3.4 Rationals and well-defined operations

Integers solve the subtraction problem: $x+b=a$ has an integer solution for every pair of integers $a,b$. Division poses the next obstruction. The equation $2x=1$ has no integer solution, so $Z$ is not closed under division by nonzero elements. We construct $Q$ so that $bx=a$ can be solved whenever $a,b\in Z$ and $b\ne0$.

The earlier construction identified pairs that encoded the same difference. Here we identify pairs that encode the same quotient. First we define equality of those pairs; then we check arithmetic and inverses; finally we define order. At each stage the same question governs the argument: does the result depend on the rational number, or merely on its chosen representative?

Why a quotient is needed

The fractions

$$\frac{1}{2}, \qquad \frac{2}{4}, \qquad \frac{-3}{-6}$$

all describe the same rational number. If we want to build $Q$ from integer data alone, then the construction must identify all such pairs automatically.

That is why we do not define a rational number to be a single ordered pair. Instead, we define a rational number to be a whole equivalence class of pairs that represent the same quotient.

Definition

Rational numbers as equivalence classes

Let

$$Y = Z \times (Z \setminus \{0\}).$$

Thus an element of $Y$ is a pair $(a, b)$ with $a \in Z$ and $b \in Z \setminus \{0\}$.

Define a relation $\sim_Q$ on $Y$ by

$$(a, b) \sim_Q (c, d) \quad\Longleftrightarrow\quad ad = bc.$$

The set of rational numbers is the quotient

$$Q = Y / \sim_Q.$$

The equivalence class of $(a, b)$ is written $[(a, b)]$, and it is represented informally by the fraction $a/b$.

The denominator is required to be nonzero because the construction is intended to model quotients. If $b = 0$, then the pair $(a, b)$ cannot represent a meaningful rational number.

Why the relation makes sense

The equation $ad = bc$ is the usual cross-multiplication test for equality of fractions. In the quotient construction, that familiar test becomes the actual definition of equality.

Theorem

The relation $\sim_Q$ is an equivalence relation

The relation defined by

$$(a, b) \sim_Q (c, d) \quad\Longleftrightarrow\quad ad = bc$$

is reflexive, symmetric, and transitive on $Y$.

Proof: $\sim_Q$ is an equivalence relation

Reflexivity is immediate because $ab = ba$, so $(a, b) \sim_Q (a, b)$.

Symmetry is also immediate: if $ad = bc$, then $cb = da$, so $(c, d) \sim_Q (a, b)$.

For transitivity, assume

$$(a, b) \sim_Q (c, d) \qquad\text{and}\qquad (c, d) \sim_Q (e, f).$$

Then

$$ad = bc \qquad\text{and}\qquad cf = de.$$

Multiply the first equation by $f$ and the second by $b$:

$$adf = bcf, \qquad bcf = bde.$$

Hence $adf = bde$. Because $d$ is nonzero and $Z$ has no zero divisors, we may cancel $d$ in $Z$ to obtain

$$af = be.$$

Therefore $(a, b) \sim_Q (e, f)$, so the relation is transitive.

Representatives and the same rational number

An equivalence class contains many representatives. That is not a defect. It is the whole point of the construction.

Worked example

Different pairs can describe the same rational number

Consider the pairs $(1, 2)$, $(2, 4)$, and $(-3, -6)$.

We have

$$1 \cdot 4 = 2 \cdot 2, \qquad 1 \cdot (-6) = 2 \cdot (-3).$$

Therefore

$$(1, 2) \sim_Q (2, 4) \qquad\text{and}\qquad (1, 2) \sim_Q (-3, -6).$$

So all three pairs belong to the same rational number:

$$[(1, 2)] = [(2, 4)] = [(-3, -6)].$$

This is the quotient-theoretic version of the elementary fact that

$$\frac{1}{2} = \frac{2}{4} = \frac{-3}{-6}.$$

It is often useful to remember that a rational number is not tied to a preferred representative. The class $[(a, b)]$ does not become a different number just because you multiply both coordinates by the same nonzero integer.

Reduced representatives and the Euclidean algorithm

The quotient construction explains why many pairs represent the same rational number. For calculation, however, it is still useful to choose a clean representative. If $a$ and $b$ have a greatest common divisor $g\gt 0$, then

$$[(a,b)]=[(a/g,b/g)].$$

The Euclidean algorithm gives a systematic way to find this $g$ by repeated division with remainder. This is not a new definition of rational numbers; it is a practical method for choosing a simpler representative inside the same equivalence class.

Worked example

Reducing a representative with the Euclidean algorithm

Consider $[(84,30)]$. The Euclidean algorithm gives

$$84=2\cdot 30+24,\qquad 30=1\cdot 24+6,\qquad 24=4\cdot 6+0.$$

So $\gcd(84,30)=6$, and

$$[(84,30)]=[(14,5)].$$

Both pairs represent the same rational number because $84\cdot 5=30\cdot 14$.

Operations on $Q$

Once the classes are defined, the next task is to define addition and multiplication on the classes themselves.

Definition

Addition, multiplication, and inverses on $Q$

For classes in $Q$, define

$$[(a, b)] + [(c, d)] := [(ad + bc, bd)],$$

and

$$[(a, b)] \cdot [(c, d)] := [(ac, bd)].$$

The additive inverse is defined by

$$-[(a, b)] := [(-a, b)].$$

If $a \neq 0$, the multiplicative inverse is defined by

$$[(a, b)]^{-1} := [(b, a)].$$

Each of these formulas is written on representatives, so each of them must be checked for well-definedness. Otherwise the formula might depend on the chosen pair rather than on the rational number itself.

Before checking representatives, check the output domain. Since $b,d\ne0$ and $Z$ has no zero divisors, the denominator $bd$ is nonzero for both sums and products. For inversion, the extra condition $a\ne0$ makes $(b,a)$ a valid pair. These are two different obligations: a formula must produce a valid object, and that object must be independent of representatives.

What "well-defined" means

A formula on equivalence classes is well-defined if changing representatives does not change the resulting class.

For addition, this means the following statement must be true:

if $(a, b) \sim_Q (a', b')$ and $(c, d) \sim_Q (c', d')$, then

$$(ad + bc, bd) \sim_Q (a'd' + b'c', b'd').$$

Theorem

Addition on $Q$ is well-defined

If $(a, b) \sim_Q (a', b')$ and $(c, d) \sim_Q (c', d')$, then

$$[(ad + bc, bd)] = [(a'd' + b'c', b'd')].$$

So the addition formula does not depend on the chosen representatives.

Proof: addition on $Q$ is well-defined

Assume

$$ab' = ba' \qquad\text{and}\qquad cd' = dc'.$$

We must show

$$(ad + bc, bd) \sim_Q (a'd' + b'c', b'd').$$

By definition of $\sim_Q$, it is enough to prove

$$(ad + bc)b'd' = (a'd' + b'c')bd.$$

Expanding the left-hand side gives

$$adb'd' + bcb'd'.$$

Using $ab' = ba'$, the first term becomes

$$adb'd' = ba'dd'.$$

Using $cd' = dc'$, the second term becomes

$$bcb'd' = bdb'c'.$$

Hence

$$(ad + bc)b'd' = ba'dd' + bdb'c' = (a'd' + b'c')bd.$$

Therefore

$$(ad + bc, bd) \sim_Q (a'd' + b'c', b'd'),$$

so the sum is well-defined.

Worked example

Compute in classes, then simplify conceptually

Let

$$x = [(1, 2)], \qquad y = [(1, 3)].$$

Then

$$x + y = [(1 \cdot 3 + 2 \cdot 1, 2 \cdot 3)] = [(5, 6)].$$

Also,

$$x \cdot y = [(1 \cdot 1, 2 \cdot 3)] = [(1, 6)].$$

If we replace $x$ by the equivalent representative $[(2, 4)]$, then the same formulas give

$$[(2, 4)] + [(1, 3)] = [(10, 12)],$$

and

$$[(2, 4)] \cdot [(1, 3)] = [(2, 12)].$$

These are the same rational numbers as $[(5, 6)]$ and $[(1, 6)]$, so the class operations behave as they should.

Multiplication and inversion require the same discipline

The addition proof displays the general method. Multiplication is shorter, but it still has to be checked explicitly because the inputs are classes rather than preferred fractions.

Why multiplication on $Q$ is well-defined

Assume

$$(a,b)\sim_Q(a',b') \qquad\text{and}\qquad (c,d)\sim_Q(c',d').$$

The hypotheses mean

$$ab'=ba' \qquad\text{and}\qquad cd'=dc'.$$

To prove that the product is independent of representatives, use the defining test for $\sim_Q$:

$$(ac)(b'd')=(bd)(a'c').$$

The left side factors as

$$(ac)(b'd')=(ab')(cd')=(ba')(dc')=(bd)(a'c').$$

Hence $(ac,bd)\sim_Q(a'c',b'd')$, and the product formula descends from pairs to rational classes.

Theorem

Every nonzero rational has one multiplicative inverse

If $q\in Q$ and $q\ne0$, there is a unique $q^{-1}\in Q$ with

$$q\cdot q^{-1}=[(1,1)].$$

Proof: existence and uniqueness of the rational inverse

Write $q=[(a,b)]$ with $b\ne0$. The condition $q\ne0$ means $a\ne0$, so $[(b,a)]$ is a valid rational class. Then

$$[(a,b)]\cdot[(b,a)]=[(ab,ba)]=[(1,1)],$$

because $ab=ba$ in $Z$.

For uniqueness, suppose $[(u,v)]$ is another inverse. The equation

$$[(u,v)]\cdot[(a,b)]=[(1,1)]$$

means $(ua,vb)\sim_Q(1,1)$, hence $ua=vb$. This is exactly the condition $(u,v)\sim_Q(b,a)$. Thus every inverse represents the same class $[(b,a)]$.

Worked example

Invert a negative rational without changing its class

Let $q=[(-3,5)]$. Since the numerator is nonzero,

$$q^{-1}=[(5,-3)].$$

Their product is

$$[(-3,5)]\cdot[(5,-3)]=[(-15,-15)]=[(1,1)].$$

The denominator in the inverse is nonzero, and the two coordinates may be multiplied by $-1$ without changing the class. The calculation therefore works even when the chosen denominator is negative.

Common mistake

Nonzero numerator is needed before inverting

The formula $[(a,b)]^{-1}=[(b,a)]$ is valid only when $a\ne0$. If $a=0$, the second coordinate of $[(b,a)]$ would be zero, which is excluded from $Y$, and zero has no multiplicative inverse.

How the construction solves division

The integers enter the new system through $j(n)=[(n,1)]$. This map is injective: equality $[(m,1)]=[(n,1)]$ means $m\cdot1=1\cdot n$, hence $m=n$. The operation formulas also give

$$j(m)+j(n)=j(m+n),\qquad j(m)j(n)=j(mn).$$

Thus integer arithmetic is preserved. More importantly, if $a,b\in Z$ and $b\ne0$, the class $x=[(a,b)]$ satisfies

$$j(b)x=[(ba,b)]=[(a,1)]=j(a).$$

The middle equality is the cross-product test. We have now answered the opening question: the quotient construction supplies a solution of $bx=a$ while keeping the original integers identifiable inside $Q$.

Not every representative formula descends to $Q$

Once you start thinking in equivalence classes, you should become suspicious of any proposed relation or operation written directly on representatives.

For example, consider the following candidate rules on classes $[(p, q)]$ and $[(m, n)]$:

1. compare $p$ and $m$;
2. compare the sign of $pn - qm$;
3. compare the sign of $(pn - mq)nq$.

The first two rules are not well-defined on $Q$, because changing a representative can change the truth value. The third rule compensates for sign changes in the denominators and is invariant under changing representatives.

Worked example

Why the denominator signs matter

Compare $[(1,2)]$ with $[(0,1)]$. The numerator rule gives $1\gt0$, and the raw cross-difference is $1\cdot1-0\cdot2=1$. Replace $(1,2)$ by $(-1,-2)$, which names the same class. The numerator comparison becomes $-1\gt0$, now false, and the raw cross-difference becomes $-1$.

Both proposed rules have changed while the rational numbers stayed fixed. The sign-corrected expression stays positive: its two values are $1\cdot2=2$ and $(-1)\cdot(-2)=2$. It compensates for the denominator sign rather than treating that sign as irrelevant.

Order: choose positive denominators

Arithmetic now belongs to rational classes. To compare those classes, choose representatives with positive denominators: if $b\lt0$, replace $(a,b)$ by $(-a,-b)$. For $b,d\gt0$, define

$$\frac{a}{b}\lt\frac{c}{d}\quad\Longleftrightarrow\quad ad\lt bc.$$

The right side uses the order already established on $Z$. Positivity matters because multiplying an inequality by a negative denominator would reverse it.

Proof that rational order is independent of representatives

Suppose $ab'=ba'$ and $cd'=dc'$, with all four denominators positive. Then

$$ad\cdot b'd'=a'd'\cdot bd, \qquad bc\cdot b'd'=b'c'\cdot bd.$$

Multiplying $ad\lt bc$ by positive $b'd'$ and substituting these equalities gives $a'd'\cdot bd\lt b'c'\cdot bd$. Cancelling positive $bd$ yields $a'd'\lt b'c'$. The reverse argument is identical, so the comparison is unchanged.

Trichotomy in $Z$ implies that exactly one of $a/b\lt c/d$, $a/b=c/d$, and $c/d\lt a/b$ holds. For transitivity, take $b,d,f\gt0$. If $ad\lt bc$ and $cf\lt de$, then

$$adf\lt bcf\lt bde.$$

Cancelling positive $d$ gives $af\lt be$, hence $a/b\lt e/f$. Thus the order laws follow from integer order, with every sign condition visible.

Explore which comparisons survive

The quickest way to understand this issue is to compare several proposed rules against different representatives of the same two rational numbers. In the panel below, the rational numbers remain $1/2$ and $1/3$; only their representatives change.

A genuine relation on $Q$ must give the same truth value after every legal change of representatives. If the output changes merely because $1/2$ was written as $(1,2)$ instead of $(2,4)$ or $(-1,-2)$, the formula has not descended to the quotient.

Checkpoint

Why does the multiplication proof use two cross-product equalities rather than decimal intuition?

Explain what can change while the rational class stays fixed.

Solution · Answer

The same rational number has many representatives. Cross-product equalities are the defining conditions that identify those representatives, so they are the data that can be substituted safely in a well-definedness proof.

Common mistakes

Common mistake

A quotient class is not one preferred fraction

The symbols $[(1, 2)]$, $[(2, 4)]$, and $[(-3, -6)]$ name the same equivalence class, hence the same rational number. The three ordered pairs $(1,2)$, $(2,4)$, and $(-3,-6)$ are distinct representatives of that class.

Common mistake

A plausible formula is not automatically well-defined

A formula on pairs may look natural and still fail on the quotient. Before accepting an operation on $Q$, check that changing to equivalent representatives preserves the equivalence class of its output. For a relation on $Q$, the truth value must remain unchanged.

Quick checks

Checkpoint

Suppose zero denominators were allowed. Use $(1,0)$, $(0,0)$, and $(0,1)$ to prove that the cross-product relation would no longer be transitive.

Test the two consecutive relations, then compare the first pair directly with the third.

Solution · Model solution

On all of $Z^2$, cross-multiplication would give $(1,0)\sim(0,0)$ and $(0,0)\sim(0,1)$, because both tests reduce to $0=0$. But $(1,0)\sim(0,1)$ would require $1=0$, which is false. Transitivity therefore fails. Excluding zero denominators is part of making the quotient construction mathematically valid, before any inverse is defined.

Checkpoint

For a rational class $[(a,b)]$ with $a\ne0$, prove that swapping the coordinates gives its multiplicative inverse. Check both the domain and the product.

Check the product with the original class.

Solution · Answer

Since $a\ne0$, the pair $(b,a)$ has a nonzero second coordinate and is valid. The inverse is $[(b,a)]$, because

$$[(a, b)] \cdot [(b, a)] = [(ab, ba)] = [(1, 1)],$$

which is the multiplicative identity in $Q$.

Exercises

Checkpoint

Why does the relation $[(p, q)] \prec [(m, n)]$ defined by $p$ greater than $m$ fail to be well-defined?

Find equivalent representatives that change the truth value.

Solution · Guided solution

Take $[(1, 2)] = [(-1, -2)]$. Compare both with $[(0, 1)]$.

Using the representative $(1, 2)$, the statement $p \gt m$ reads $1 \gt 0$, which is true. Using the equivalent representative $(-1, -2)$, it reads $-1 \gt 0$, which is false.

So the rule depends on the chosen representative and therefore does not define an order relation on $Q$.

Checkpoint

Let $a,b,k,c\in Z$, $b>0$, $a=bk+c$, and $0\le c<b$. Prove that $\gcd(a,b)=\gcd(b,c)$.

Show that the two pairs have exactly the same common divisors.

Solution · Guided solution

Let $d$ be a common divisor of $a$ and $b$. Since $c=a-bk$, the same $d$ divides $c$, so $d$ is a common divisor of $b$ and $c$.

Conversely, if $d$ divides both $b$ and $c$, then $d$ divides $bk+c=a$, so $d$ is a common divisor of $a$ and $b$.

Thus the two pairs have the same common divisors, and therefore the same greatest common divisor:

$$\gcd(a,b)=\gcd(b,c).$$

Checkpoint

Prove the multiplication formula on $Q$ is well-defined: if $(a,b)\sim_Q(a',b')$ and $(c,d)\sim_Q(c',d')$, then $(ac,bd)\sim_Q(a'c',b'd')$.

Translate each equivalence into a cross-product equality.

Solution · Guided solution

From $(a,b)\sim_Q(a',b')$ and $(c,d)\sim_Q(c',d')$, we know

$$ab'=ba', \qquad cd'=dc'.$$

To prove $(ac,bd)\sim_Q(a'c',b'd')$, we must show

$$(ac)(b'd')=(bd)(a'c').$$

But

$$(ac)(b'd')=(ab')(cd')=(ba')(dc')=(bd)(a'c').$$

So multiplication does not depend on the chosen representatives.

Related notes

Read 3.3 Integers from equivalence classes for the previous quotient construction, and 3.5 Gaps in Q and why sqrt(2) is not rational for the next point where the rational number system shows its limitations.