The Axiom of Completeness for the real numbers says: Every...

Name: Problem 83, Chapter 0: Calculus
Uploaded: 2021-06-11T17:30:19-08:00
Duration: 3 min 55 s
Channel: Nick Johnson
Description: The Axiom of Completeness for the real numbers says: Every set of real numbers that has an upper bound has a least upper bound that is a real number. (a) Show that the italicized statement is false if the word real is replaced by rational. (b) Would the italicized statement be true or false if the word real were replaced by natural?

The Axiom of Completeness for the real numbers says: Every set of real numbers that has an upper bound has a least upper bound that is a real number. (a) Show that the italicized statement is false if the word real is replaced by rational. (b) Would the italicized statement be true or false if the word real were replaced by natural?

The Axiom of Completeness for the real numbers says: Every set of real numbers that has an upper bound has a least upper bound that is a real number.
(a) Show that the italicized statement is false if the word real is replaced by rational.
(b) Would the italicized statement be true or false if the word real were replaced by natural?

Key Concepts

Completeness Axiom

This axiom states that every nonempty set in an ordered field that is bounded above must have a least upper bound (or supremum) within that field. It is a defining property of the real numbers and ensures there are no gaps in the number system.

Supremum (Least Upper Bound) Property

This concept asserts that for any nonempty set that is bounded above, there exists a smallest number among all upper bounds. This smallest number is called the supremum and guarantees that the set 'approaches' a specific bound within the number system.

Incompleteness of the Rationals

When the least upper bound property is applied to the rational numbers, it fails because there are bounded sets of rationals whose supremum is irrational. This shows that the rationals have gaps, meaning not every bounded set has a least upper bound within the rationals.

Discrete Order of the Natural Numbers

Due to the discrete nature of the natural numbers, any subset of natural numbers that is bounded above is finite, and consequently always has a maximum element. This maximum element serves as the least upper bound, making the completeness axiom hold in a trivial way for the natural numbers.

Transcript

00:01 Okay, so here for part a, we can prove that this statement is false by finding an example of a set of rational numbers that has an upper bound, but that does not have a least upper bound, or that has an upper bound that is not a rational number.

00:20 So we can start by creating an infinite set comprised entirely of just rational numbers, and then the infinite set we choose can be comprised.

00:30 Of truncations of an irrational number.

00:36 For instance, the square root of 2.

00:40 So let's go ahead.

00:42 Let's let our set s be equal to the set of, well, truncations here.

00:51 So we have, let's say, 1 .4 and then 1 .41, and then 1 .4 .1, and then 1 .4 .1.

01:01 414, and then we'd have 1 .442, and so on, this truncations of the square root of 2.

01:13 And then since every member of this set is a terminating decimal, we know that each is going to definitely be a rational number.

01:22 So we can certainly obtain one or more valid upper bounds for this set.

01:27 For example, five or three, right, would be upper bounds.

01:36 So the least upper bound of this set, the least upper bound, least upper bound of this set is going to be the irrational number square root of two.

01:48 So here we have a set comprised entirely of rational numbers that has an upper bound, but has a least upper bound here that is not a rational number.

02:03 So this is going to then disprove the original statement...

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