## Sunday, September 20, 2009

### 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.

(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):

(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

amirmahallati said...

Dear Sir,

I don't understand how can the set of rational numbers form a field, as the Property of Reciprocals doesn't hold true, because the reciprocal of 9/10 is an irrational number. Could you tell me whether or not you disagree, and please explain why?

Thank you,
Amir Mahallati

Larry Freeman said...

Hi Amirmahallati,

The reciprocal of a nonzero rational number is also rational.

The reciprocal of 9/10 is 10/9.

Remember 1/(a/b) = (b/a).

-Larry

willson said...

This blog is full of knowledge.Rational numbers is a very important topic of mathematics.Any student can understand the application of rational numbers from this blog.board of gujarat syllabus

Prosper Ablordeppey said...

Hi, Amirmahallati, the Rationals are all fractions that terminate or have recurring decimals. the ones which don't recur and also do not terminate are irrational.