📐 Today’s Lesson: Section 3 From: Principles of Analysis (Rudin) — Section 3

The Real Field

We now state the existence theorem which is the core of this chapter. The proof is rather long and a bit tedious, so it is presented in an Appendix to Chapter 1. The proof actually constructs R from Q.

📐 Theorem — 1.19 - Existence of the Real Field

The proof is presented in the Appendix to Chapter 1, where R is actually constructed from Q using Dedekind cuts.

There exists an ordered field R which has the least-upper-bound property.

Moreover, R contains Q as a subfield.

The second statement means that Q ⊂ R and that the operations of addition and multiplication in R, when applied to members of Q, coincide with the usual operations on rational numbers; also, the positive rational numbers are positive elements of R.

The members of R are called real numbers.

The next theorem could be extracted from the Dedekind cut construction with very little extra effort. However, we prefer to derive it from Theorem 1.19 since this provides a good illustration of what one can do with the least-upper-bound property.

📐 Theorem — 1.20 - The Archimedean Property and Density of Q

(a) Let A be the set of all nx, where n runs through the positive integers. If (a) were false, then y would be an upper bound of A. But then A has a least upper bound in R. Put α = A. Since x > 0, α - x < α, and α - x is not an upper bound of A. Hence α - x < mx for some positive integer m. But then α < (m + 1)x ∈ A, which is impossible, since α is an upper bound of A.

(b) Since x < y, we have y - x > 0, and (a) furnishes a positive integer n such that

n(y - x) > 1

Apply (a) again, to obtain positive integers m₁ and m₂ such that m₁ > nx, m₂ > -nx. Then

-m₂ < nx < m₁

Hence there is an integer m (with -m₂ ≤ m ≤ m₁) such that

m - 1 ≤ nx < m

If we combine these inequalities, we obtain

nx < m ≤ 1 + nx < ny

Since n > 0, it follows that x < m/n < y. This proves (b), with p = m/n.

(a) If x ∈ R, y ∈ R, and x > 0, then there is a positive integer n such that

nx > y

(b) If x ∈ R, y ∈ R, and x < y, then there exists a p ∈ Q such that x < p < y.

Part (a) is usually referred to as the archimedean property of R.

Part (b) may be stated by saying that Q is dense in R: Between any two real numbers there is a rational one.

We shall now prove the existence of nth roots of positive reals. This proof will show how the difficulty pointed out in the Introduction (the irrationality of √(2)) can be handled in R.

📐 Theorem — 1.21 - Existence of nth Roots

That there is at most one such y is clear, since 0 < y₁ < y₂ implies y₁ⁿ < y₂ⁿ.

Let E be the set consisting of all positive real numbers t such that tⁿ < x.

If t = x/(1 + x) then 0 ≤ t < 1. Hence tⁿ ≤ t < x. Thus t ∈ E, and E is not empty.

If t > 1 + x then tⁿ ≥ t > x, so that t ∉ E. Thus 1 + x is an upper bound of E.

Hence Theorem 1.19 implies the existence of

y = E

To prove that yⁿ = x we will show that each of the inequalities yⁿ < x and yⁿ > x leads to a contradiction.

The identity bⁿ - aⁿ = (b - a)(bⁿ⁻¹ + bⁿ⁻²a + ⋯ + aⁿ⁻¹) yields the inequality

bⁿ - aⁿ < (b - a)nbⁿ⁻¹

when 0 < a < b.

Assume yⁿ < x. Choose h so that 0 < h < 1 and

h < (x - yⁿ)/(n(y + 1)ⁿ⁻¹)

Put a = y, b = y + h. Then

(y + h)ⁿ - yⁿ < hn(y + h)ⁿ⁻¹ < hn(y + 1)ⁿ⁻¹ < x - yⁿ

Thus (y + h)ⁿ < x, and y + h ∈ E. Since y + h > y, this contradicts the fact that y is an upper bound of E.

Assume yⁿ > x. Put

k = (yⁿ - x)/nyⁿ⁻¹

Then 0 < k < y. If t ≥ y - k, we conclude that

yⁿ - tⁿ ≤ yⁿ - (y - k)ⁿ < knyⁿ⁻¹ = yⁿ - x

Thus tⁿ > x, and t ∉ E. It follows that y - k is an upper bound of E.

But y - k < y, which contradicts the fact that y is the least upper bound of E. Hence yⁿ = x, and the proof is complete.

For every real x > 0 and every integer n > 0 there is one and only one positive real y such that yⁿ = x.

This number y is written ⁿ√(x) or x¹^/ⁿ.

📐 Theorem — Corollary - Product Rule for Roots

Put α = a¹^/ⁿ, β = b¹^/ⁿ. Then

ab = αⁿβⁿ = (αβ)ⁿ

since multiplication is commutative. [Axiom (M2) in Definition 1.12.] The uniqueness assertion of Theorem 1.21 shows therefore that

(ab)¹^/ⁿ = αβ = a¹^/ⁿb¹^/ⁿ

If a and b are positive real numbers and n is a positive integer, then

(ab)¹^/ⁿ = a¹^/ⁿb¹^/ⁿ

We conclude this section by pointing out the relation between real numbers and decimals.

Let x > 0 be real. Let n₀ be the largest integer such that n₀ ≤ x. (Note that the existence of n₀ depends on the archimedean property of R.) Having chosen n₀, n₁, …, nₖ₋₁, let nₖ be the largest integer such that

n₀ + n₁/10 + ⋯ + nₖ/10ᵏ ≤ x

Let E be the set of these numbers. Then x = E. The decimal expansion of x is

n₀.n₁n₂n₃⋯

Conversely, for any infinite decimal, the set E of numbers as above is bounded above, and the decimal is the decimal expansion of E.

Key Takeaways

• There exists an ordered field R with the least-upper-bound property • Q is contained in R as a subfield • The Archimedean property: for any x > 0 and any y, there exists n with nx > y • Q is dense in R: between any two reals there is a rational • Every positive real has a unique positive nth root • Real numbers correspond to infinite decimal expansions

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