## Axioms of Real Numbers Essay

A similar argument starting with i &lt; 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 &gt; on R is transitive. That is, if a &gt; b and b &gt; c then a &gt; c. Proof a &gt; b if and only if a – b &gt; b – b = 0 by Axiom II c) a &gt; c if and only if a – c &gt; c – c = 0 Hence (a – b) + (a – c) &gt; 0 and so a – c &gt; 0 and we have a &gt; c. The thing which distinguishes R from Q (and from other subfields) is the Completeness Axiom. Definitions

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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 &lt; 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 &gt; 0 in R, then for some n N we have 1/n &lt; a. Equivalently: Given any x R, for some n N we have n &gt; 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 &gt; M – 1.

But then n + 1 &gt; 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 &lt; b. Choose n so that 1/n &lt; b – a. Then look at multiples of 1/n. Since these are unbounded, we may choose the first such multiple with m/n &gt; a. We claim that m/n &lt; b.

If not, then since (m-1)/n &lt; a and m/n &gt; b we would have 1/n &gt; 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 &lt; 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 &lt; 2 and b2 &gt; 2 both lead to contradictions and so we must have b2 = 2 (by the Trichotomy rule). So suppose that b2 &gt; 2. Look at (b – 1 /n )2 = b2 – 2b /n + 1 /n2 &gt; b2 – 2b/n. When is this &gt; 2 ? Answer: When b2 – 2b/n &gt; 2 which happens if and only if b2 – 2 &gt; 2b/n or 1/n &lt; (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 &lt; 2 then (b + 1/n)2 = b2 + 2b/n + 1/n2 &gt; b2 + 2b/n. Can this be &lt; 2 ?

Answer: Yes, when b2 + 2b/n &lt; 2 which happens if and only if 2 – b2 &gt; 2 b/n or 1/n &lt; (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.