# Math Help - Archimedes Postulate

1. ## Archimedes Postulate

If a & b are any positive real numbers, integer n exists st na > b.

Proof:
Rational positive numbers ra and rb exist st ra < a and rb > b by rational cut definition of real numbers.
Then n > rb/ra = Nb/Na → na > b.*
Nb/Na = (1/Na)Nb <= Nb < Nb+1 (Taylor)
n = Nb+1

*na > nra, & n > rb/ra → nra > rb > b, → na > b.

2. ## Re: Archimedes Postulate

Originally Posted by Hartlw
If a & b are any positive real numbers, integer n exists st na > b.
Proof:
Rational positive numbers ra and rb exist st ra < a and rb > b by rational cut definition of real numbers.
Then n > rb/ra = Nb/Na → na > b.*
Nb/Na = (1/Na)Nb <= Nb < Nb+1 (Taylor)
n = Nb+1

*na > nra, & n > rb/ra → nra > rb > b, → na > b.
If you know that $\mathbb{N}$ is not bounded above then $\exists n\in\mathbb{N}$ such that $n>\frac{b}{a}$.

3. ## Re: Archimedes Postulate

Originally Posted by Plato
If you know that $\mathbb{N}$ is not bounded above then $\exists n\in\mathbb{N}$ such that $n>\frac{b}{a}$.
But you don't know that b/a is bounded above. And furthermore, I didn't assume a definition of b/a, which is a big step up in complexity.

If you want to assume that the real numbers are defined as a field by rational cuts, then:

For any real number, there is a rational number Np/Nq st Np/Nq > x.
But Np/Nq = (1/Nq) Np <= Np < Np+1. So for n = Np+1, n>x.

Then n > b/a.

4. ## Re: Archimedes Postulate

Originally Posted by Hartlw
But you don't know that b/a is bounded above.

Once again you have totally missed the point.

$\frac{b}{a}$ is not an upper bound for $\mathbb{N}$ which has no upper bound.

So $(\exists n\in\matbb{N})\left[n>\frac{b}{a}\right].$

5. ## Re: Archimedes Postulate

b/a and N are both potentially unbounded, so your point makes no sense. It only makes sense if all b/a are bounded, which is not the case. Perhaps you missed my post, to which I added an edit showing the correct proof.

Originally Posted by Hartlw
But you don't know that b/a is bounded above. And furthermore, I didn't assume a definition of b/a, which is a big step up in complexity.

If you want to assume that the real numbers are defined as a field by rational cuts, then:

For any real number, there is a rational number Np/Nq st Np/Nq > x.
But Np/Nq = (1/Nq) Np <= Np < Np+1. So for n = Np+1, n>x.

Then n > b/a.
EDIT: The fact that N is unbounded means that given any n an N exists st N > n.

6. ## Re: Archimedes Postulate

If a & b are any positive real numbers, integer n exists st na > b.
Proof:
Rational positive numbers ra and rb exist st ra < a and rb > b by rational cut definition of real numbers.
Then n > rb/ra = Nb/Na → na > b.*
Nb/Na = (1/Na)Nb <= Nb < Nb+1 (Taylor)
n = Nb+1

*na > nra, & n > rb/ra → nra > rb > b, → na > b.[/QUOTE]

That proof is incorrrect.

Originally Posted by Hartlw
b/a and N are both potentially unbounded, so your point makes no sense. It only makes sense if all b/a are bounded, which is not the case. Perhaps you missed my post, to which I added an edit showing the correct proof.
I understand how much want to understand these concepts.
But you do not. You don't even know what bounded means.

7. ## Re: Archimedes Postulate

Referring to your last post, the proof is correct, and the bounded statement is correct. What is wrong with the proof or the unbounded statement?

Apparently mathematicians since Archimedes have missed your proof that N > b/a because N is unbounded.

EDIT: And I wasn't asking a question, thank you, which is why I posted in this forum. Constructive criticism (excluding insults) is welcome.

The proofs I have seen used a second cut to create a contradiction, and I thought one cut was complication enough, which is why I did it this way and decided to pass it along because it was simpler. After you assume rational cut definition of real numbers, the rest is simple algebra,

8. ## Re: Archimedes Postulate

Originally Posted by Plato
If you know that $\mathbb{N}$ is not bounded above then $\exists n\in\mathbb{N}$ such that $n>\frac{b}{a}$.
Incorrect proof because b/a is also not bounded above. To the extent it is true, it is ultimateley true because every integer has a successor.

9. ## Re: Archimedes Postulate

Originally Posted by Hartlw
Incorrect proof because b/a is also not bounded above. To the extent it is true, it is ultimately true because every integer has a successor.
I know that you must really be enjoying your trolling today.

10. ## Re: Archimedes Postulate

Originally Posted by Hartlw
Incorrect proof because b/a is also not bounded above. To the extent it is true, it is ultimateley true because every integer has a successor.
"To the extent it is true.." because, if I challenge Plato to give me a number larger than n he comes back with n+1, then I give him n+2 and he comes back with n+3, ............

EDIT: By the principle of mathematical induction, neither side wins. I can always come up with n+1 and so can Plato. It is truly a paradox.

11. ## Re: Archimedes Postulate

Originally Posted by Hartlw
"To the extent it is true.." because, if I challenge Plato to give me a number larger than n he comes back with n+1, then I give him n+2 and he comes back with n+3, ............

EDIT: By the principle of mathematical induction, neither side wins. I can always come up with n+1 and so can Plato. It is truly a paradox.
Once again missing the point. Correct me if I'm wrong, but you are given a and b. So you can form the rational number (a neq 0) b/a. Now, $\mathbb{N}$ is unbounded, which means we can always pick a number $n > b/a$ for any given a and b.

Example. We are given a = 10 and b = 374. Thus b/a is 37.4. We may now pick a number n = 40 which is larger than that.

@hartlw: Why is this so hard for you to see?

-Dan

12. ## Re: Archimedes Postulate

OK, I give you 41. This is going to be a very long thread.

Edit: You would be correct if the postulate stated b/a was bounded. It does not, it is ANY b/a.

13. ## Re: Archimedes Postulate

Originally Posted by Hartlw
Edit: You would be correct if the postulate stated b/a was bounded. It does not, it is ANY b/a.

Look at the original post.
Originally Posted by Hartlw
If a & b are any positive real numbers, integer n exists st na > b.

The title of the thread is misleading. The words axiom and postulate are used interchangeably. We do not prove axioms. The statement "If a & b are any positive real numbers, integer n exists st na > b." is almost universality known as The Archimedean Principle( or property). It can be proven from one simple property of the natural numbers: the natural numbers are not bounded.

The importance of this principle is that given any two positive real numbers a & b no matter how small a is nor no matter how large b is, then the number a can be added to itself enough times to form a sum that is greater than b. Thus every infinite series of constants diverges.

14. ## Re: Archimedes Postulate

Archimedes postulate is also called axiom, principle. Postulates (axioms, postulates) can be disproved. I have done so. I note that it is the opinion of others in this thread that I have not. Simply rewording the postulate does not prove I am wrong.

15. ## Re: Archimedes Postulate

Originally Posted by Hartlw
OK, I give you 41. This is going to be a very long thread.

Edit: You would be correct if the postulate stated b/a was bounded. It does not, it is ANY b/a.
Yes, but we are given a and b from the start. Given appropriate values of a and b we can always find such an n. This is what the postulate (axiom, whatever) is talking about.

-Dan

PS
Okay, let's start with Plato's first post. Let's take a microstep here. Do you know what it means for the set $\mathbb{N}$ to be countably infinite?

-Dan

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