Axioms of Real Numbers Essay

AXIOMS OF REAL NUMBERS Field Axioms: there exist notions of addition and multiplication, and additive and multiplicative identities and inverses, so that: • (P1) (Associative law for addition): a + (b + c) = (a + b) + c • (P2) (Existence of additive identity): 9 0 : a + 0 = 0 + a = a • (P3) (Existence of additive inverse): a + (? a) = (? a) + a = 0 • (P4) (Commutative law for addition): a + b = b + a • (P5) (Associative law for multiplication): a · (b · c) = (a · b) · c • (P6) (Existence of multiplicative identity): 9 1 6= 0 : a · 1 = 1 · a = a • (P7) (Existence of multiplicative inverse): a · a? = a? 1 · a = 1 for a 6= 0 • (P8) (Commutative law for multiplication): a · b = b · a • (P9) (Distributive law): a · (b + c) = a · b + a · c Order Axioms: there exists a subset of positive numbers P such that • (P10) (Trichotomy): exclusively either a 2 P or ? a 2 P or a = 0. • (P11) (Closure under addition): a, b 2 P ) a + b 2 P • (P12) (Closure under multiplication): a, b 2 P ) a · b 2 P Completeness Axiom: a least upper bound of a set A is a number x such that x _ y for all y 2 A, and such that if z is also an upper bound for A, then necessarily z _ x. (P13) (Existence of least upper bounds): Every nonempty set A of real numbers which is bounded above has a least upper bound. We will call properties (P1)–(P12), and anything that follows from them, elementary arithmetic. These properties imply, for example, that the real numbers contain the rational numbers as a subfield, and basic properties about the behavior of ‘>’ and ‘ on R. (That is, given any pair a, b then a > b is either true or false). It satisfies: a) Trichotomy: For any a R exactly one of a > 0, a = 0, 0 > a is true. ) If a, b > 0 then a + b > 0 and a. b > 0 c) If a > b then a + c > b + c for any c Something satisfying axioms I and II is called an ordered field. Examples 1. The field Q of rationals is an ordered field. Proof Define a/b > c/d provided that b, d > 0 and ad > bc in Z. One may easily verify the axioms. 2. The field C of complex numbers is not an ordered field under any ordering. Proof Suppose i > 0. Then -1 = i2 > 0 and adding 1 to both sides gives 0 > 1. But squaring both sides gives (-1)2 = 1 > 0 and so we get a contradiction.

A similar argument starting with i < 0 also gives a contradiction. The above two groups of axioms can be used to deduce any algebraic or order properties of R. Example The ordering > on R is transitive. That is, if a > b and b > c then a > c. Proof a > b if and only if a – b > b – b = 0 by Axiom II c) a > c if and only if a – c > c – c = 0 Hence (a – b) + (a – c) > 0 and so a – c > 0 and we have a > c. The thing which distinguishes R from Q (and from other subfields) is the Completeness Axiom. Definitions

We Will Write a Custom Essay Specifically
For You For Only $13.90/page!

order now

An upper bound of a non-empty subset A of R is an element b R with b a for all a A. An element M R is a least upper bound or supremum of A if M is an upper bound of A and if b is an upper bound of A then b M. That is, if M is a lub of A then (b R)(x A)(b x) b M A lower bound of a non-empty subset A of R is an element d R with d a for all a A. An element m R is a greatest lower bound or infimum of A if m is a lower bound of A and if d is an upper bound of A then m d. We can now state: III The Completeness Axiom If a non-empty set A has an upper bound, it has a least upper bound.

Something which satisfies Axioms I, II and III is called a complete ordered field. Remark In fact one can prove that up to “isomorphism of ordered fields”, R is the only complete ordered field. Note that the ordered field Q is not complete For example, the set {q Q | q2 < 2} is bounded but does not have a least upper bound in Q. We will see why in a little while. Some consequences of the completeness axiom. 1. A subset A which has a lower bound has a greatest lower bound. Proof Let B = {x R | -x A}. Then B is bounded above by -(the lower bound of A) and so has a least upper bound b say.

It is then easy to check that -b is a greatest lower bound of A. 2. The Archimedean property of the Reals If a > 0 in R, then for some n N we have 1/n < a. Equivalently: Given any x R, for some n N we have n > x. Proof This last statement is equivalent to saying that N is not bounded above. This seems like a very obvious fact, but we will prove it rigorously from the axioms. Suppose N were bounded above. Then it would have a least upper bound, M say. But then M – 1 is not an upper bound and so there is an integer n > M – 1.

But then n + 1 > M contradicting the fact that M is an upper bound. Remark This result has been attributed to the great Greek mathematician (born in Syracuse in Sicily) Archimedes (287BC to 212BC) and appears in Book V of The Elements of Euclid (325BC to 265BC). From this we can deduce : 3. Between any two real numbers is an rational number. Proof Let a b be real numbers with (say) a < b. Choose n so that 1/n < b – a. Then look at multiples of 1/n. Since these are unbounded, we may choose the first such multiple with m/n > a. We claim that m/n < b.

If not, then since (m-1)/n < a and m/n > b we would have 1/n > b – a. Remark A set A with the property that an element of A lies in every interval (a, b) of R is called dense in R. We have just proved that the rationals Q are dense in R. In fact, the irrationals are also dense in R. We can now prove the result we stated earlier. 4. The real number 2 exists. Proof We will get 2 as the least upper bound of the set A = {q Q | q2 < 2 }. We know that A is bounded above (by 2 say) and so its least upper bound b exists by Axiom III.

We now prove that b2 < 2 and b2 > 2 both lead to contradictions and so we must have b2 = 2 (by the Trichotomy rule). So suppose that b2 > 2. Look at (b – 1 /n )2 = b2 – 2b /n + 1 /n2 > b2 – 2b/n. When is this > 2 ? Answer: When b2 – 2b/n > 2 which happens if and only if b2 – 2 > 2b/n or 1/n < (b2 -2)/2b and we can choose such an n by the Archmedean property. Thus b – 1/n is an upper bound, contradicting the assumption that b was the least upper bound. Similarly, if b2 < 2 then (b + 1/n)2 = b2 + 2b/n + 1/n2 > b2 + 2b/n. Can this be < 2 ?

Answer: Yes, when b2 + 2b/n < 2 which happens if and only if 2 – b2 > 2 b/n or 1/n < (2 – b2)/2b and we can choose an n satisfying this, leading to the conclusion that b would not be an upper bound. 5. Real numbers can be defined by decimal expansions. Proof Given the decimal expansion (say) 0. a1a2a3… consider the set (of rationals) {0. a1 , 0. a1a2 , 0. a1a2a3 , … } = { a1/10 , (10 a1 + a2)/100 , (100 a1 + 10 a2 + a3)/1000 , … }. This is bounded above — say by (a1+ 1)/10 or by (10 a1+a2+ 1)/100 etc. and so it has a least upper bound. This is the real number defined by the decimal expansion.