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- We saw before that the Real numbers
- How do we know that √2 exists? In other words how can we be sure that there is some real number whose square is 2?

It is easy to convince yourself that (say) 2 + 3 = 3 + 2. How about √2 + √3 = √3 + √2 or*e*+π = π +*e*? - One's intuition about what should be true works pretty well for
**N**or**Z**or even for**Q**. Things don't get hard until we are forced (like the Pythagoreans) to admit the existence of irrationals.There are constructive methods for making the full set

**R**from**Q**and hence starting with**N**. The first rigorous construction was given by Richard Dedekind (1831 to 1916) in 1872.You can see more about Dedekind's construction.

However, for the moment we will simply give a set of axioms for the Reals and leave it to intuition that there is something that satisfies these axioms.

We start with a set, which we'll call

**R**and a pair**+ .**of binary operations.**The Axioms** - These are divided into three groups.
**I The algebraic axioms****R***is a field under***+***and***.**

This means that (**R**, +) and (**R**, .) are both abelian groups and the distributive law (*a*+*b*)*c*=*ab*+*ac*holds.**II The order axioms**- There is a relation > 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.b) If

*a*,*b*> 0 then*a*+*b*> 0 and*a*.*b*> 0c) If

*a*>*b*then*a*+*c*>*b*+*c*for any*c*Something satisfying axioms

**I**and**II**is called an**ordered field**. **Examples**

**Examples**

*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.

*The field***C***of complex numbers is not an ordered field under any ordering.***Proof**

Suppose*i*> 0. Then -1 =*i*^{2}> 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** - An
**upper bound**of a non-empty subsetof*A***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**|*q*^{2}< 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.**

*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*.

**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 :

*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.

*The real number*√2*exists.***Proof**

We will get √2 as the least upper bound of the set*A*= {*q*∈**Q**|*q*^{2}< 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*b*^{2}< 2 and*b*^{2}> 2 both lead to contradictions and so we must have*b*^{2}= 2 (by the Trichotomy rule).

So suppose that*b*^{2}> 2. Look at (*b*- 1 /*n*)^{2}=*b*^{2}- 2*b*/*n*+ 1 /*n*^{2}>*b*^{2}- 2*b*/*n*.

When is this > 2 ?

Answer: When*b*^{2}- 2*b*/*n*> 2 which happens if and only if*b*^{2}- 2 > 2*b*/*n*or 1/*n*< (*b*^{2}-2)/2*b*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*b*^{2}< 2 then (*b*+ 1/*n*)^{2}=*b*^{2}+ 2*b*/*n*+ 1/*n*^{2}>*b*^{2}+ 2*b*/*n*.

Can this be < 2 ?

Answer: Yes, when*b*^{2}+ 2*b*/*n*< 2 which happens if and only if 2 -*b*^{2}> 2*b*/*n*or 1/*n*< (2 -*b*^{2})/2*b*and we can choose an*n*satisfying this, leading to the conclusion that*b*would not be an upper bound.

*Real numbers can be defined by decimal expansions.***Proof**

Given the decimal expansion (say) 0.*a*_{1}*a*_{2}*a*_{3}... consider the set (of rationals) {0.*a*_{1}, 0.*a*_{1}*a*_{2}, 0.*a*_{1}*a*_{2}*a*_{3}, ...} = {*a*_{1}/10 , (10*a*_{1}+*a*_{2})/100 , (100*a*_{1}+ 10*a*_{2}+*a*_{3})/1000 , ... }.

This is bounded above -- say by (*a*_{1}+ 1)/10 or by (10*a*_{1}+*a*_{2}+ 1)/100 etc. and so it has a*least upper bound*.

This is the real number defined by the decimal expansion.

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