Integers and Rationals

Integers arise as equivalence classes of natural pairs with

(a,b)\sim(c,d)\iff a+d=b+c

, and rationals as classes of integer pairs with

(a,b)\sim(c,d)\iff ad=bc

. The "numbers as equivalence classes" trick returns when the reals are built from the rationals.

Examples & Exercises

Assume $(a,b)\sim(a',b')$ and $(c,d)\sim(c',d')$ . Prove that $(a+c,b+d)\sim(a'+c',b'+d')$ . (This says the addition $[(a,b)]+[(c,d)]=[(a+c,b+d)]$ is well-defined, independent of the chosen representatives.)

(a,b)\sim(a',b'),\ (c,d)\sim(c',d')

(a+c,b+d)\sim(a'+c',b'+d')

Integers mode — each dot is a class of pairs, and addition is the same whichever representatives are chosen.

Show proof

By definition of the relation, $(a,b)\sim(a',b')$ means $a+b'=a'+b$ and $(c,d)\sim(c',d')$ means $c+d'=c'+d$ .
Adding the two equations side by side gives $(a+b')+(c+d')=(a'+b)+(c'+d)$ . Rearranging with the associativity and commutativity of natural-number addition yields $(a+c)+(b'+d')=(a'+c')+(b+d)$ .
This is exactly the definition of $(a+c,b+d)\sim(a'+c',b'+d')$ . So the value of $[(a,b)]+[(c,d)]$ never depends on which representatives of the two classes are chosen — addition on $\mathbb{Z}$ is well-defined.

For rationals $p<q$ , prove that $m=\frac{p+q}{2}$ is rational and $p<m<q$ .

p<q \implies p<\frac{p+q}{2}<q

Rationals mode, max denominator 4 — between any two rationals sits another dot.

Show proof

Write $p=\frac{a}{b}$ and $q=\frac{c}{d}$ with $b>0,\ d>0$ . Then $m=\frac{p+q}{2}=\frac{ad+bc}{2bd}$ : numerator and denominator are integers and the denominator is nonzero, so $m$ is rational.
Since $p<q$ , adding $p$ to both sides gives $2p<p+q$ ; dividing by the positive $2$ preserves the inequality, so $p<\frac{p+q}{2}=m$ .
Likewise, adding $q$ to both sides of $p<q$ gives $p+q<2q$ , i.e. $m<q$ . Hence $p<m<q$ — between any two rationals there is always another rational, so the rationals are dense. Integers lack this property (no integer lies strictly between consecutive integers).

For a nonzero rational $\frac{a}{b}$ , prove that $\frac{b}{a}$ is its multiplicative inverse, and explain why this makes $\mathbb{Q}$ a field.

\frac{a}{b}\cdot\frac{b}{a}=1\qquad(a\ne 0)

Rationals mode, max denominator 4 — every nonzero dot has a multiplicative inverse.

Show proof

$\frac{a}{b}\ne 0$ means $a\ne 0$ (and $b\ne 0$ is part of being a rational). So the fraction $\frac{b}{a}$ , whose denominator $a\ne 0$ , is a legitimate rational.
By the definition of fraction multiplication, $\frac{a}{b}\cdot\frac{b}{a}=\frac{ab}{ba}$ . Commutativity gives $ab=ba$ , so this fraction equals $\frac{ab}{ab}=1$ . Hence $\frac{b}{a}$ is the multiplicative inverse of $\frac{a}{b}$ .
Since every nonzero rational has an inverse, division by a nonzero number is always possible in $\mathbb{Q}$ — together with the well-defined addition and multiplication, $\mathbb{Q}$ is a field. The field axioms and their structure are covered in the groups-fields scene.

Part of these learning paths

Examples & Exercises