Reals from Oder Axioms

definiton of ordered field

is a field S that contains a set P satisfying specific properties

O1) a, b ∈ P =>

a + b ∈ P

O2) a, b ∈ P =>

ab ∈ P

O3) for any a ∈ S one adn only one of the three cases occurs

a ∈ P, -a ∈ P, a = 0

the set P is called the set of

positive elements of S

the set -P = {a ∈ S: -a ∈ P} is the set of

negative elements of S

easily have that a > b <=> b < a

Proof: a - b > 0 <=> -(b-a) > 0 <=> b < a

Prop. if S is an ordered field then for any a, b, c ∈ S: a > b and b > c =>

a > c

Proof: we know a - b > 0 and b - c > 0, we need a - c > 0. BUt by axoim O1. (a -b) + (b -c) > 0, hence a - c > 0

Prop. if S is an ordered field then for any a, b, c ∈ S: a > b <=>

a + c > b + c

Proof: a + c > b + c os equivalent to (a + c) - (b + c) > 0 i.e. a - b > 0

Prop. if S is an ordered field then for any a, b, c ∈ S: a > b and a’ > b’ =>

a + a’ > b + b’

Proof:

Prop. if S is an ordered field then for any a, b, c ∈ S: if c > 0 then a > b <=>

ac > bc

Proof: suppose c > 0 then a - b > 0 implies c(a - b) > 0 hence ca - cb > 0 which is ca > cb. Vice versa, suppose that a ≤ b: if a = b then ac = bc, and if a < b then by the previsous implication (wiht a, b switched) we have ac < bc. In eihter case it is false that ac > bc

Prop. if S is an ordered field then for any a, b, c ∈ S: if c < 0 then a > b <=>

ac < bc

Proof: if a - b > 0 and -c > 0 then (-c)(a -b) > 0, but (-c)(a - b) = -c(a - b) = -ca - c(-b) = -ac - (-bc) = -ac + bc > 0. THe converse os smiler to the converse in similar prop

Corollary: in an ordered field S define a² = a • a, for any a ∈ S: if a ≠ 0 then

a² > 0

Proof: follows form another prop. (if S is an ordered field then for any a, b, c ∈ S) if a > 0 just multiply both sides by c = a > 0, whereas if a < 0. multplying by c = a < 0 still gives a • a > 0

Corollary: in an ordered field S define a² = a • a, for any a ∈ S: 1 > 0

Proof: follws from if a ≠ 0 then a² > 0 from the fact that 1 ≠ 0, adn 1 = 1² > 0

the natural numbers in an ordered field

inductive set

Def: call a subset X of S inductive if

it satifies these properties

I1) 1 ∈ X

I2) ∀x ∈ X, x + 1 ∈ X (i.e. for any x, if x is in X then x + 1 is also in X)

Def: the set of natural numbers on an ordered field S is defined as

the intersection of all subsets X of S that satisfy I1 and I2

formal: N = {n ∈ S: n ∈ X for all X satisfying I1 and I2}

Prop. the set N is inductive and it is the smallest inductive subset of S in the sense that

if X is ductive then N ⊆ X

Proof: proof is immediate since by defintion 1 ∈ X for all inductive sets, hence 1 ∈ N. ALso if n ∈ N then n ∈ X for all inductive sets X, so n + 1 ∈ X for all such sets. THis means n + 1 ∈ N. The fact that N ⊆ X for any X inductive follows from the very defintion of N

Prop. for every n ∈ N we have

n ≥ 1

Proof: this is bst proved by using induction but can also be proved by considing the set X = {n ∈ N: n ≥ 1} ⊆ N. Certain 1 ∈ X, and of n ∈ X then n ≥ 1, hence n + 1 ≥ 1 + 1 = 2 > 1, so n + 1 ∈ X. Hence X satisfies the inductive axioms I1 and I2, so N ⊆ X, hence X = N which means that all n ∈ N greater or equal 1.

intuitive property: sums and multiplications of elements of N are

still in N

Prop: for any m, n ∈ N we have

m + n ∈ N and mn ∈ N

Proof: intutiutvely, if n ∈ N then n + 1 ∈ N, n + 2 ∈ N, … anytime we want until we reach any other natural m > n. To make this precise need once again inductive propoerty of N. The proof for the product is the same, if n ∈ N then 2n = n + n ∈ N, 3n ∈ N, …

Prop: Reverse Inductive Property, for every n ∈ N, if n > 1 then

n - 1 ∈ N

Proof: using the defintion of N ad the smallest inductive set. Let X. = {1} ∪ {n ∈ N : n > 1 and n − 1 ∈ N} ⊆ N. Want to show that N ⊆ X, which shows X = N. Clearly 1 ∈ X and if n ∈ X, then either x = 1, in which n + 2 = 1 + 1 = 2 ∈ X since 2 > 1 and 2 - 1 =1 1 ∈ N, or else n > 1, in which case n + 1 > 1 and n + 1 -1 = n ∈ N. Hence X is inductive and so N ⊆ X.

Prop: Reverse Inductive Property, more generally if n,m ∈ N and n > m then

n - m ∈ N

Proof: fix n ∈ N and let X_n = { m ∈ N: n > m => n - m ∈ N} ⊆ N. Once again we want to prove that X_n is inductive so that N ⊆ X_n, whihc shows that N = X_n so that the statment follows. Clearly 1 ∈ X_n since we proved it. Suppose that m ∈ X_n and let n > m +1, Then n > m so that n - m ∈ N. Since n - m > 1 then n - (m + 1) = (n0 - m) - 1 ∈ N, which proves that m + 1 ∈ X_n so X_n is inductive

Corollary: for any m ∈ N there is no n ∈ N such that

m < n < m + 1

Proof: if that were the case then there would exist m, n ∈ N such that m < n < m + 1, which implies n - m ∈ N, so n - m ≥ 1, contradicting n < m + 1

Prop: in an ordered field S we have: a > 0 =>

a^-1 > 0

Proof:

Prop: in an ordered field S we have: a > b > 0 =>

b^-1 > a^-1 > 0

Proof:

Prop: in an ordered field S we have: a < b =>

a < (a+b) / 2 < b

Proof: since 2 > 0 then a < (a+b) / 2 <=> 2a < 2 • ((a+b) / 2) = 2 • (a+b) • 2^-1 = a + b however 2a = (1 + 1)a = a + a hence a < (a+b) / 2 <=> a + a < a + b <=> a < b similarly b > (a+b) / 2 <=> b + b > b + a <=> b > a

Prop: in an ordered field S we have: if a, b > 0 then

a < b <=> 0 < a² < b²

Proof: use if c > 0 then a > b <=> ac > bc and note that b² - a² = (b - a)(b + a) and since a + b > 0 this product is positive if anf only if b - a > 0

Corollary: in an ordered field S there is no minimum postive number, I.e. if a > 0 then

there is b > 0 with b > a

Proof: sinc e0 < a we cna take avergae b = (0+a) / 2 which si also between 0 and a using a < b => a < (a+b) / 2 < b : 0 < a => 0 < (0+a) / 2 < a

Corollary: in ordered field S, suppose that a,b ∈ S and that for all ϵ > 0 we have a ≤ b + ϵ then

we have a ≤ b

Proof: we want to prove (∀ϵ > 0, a ≤ b + ϵ) =⇒ a ≤ b lets proce it by proving the contrpositive a > b =⇒ ∃ϵ > 0 : a > b + ϵ. IF a > b then a - b > 0. Take ϵ = ½ (a-b). Such ϵ is positive, and b + ϵ = (a+b) / 2 < a

Def: for a ∈ S an ordered field, define the absolute value of a the number |a| =

{a if a > 0, 0 if a = 0, -a if a < 0

Prop: for a, b ∈ S we have |ab| =

|a| |b|

Proof: if either a or b = 0 (or both) then the reusl is obvious. Suppose ab ≠ 0, if a > 0 and b > 0, then ab > 0 and |ab| = ab = |a| |b|. IF a > 0, b < 0 then ab < 0 and |ab| = -ab = a(-b) = |a| |b|. if both a < 0, b < 0 then ab > 0 and |ab| = ab | (-a)(-b) = |a| |b|

Prop: for a, b ∈ S we have |a|² =

a²

Proof: obivous if a = 0. if a ≠ 0 we have a² > 0 so |a²| = a²

Prop: for a, b ∈ S we have if c ≥ 0 then |a| ≤ c <=>

-c ≤ a ≤ c, and |a| ≥ c <=> a ≤ -c or a ≥ c

Proof: suppose c ≥ 0 and |a| ≤ c. if a = 0 then clearly -c ≤ 0 ≤ c. If a > 0 then 0 < |a| = a ≤ c so -c ≤ a ≤ c, if a < 0 then 0 < |a| = -a ≤ c or -c ≤ a < 0 ≤ c.

Prop: for a, b ∈ S we have -|a|…

≤ a ≤ |a|

Proof: follows from property |a| ≤ c <=> …, with c |a| or note that either |a| = a or |a| = -a for any a

triangle inequality, for a,b ∈ S we have |a + b| ≤

|a| + |b|

Proof: from property -|a| ≤ a ≤ |a| adding we obtain −|a| − |b| ≤ a + b ≤ |a| + |b| which si the same as |a + b| ≤ |a| + |b|, still from -|a| ≤ a ≤ |a| (with a + b in place of a and |a| + |b| in place of c)

Corollary: for any a,b ∈ S we have

1) |a − b| ≤ |a − x| + |x − b| ∀x ∈ S

2) |a − b| ≥ ||a| − |b|| (Reversed triangle inequality)

Def: let S be an ordered field ad deifne the set of intergers as

Z = {n ∈ S : n ∈ N, or n = 0, or − n ∈ N}

Prop: for any m,n ∈ Z, we have m + n ∈ Z and mn ∈ Z

also for n ∈ Z we have -n ∈ Z

Proof: this follows ohter props. (For any m, n ∈ N we have m + n ∈ N and mn ∈ N and or every n ∈ N, if n > 1 then n − 1 ∈ N) by distinguishing cases using the defintion of Z

Prop: every interger is either evne or add

Z = E ∪ O, moreover E ∩ O = ∅

Proof: clearly each even adn eahc odd is an integer, so we really need to prove Z ⊆ E∪O

Prop: the follwing hold, n is even <=>

n² is even

Prop: the follwing hold, m,n odd =>

mn odd, and m + n even