How is my proof of the Archimedean property?

In summary: And since M is an upper bound, it must be that M≥b. Therefore, M=b. This is not a valid argument. Just because there is an upper bound does not mean there is a least upper bound. Consider the set S={x∈R | 0≤x<1}. This set has an upper bound, 1, but it does not have a least upper bound. In order to prove that b is the least upper bound, you must show that it is an upper bound and that it is the smallest upper bound. This requires a more detailed proof. In summary, the conversation discusses how to prove the Archimedean property and how to correctly phrase and understand the hypothesis. The correct hypothesis
  • #1
Jamin2112
986
12

Homework Statement



Prove the Archimedean property

Homework Equations



Know what a least upper bound is

The Attempt at a Solution



Assume that if a and b are positive real numbers, na≤b for all natural numbers n. Then the set S of all numbers na, where n is a natural number, has b as its least upper bound.

Let n' be a natural number such that b-∂ < ∂n' ≤ b. Then b < ∂(n'+1). Since n' is a natural number, n'+1 is a natural number, and so ∂(n'+1) is an element of S. But since b < ∂(n'+1), S cannot have b as its least upper bound, and we have a contradiction.
 
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  • #2
You have the right basic idea, but several details and phrasings are wrong.

Jamin2112 said:
Assume that if a and b are positive real numbers, na≤b for all natural numbers n.
This hypothesis has the quantifier in the wrong place; what you said is "suppose that, for all positive real numbers [tex]a[/tex] and [tex]b[/tex] and all natural numbers [tex]n[/tex], we have [tex]na \leq b[/tex]". That's not the negation of the archimedean property. The correct hypothesis is "suppose that [tex]a[/tex] and [tex]b[/tex] are [specific] positive real numbers such that, for all natural numbers [tex]n[/tex], we have [tex]na \leq b[/tex]."

Jamin2112 said:
Then the set S of all numbers na, where n is a natural number, has b as its least upper bound.

This is a false deduction. You cannot conclude that [tex]\sup S = b[/tex], only that [tex]\sup S \leq b[/tex], because [tex]b[/tex] is an upper bound for [tex]S[/tex].

Jamin2112 said:
Let n' be a natural number such that b-∂ < ∂n' ≤ b. Then b < ∂(n'+1). Since n' is a natural number, n'+1 is a natural number, and so ∂(n'+1) is an element of S. But since b < ∂(n'+1), S cannot have b as its least upper bound, and we have a contradiction.

I'm not sure where the [tex]\partial[/tex] symbol came from. But if you substitute [tex]\sup S[/tex] for [tex]b[/tex], and [tex]a[/tex] for [tex]\partial[/tex], the rest of the argument is correct.
 
  • #3
Hmmmm ... Here's how my handout defines "upper bound" and "least upper bound":

Suppose that F is an ordered field and S is a nonempty subset of F. An upper bound for S is any element M of F so that x≤M for all x in S. A least upper bound for S is any element L of F which is an upper bound for S and which also has the property that every a<L is not and upper bound for S: L is the smallest upper bound for S.



So if I start with the assumption (rewritten the correct way)

"Suppose for all integers n, an≤b for some integers a and b."

Looking at the definitions, I see that any M≥b is an upper bound for the set S={na for all integers n and some positive real number a}. The least upper bound, of course, would be b.

Right?
 
  • #4
Jamin2112 said:
So if I start with the assumption (rewritten the correct way)

"Suppose for all integers n, an≤b for some integers a and b."

This is also not right. Perhaps it's easier to see if you write it in symbols. Abbreviate the set of positive reals as [tex]\mathbb{R}_{>0}[/tex].

Your first attempt at the hypothesis was: [tex]\forall a \in \mathbb{R}_{>0}.\, \forall b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}.\, an \leq b[/tex].

Your second attempt translates to: [tex]\forall n \in \mathbb{N}. \, \exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, an \leq b[/tex]. You can tell this is not the right hypothesis, because it's true: for any natural number [tex]n > 0[/tex], one may choose [tex]a = \frac1{2n}, b = 1[/tex], and then [tex]an = \frac12 < 1 = b[/tex]. (For [tex]n = 0[/tex] just choose [tex] a = b = 1[/tex].)

The correct hypothesis is the negation of the archimedean property. The archimedean property is [tex]\forall a \in \mathbb{R}_{>0}. \, \forall b \in \mathbb{R}_{>0}. \, \exists n \in \mathbb{N}. \, an > b[/tex], so its negation is [tex]\exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}. \, an \leq b[/tex]. If you don't know how to negate statements with strings of quantifiers like this, you need to learn before you go much further in analysis.

Jamin2112 said:
Looking at the definitions, I see that any M≥b is an upper bound for the set S={na for all integers n and some positive real number a}. The least upper bound, of course, would be b.

Right?

No. Just because [tex]b[/tex] is the smallest upper bound you see in front of your face does not mean it is actually the least upper bound. You must prove that any other upper bound is at least [tex]b[/tex], and you can't do that from the hypotheses at hand.
 
  • #5
ystael said:
No. Just because [tex]b[/tex] is the smallest upper bound you see in front of your face does not mean it is actually the least upper bound. You must prove that any other upper bound is at least [tex]b[/tex], and you can't do that from the hypotheses at hand.

Hmmm ... but since there is an upper bound b, there must a least upper bound M≤b.
 

Related to How is my proof of the Archimedean property?

1. What is the Archimedean property?

The Archimedean property is a mathematical principle that states that for any two real numbers, there exists a natural number that is greater than their product.

2. How is the Archimedean property used in mathematics?

The Archimedean property is used in various areas of mathematics, including calculus, real analysis, and number theory. It helps to provide a basis for many important proofs and theorems.

3. How is the Archimedean property proven?

The proof of the Archimedean property is typically done using a contradiction argument, assuming that the property does not hold and then showing that it leads to a contradiction. It can also be proven using the completeness axiom in the real numbers.

4. Can the Archimedean property be generalized to other number systems?

Yes, the Archimedean property can be generalized to other number systems, such as the rational numbers, complex numbers, and even certain infinite-dimensional vector spaces. However, the exact formulation may vary slightly depending on the number system.

5. Why is the Archimedean property important in mathematics?

The Archimedean property is important in mathematics because it helps to establish the order and structure of the real numbers. It also plays a crucial role in many foundational theorems and concepts, such as the intermediate value theorem and the completeness axiom.

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