Defn.The real numbers are defined to be a set IR with two binary operations (+,·) which satisfy the following properties: Given any a,b,c in IR

1. a+(b+c) = (a+b)+c.
2. a+b = b+a.
3. $0 Î IR ' a+0 = a, "a Î IR.
4. for each a Î IR, $(-a) Î IR, so that a+(-a) = 0.
5. a·(b·c) = (a·b)·c.
6. a·b = b ·a.
7. $ 1 Î IR ' 1 ¹ 0 and a·1 = a, "a Î IR.
8. for each a Î IR with a ¹ 0, $a^-1 Î IR, so that a·a^-1 = 1.
9. a·(b+c) = (a·b) + (a·c).

a. a+b Î IP,

b. a ·b Î IP,

c. For each a in IR, exactly one of the following properties holds:

i)

a Î IP,

ii)

-a Î IP,

iii)

a = 0.

Notation:

`b-a' is defined as  b+(-a).

`a < b' means that  b-a Î IP.

`a £ b' means    either a < b  or  a = b.

The fraction ` [a/b] ' means   a·b^-1.

(i.) (-a) = (-1)·a.

(ii.) 0 < 1.

(iii.) if 0 < a, then (-a) < 0.

Pf. Since

it follows from Homework Problem 2.3 that a: = (-1)·a is an additive inverse for a. But additive inverses are unique (from your Homework Problem 2.2), so the conclusion of part (i) follows.

To prove part (ii), we assume to the contrary, i.e. that 1 does not belong IP. By the definition, 1 ¹ 0, so (-1) Î IP and therefore 0 < -1. But (-1)·(-1) = -(-1) = 1 by part (i) and the HW Problem that additive inverses are unique. This shows that 1 Î IP by property (b) of the positive cone and the assumption that (-1) Î IP. Contradiction, by the tricotomy property (c).

For part (iii), observe that 0 < a means a Î IP. Since additive inverses are unique, then -(-a) = a, and so (0-(-a)) = -(-a) Î IP. This is equivalent to the statement (-a) < 0. ^[¯]

1. Show that the additive (or multiplicative) identity is unique.
2. Show that additive (or multiplicative) inverses are unique.
3. Prove that a·0 = 0 for each a in IR.
4. Prove that a < b and 0 < c implies that a·c < b·c.
5. For each pair of real numbers a,b, prove that exactly one of the following properties hold:

i)

a < b,

ii)

b < a,

iii)

a = b.

6. Prove that if 0 < a < b, then 0 < [1/b] < [1/a].
7. Prove that the interval (a,b) is uncountable.
8. Prove that -(a+c) = (-a)+ (-c).

Defn. A real number a is said to be an upper bound for A Í IR if

1. M is an upper bound for A
2. if a is any upper bound for A, then M £ a.

Lemma. Suppose that A is a nonempty subset of IR, with least upper bound M, then for every e > 0, there exists a Î A such that

Pf.  Since 0 < e, then M-e < M. This shows that M-e cannot be an upper bound for A. Hence there is a member of A, call it a, so that M-e < a. ^[¯]

Theorem. (Archimedean Property)  Suppose a,b are positive real numbers, then there exists n Î IN such that b < n·a.
 

Pf.  Suppose to the contrary that na < b for all n Î IN, then it follows that a: = b/a is an upper bound for the natural numbers. Let M be the least upper bound. By the lemma, 1/2 > 0, so there exists a natural number N so that M-1/2 < N. But then, M < N+1/2 < N+1, which shows that M is not an upper bound for IN. Contradiction.   ^[¯]

Corollary. The natural numbers are not bounded.

Corollary. Given any e > 0, there exists n Î IN such that 1/n < e.

1. Prove that least upper bounds are unique.
2. Prove the first corollary to the Archimedean Property.
3. Prove the greatest lower bound property for the real numbers, i.e., each nonempty subset of IR which has a lower bound, has a greatest lower bound.

Notation: Next we define intervals of real numbers.

(a,b) : = {x Î IR| a < x < b}   is called the open interval with endpoints a,b.

[a,b] : = {x Î IR| a £ x £ b}   is called the closed interval with endpoints a,b.

(a,b] : = {x Î IR| a < x £ b} and [a,b) : = {x Î IR| a £ x < b}   are called the half open intervals with endpoints a,b.

Pf. Define the length of I  by L: = b-a. By the previous corollary, there exists n_o Î IN such that 0 < 1/n_o <L. Let A : = {k |  k  an integer and  k/n_o < a}. A is nonempty, since the negative integers are not bounded from below. Let k_o belong to A. Set B : = {k|  k  an integer and  k ³ k_o }ÇA. Also, A is bounded from above by a·n_o, which shows that B is in fact a finite set of integers. Let K be the largest member of B and therefore of A, then K+1 does not belong to A. Let r : = (K+1)/n_o, then

            a  <  (K+1)/n_o < K/n_o + L £ a + (b-a)

which shows that the rational r Î (a,b) Í I. ^[¯]

Corollary. Each interval with nonzero length contains an infinite number of rationals.

Defn. A real number is said to be irrational if it is not rational.

Corollary. Each interval with nonzero length contains an uncountably infinite number of irrationals.

We establish a few other facts about irrational numbers and also prove directly that each interval of positive length contains an infinite number of irrationals.

Lemma. The product of a nonzero rational with an irrational is irrational.
 

Pf. Suppose that q₁ ·a = q₂, where q₁,q₂ are rational and a is irrational. Since q₁ ¹ 0, then a = q₂/q₁ and it follows that a is rational. Contradiction. ^[¯]

Pf. If m is odd, then there exists an integer k such that m = 2k+1. In this case m² = 2(2k² +2k)+1.  ^[¯]

Lemma.Ö2 is irrational.
 

Pf. Suppose that Ö2 = m/n where m,n are integers with n > 0. We may assume that the rational is in lowest terms (i.e. m and n have no common factors). Squaring the equation and multiplying by n², we obtain that m² = 2 n². This shows that m² is even. By the lemma m must be even and equivalently that it contains 2 as a factor. This shows 4 k² = 2 n² for some integer k. Consequently, n is even and 2 appears as one of its factors. Contradiction, since m/n was supposed to be in lowest terms. ^[¯]

Pf. (Another proof of the `density' of the irrationals) Let a,b be the endpoints of the interval I. Consider the interval (a/ Ö2, b/ Ö2). It has length (b-a)/Ö2 > 0, and so contains a nonzero rational number q. It follows that qÖ2 is between a and b and hence belongs to I.   ^[¯]

Defn. The absolute value of a real number a is defined by

Lemma. The absolute value function has the following properties:

1. |a|³ 0 for all a Î IR.
2. ±a £|a|
3. |-a| = |a|
4. |a·b| = |a|·|b|
5. |a+b|£|a| + |b|
6. | |b| - |a| |£|b-a|


Pf. To prove property 1, `case it out.' Either a ³ 0 or a < 0. In the first case |a| = a ³ 0. In the second case, |a| = -a > 0, since a < 0. Property 2 follows similarly. Properties 3 and 4 are left for the student.

To prove property 5, recall that either c : = a+b satisfies c ³ 0 or c < 0. In the first case, |c| = c = a+b £ |a| + |b|. In the second case, |c| = -c = -a + (-b) £ |a| + |b|. This establishes property 5.

To prove property 6, use the fact that |b| = |(b-a) + a| £|b-a| + |a| and subtract |a| from each side. This shows that |b| - |a| £|b-a|. By symmetry in a and b, one proves that |a| - |b| £|a-b| = |b-a|.  ^[¯]

1. Work problem 1.46 in the text.
2. Work problem 1.54 in the text.

1. Prove 0 < a < b that implies that a² < b².
2. Prove that the number a: = sup{x Î IR| x²< 2 } is the unique positive number that satisfies the equation a² = 2.
3. Give a definition for square roots of nonnegative real numbers.
4. Prove that |a| = [Ö(a²)] for all a Î IR.