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==Algebraic Properties==

==Algebraic Properties==

The algebraic implications of the rational numbers are both interesting and far reaching. The rational numbers represent a specific example of a [[field (mathematics)|Field of Fractions]] over the [[Integers]]. The [[integers]] constitute a [[ring]], there are two defined operations - addition and multiplication, which follow a certain set of rules (or as Mathematicians would say - have a certain structure). Specifically, the integers are an integral domain, a special type of ring with no zero divisors (There is no integer <math>a,b\neq 0</math> such that <math>a*b = 0</math>) and where the multiplication is commutative. The integers, however, are not a [[field (mathematics)|field]] because not every integer has an integer multiplicative inverse (in fact, only the number 1 and -1 do) .  

The algebraic implications of the rational numbers are both interesting and far reaching. The rational numbers represent a specific example of a [[field (mathematics)|Field of Fractions]] over the [[Integers]]. The [[integers]] constitute a [[ring (mathematics)|ring]], there are two defined operations - addition and multiplication, which follow a certain set of rules (or as Mathematicians would say - have a certain structure). Specifically, the integers are an integral domain, a special type of ring with no zero divisors (There is no integer <math>a,b\neq 0</math> such that <math>a*b = 0</math>) and where the multiplication is commutative. The integers, however, are not a [[field (mathematics)|field]] because not every integer has an integer multiplicative inverse (in fact, only the number 1 and -1 do) .  

As mentioned above, we can define an [[equivalence class|equivalence relation]] on (<math>\mathbb{Z}\times\mathbb{Z}\backslash\{0\}</math>) (A set of pairs of integers, where the second number in the pair cannot be 0 - thus avoiding the issue of dividing by zero) where two tuples <math>(a,b)</math> and <math>(c,d)</math>  (which can be written in the more familiar form <math>\frac{a}{b}</math> and <math>\frac{c}{d}</math>) are equivalent if <math>ad-bc = 0</math>. For example, the two fractions <math>\frac{1}{4}</math> and <math>\frac{2}{8}</math> are equivalent because <math>2*4 - 1*8 = 0</math>. When two pairs of numbers (or fractions) are equivalent we call them members of the same equivalence class. The Field of fractions is the [[set]] of these equivalence classes, it is a simple exercise to show that they constitute a field. Along with the equivalence class, we define the operations of addition (<math>\frac{a}{b} + \frac{c}{d} = \frac{ad+cb}{bd}</math>) and multiplication (<math>\frac{a}{b}*\frac{c}{d} = \frac{ac}{bd}</math>), these operations are intentionally designed to mimic the addition and multiplication of fractions often taught in gradeschool, however keep in mind that the underlying ring does not necessarily have to be the integers (this same formulation constructs a field for ''any'' ring). Since the [[integers]] are ''naturally embedded''  in its field of fractions (we define the trivial mapping that <math>a\in \mathbb{Z} \rightarrow (a,1) \equiv \frac{a}{1} \in \mathbb{Z}\times \mathbb{Z}\backslash\{0\}</math> we only need to show the existence of a multiplicative inverse for each member of the set (all the other rules are satisfied by the ring and integral domain structure of the integers). Since for every fraction <math>\frac{a}{b}</math> we can construct a multiplicative inverse <math>\frac{b}{a}</math>, we satisfy all the conditions for a [[field]] and complete the proof. If the field of fractions is taken over a bona-fide field we will just get back the same field (technically, a new field but which is isomorphic to the original) - so the field of fractions over the rationals is just the rationals again.  

As mentioned above, we can define an [[equivalence class|equivalence relation]] on (<math>\mathbb{Z}\times\mathbb{Z}\backslash\{0\}</math>) (A set of pairs of integers, where the second number in the pair cannot be 0 - thus avoiding the issue of dividing by zero) where two tuples <math>(a,b)</math> and <math>(c,d)</math>  (which can be written in the more familiar form <math>\frac{a}{b}</math> and <math>\frac{c}{d}</math>) are equivalent if <math>ad-bc = 0</math>. For example, the two fractions <math>\frac{1}{4}</math> and <math>\frac{2}{8}</math> are equivalent because <math>2*4 - 1*8 = 0</math>. When two pairs of numbers (or fractions) are equivalent we call them members of the same equivalence class. The Field of fractions is the [[set]] of these equivalence classes, it is a simple exercise to show that they constitute a field. Along with the equivalence class, we define the operations of addition (<math>\frac{a}{b} + \frac{c}{d} = \frac{ad+cb}{bd}</math>) and multiplication (<math>\frac{a}{b}*\frac{c}{d} = \frac{ac}{bd}</math>), these operations are intentionally designed to mimic the addition and multiplication of fractions often taught in gradeschool, however keep in mind that the underlying ring does not necessarily have to be the integers (this same formulation constructs a field for ''any'' ring). Since the [[integers]] are ''naturally embedded''  in its field of fractions (we define the trivial mapping that <math>a\in \mathbb{Z} \rightarrow (a,1) \equiv \frac{a}{1} \in \mathbb{Z}\times \mathbb{Z}\backslash\{0\}</math> we only need to show the existence of a multiplicative inverse for each member of the set (all the other rules are satisfied by the ring and integral domain structure of the integers). Since for every fraction <math>\frac{a}{b}</math> we can construct a multiplicative inverse <math>\frac{b}{a}</math>, we satisfy all the conditions for a [[field]] and complete the proof. If the field of fractions is taken over a bona-fide field we will just get back the same field (technically, a new field but which is isomorphic to the original) - so the field of fractions over the rationals is just the rationals again.