The Set of Rational Numbers

In today's blog, I show how to formally construct the set of rational numbers from the set of integers.

I will then give a proof that the set of rational numbers forms a field. If you need a review of fields, check out here.

Definition 1: Set of rational numbers

We can define the set of rational numbers as the ordered pair of integers (a,b) where a,b are integers and b ≠ 0.


Definition 2: Addition of rationals

(a,b) + (c,d) = (ad + bc, bd)


Definition 3: Multiplication of rationals

(a,b) * (c,d) = (ac,bd)


Definition 4: Equality of rationals

Two rational numbers (a,b) and (c,d) are equal if and only if ad=bc.


Definition 5: Comparison of rationals

(a,b) is less than (c,d) if and only if abd² is less than b²cd.


Lemma 1: All integers can be represented as rational numbers

(1) Any integer x can be represented as (x,1) [see definition 1 above]

(2) This correspondence holds over addition

x + y = z if and only if (x,1) + (y,1) = (x*1+y*1,1*1) = (x+y,1) = (z,1)

(3) This correspondence holds over multiplication

x*y = z if and only if (x,1)*(y,1) = (xy,1*1) = (xy,1) = (z,1)

QED


Lemma 2: The set of rational numbers is closed on addition

Proof:

(1) The integers are closed on addition [see Lemma 1, here] and multiplication [see Lemma 2, here].

(2) So, it follows from Definition 2 above that the rational numbers are closed on addition.

QED


Lemma 3: The set of rational numbers is closed on multiplication

Proof:

The follows directly from Definition 3 and the fact that the integers are closed on multiplication [see Lemma 2, here].

QED


Lemma 4: The set of rational numbers satisfy the Commutative Rule for Addition

Proof:

(1) From Definition 2 above:

(a,b) + (c,d) = (ad + bc, bd)

(2) Since integers are commutative by addition (see Lemma 7, here):

(ad + bc, bd) = (bc + ad, bd)

(3) From Definition 2 again, we get:

(bc + ad,bd) = (c,d) + (a,b)

QED


Lemma 5: The set of rational numbers satisfy the Associative Rule for Addition

Proof:

From Definition 2 above:

[(a,b) + (c,d)] + (e,f) = (ad+bc,bd) + (e,f) = (adf+bcf + bde,bdf) = (a,b) + (cf + de,df) =

= (a,b) + [(cf + de,df)] = (a,b) + [(c,d) + (e,f)]

QED


Lemma 6: The set of rational numbers has an Additive Identity for all elements.

Proof:

(1) Using Definition 2 above, we have:

(a,b) + (0,c) = (a*c + 0*b,b*c) = (ac,bc)

(2) Using Definition 4 above, we note that:

(ac,bc) = (a,b) since acb = bca [using the commutative property multiplication for integers, see Lemma 8, here]

QED


Lemma 7: The set of rational numbers has an Additive Inverse for all elements.

Proof:

(1) Let (a,b) be a rational number.

(2) Then, it's additive inverse is (-a,b) since:

(a,b) + (-a,b) = (a*b + -a*b,b*b) = (ab-ab,b*b) = (0,b*b)

QED


Lemma 8: The set of rational numbers supports the Associative Rule for Multiplication

Proof:

Using Definition 3 above, we have:

[(a,b)*(c,d)]*(e,f) = [(ac,bd)]*(e,f) = (ace,bdf) = (a,b)*[(ce,df)] = (a,b)*[(c,d)*(e,f)]

QED


Lemma 9: (c,c) = (1,1)

Proof:

The follows directly from definition 4 above since:

c*1 = 1*c

QED


Lemma 10: The set of rational numbers supports the Distributive Rule

Proof:

(1) Using Definition 2 and Definition 3 above, we have:

(a,b)[(c,d) + (e,f)] = (a,b)*(cf+de,df) = (acf + ade,bdf)

(2) Using Definition 3 above and Lemma 8 above, we have:

(acbf + aebd,bdbf) = (b,b)*(acf + aed,bdf) = (1,1)*(acf + aed,bdf) = (acf + aed,bdf)

(3) Using Definition 2 above, we have:

(ac,bd) + (ae,bf) = (acbf + aebd,bdbf)

QED


Lemma 11: The set of rational numbers supports the Commutative Rule for Multiplication

Proof:

(1) Using Definition 3 above, we have:

(a,b)*(c,d) = (ac,bd)

(2) Using the Commutative Property of Multiplication for Integers (see Lemma 8, here):

(ac,bd) = (ca,db)

(3) Using Definition 3 above, we have:

(ca,db) = (c,d)*(a,b)

QED


Lemma 12: The set of rational numbers has a Multiplicative Identity

Proof:

For any rational number (a,b), we have (see Definition 3 above):

(a,b)*(1,1) = (a*1,b*1) = (a,b)

QED


Lemma 13: For every nonzero element, the set of rational numbers has a Multiplicative Inverse

Proof:

(1) Let (a,b) be any rational number.

(2) Then (b,a) will be its multiplicative inverse since:

(a,b)*(b,a) = (ab,ba)

(3) Using the Commutative Property of Multiplication of Integers (see Lemma 8, here), we have:

(ab,ba) = (ab,ab)

(4) Using Lemma 8 above, we have (ab,ab)=1.

QED


Theorem 14: The set of rational numbers forms a field

Proof:

This follows directly from Lemma 2 through Lemma 13 above and from Definition 3, here.

QED

References

• "Rational Numbers", Wikipedia.Org
• Bruce E. Meserve, Fundamental Concepts of Algebra, Dover Books, 1981.