\) and \[0^m\cdot 0^n = 0\cdot 0=0=0^{(m+n)}.\]
  • [Case 2:] If \(p\neq 0\), we let \(m,n\in \mathbb{Z}\). If \(m\) and \(n\) are both non-negative, then we have the property given above. If \(m\) and \(n\) are both negative, then \[\left(\frac{p}{q}\right)^{m} \cdot \left(\frac{p}{q}\right)^{n} = \left(\frac{q}{p}\right)^{-m} \cdot \left(\frac{q}{p}\right)^{-n} = \left(\frac{q}{p}\right)^{(-m-n)} = \left(\frac{p}{q}\right)^{m+n}.\] If one is negative, \(m\), and the other is non-negative, \(n\), then \[ \left(\frac{p}{q}\right)^{m} \cdot \left(\frac{p}{q}\right)^{n} = \frac{p^m}{q^m} \cdot \frac{q^{-n}}{p^{-n}} = \frac{p^{m+n}}{q^{m+n}} = \left(\frac{p}{q}\right)^{m+n}. \]

Therefore, we have that for all \(a\in \mathbb{Q}\) and for all \(m,n\in \mathbb{Z}\) that \(a^m\cdot a^n = a^{m+n}\).

One can make similar arguments to extend the remainder of Theorem \(\ref{thm:exponents-integers}\) so that \((ab)^n=a^n\cdot b^n\) for each \(n\in \mathbb{Z}\) and \((a^m)^n = a^{mn}\) for each \(m,n\in \mathbb{Z}\).

Since the rational numbers include values between integers, it makes sense to explore the relationships between exponents and the order of the rational numbers given by inequalities. If \(a\in \mathbb{Q}\) with \(a>1\), then we can write \(a\) as \(\frac{p}{q}\) for some integers \(p\) and \(q\) with \(0<q<p\), and from Theorem \(\ref{thm:exponents-integers}\) we have \(0<q^n<p^n\) for any integer \(n\geq 1\). So we can conclude that \(a^n>1\). If \(n\) is a negative integer, then \(0<q^{-n}<p^{-n}\) and so \(a^{-n} >1\), and hence \(a^n<1\).

If \(a\) and \(b\) are rational numbers with \(0<a<b\), then \(\frac{a}{b}\in \mathbb{Q}\) with \(0<\frac{a}{b}<1\) so \(0<\left(\frac{a}{b}\right)^n<1\) and \(0<a^m<b^m\).

We combine all of the results into the following theorem.

Theorem 4.6 Let \(a\) and \(b\) be rational numbers.

  1. \(a^0=1\) and \(a^1=a\)

  2. If \(a>1\), then \(a^n >1\) for all integers \(n>0\), and if \(a<1\), then \(0<a^n<1\) for all integers \(n<0\)

  3. \(a^{-n} = \frac{1}{a^n}\), \(a^m\cdot a^n = a^{m+n}\), \((ab)^n=a^n\cdot b^n\), and \((a^m)^n = a^{mn}\) for each \(m,n\in \mathbb{Z}\)

  4. If \(0<a<b\) and \(m\in \mathbb{N}\), then \(0<a^m<b^m\).

  5. If \(a>1\) and \(m<n\), then \(a^m<a^n\).

4.3.5 Exercises

  1. Prove that there exists a rational number between any two rational numbers. (If \(a,b\in \mathbb{Q}\) with \(a<b\), then there exists a \(c\in \mathbb{Q}\) such that \(a<c<b\).)

  2. Let \(x,y\in \mathbb{Q}\) such that \(0<x<y\). Prove that \(0<\frac{1}{y} < \frac{1}{x}\).

  3. Look graphically at \(\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})\), which is a different representation of \(\widehat{\mathbb{Q}}\) defined at the beginning of this section.

    1. How would you graphically describe the equivalence classes of the relation \[(a,b) \sim (c,d) \Leftrightarrow ad=bc?\]
    2. How does this interact with the standards in Grades 7 and 8 regarding ratio and direct proportions?

  4. Compare the definition of order of rational numbers in this section with other techniques of determining the order of rational numbers (i.e. cross-multiplying, finding lowest common denominator, `butterfly method’, etc.).

    1. What types of (mis)conceptions arise with each of these methods?
    2. Is there an order to presenting these various methods for determining the order of rational numbers that improves student understanding?

  5. Write your own definition of a rational number.