Login

**The Grand Nudger** · (This post was last modified: July 5, 2019 at 2:02 pm by The Grand Nudger.)

I have no clue about that, but I do know that even rational numbers can have an infinite amount of numerals in decimal representation, it’s just that they repeat.

(Point being, that while it’s an interesting question, it may not shed light on the issue one way or another)

polymath257 · July 7, 2019 at 3:38 am

(July 5, 2019 at 1:33 am)A Toy Windmill Wrote:
(July 4, 2019 at 6:46 pm)Jehanne Wrote: The proof of the irrationality of the square root of 2 by contradiction is what I had in mind, anything but finitism in my opinion.
That proof looks straightforwardly finitistic to me. Where do you think it requires quantification over all the natural numbers?

Well, you need to know that the square of every even number is even, that the square of every odd number is odd, that every natural number is either even or odd, and that irrationality is the non-existence of two natural numbers x,y with x^2 =2 y^2.

The last quantification is probably the one that breaks the proof since I don't see how to get by with just bounded quantification in it.

Another aspect of this is that the existence of sqrt(2) is not guaranteed. All this proof does is show that it cannot be rational.

I'm not sure a purely finitistic proof of the existence of sqrt(2) is possible. Anyone?

A Toy Windmill · (This post was last modified: July 7, 2019 at 6:33 am by A Toy Windmill.)

(July 7, 2019 at 3:38 am)polymath257 Wrote: Well, you need to know that the square of every even number is even, that the square of every odd number is odd, that every natural number is either even or odd, and that irrationality is the non-existence of two natural numbers x,y with x^2 =2 y^2.

The first three claims can be expressed as simple equations on a primitive recursive quotient function.

Quote:The last quantification is probably the one that breaks the proof since I don't see how to get by with just bounded quantification in it.

Yes. Here's as much as I'm willing to claim:

(July 4, 2019 at 11:27 am)A Toy Windmill Wrote:
(July 4, 2019 at 7:33 am)Jehanne Wrote: Do finistists accept the existence of irrational numbers?
I don't think so, at least not in the sense that everyone else does. The classic proof that a square's diagonal is incommensurable with its side looks finitistic to me, but that's not what a modern mathematician means by "the square root of 2 exists."

The classic proof is just that the square of a rational cannot be 2. This was the only commitment of the classical Greeks and those for followed them, and I don't think they're bad company.

Quote:I'm not sure a purely finitistic proof of the existence of sqrt(2) is possible. Anyone?

I'm confident that the standard rational sequence whose squares converge to 2 is finitistically provably such. Again, this isn't as much as a classical mathematician can assert today, which is that there is a unique positive object which squares to 2 in any interpretation of the axioms of closed ordered fields, of which Dedekinds cuts are an example.

Jehanne · July 8, 2019 at 7:05 am

(July 7, 2019 at 6:22 am)A Toy Windmill Wrote: I'm confident that the standard rational sequence whose squares converge to 2 is finitistically provably such. Again, this isn't as much as a classical mathematician can assert today, which is that there is a unique positive object which squares to 2 in any interpretation of the axioms of closed ordered fields, of which Dedekinds cuts are an example.

Your ticket to fame, perhaps? On the other hand, Mike Pence once said on the floor of the United States Senate that he believed that Science would someday vindicate ID. "Someday" is, of course, a long, long time.

A Toy Windmill · July 8, 2019 at 8:19 am

(July 8, 2019 at 7:05 am)Jehanne Wrote: Your ticket to fame, perhaps? On the other hand, Mike Pence once said on the floor of the United States Senate that he believed that Science would someday vindicate ID. "Someday" is, of course, a long, long time.

You make me sad.

Take the following primitive recursive function

x_0 = 1
x_{n+1} = 1 + 1 / (1 + x_n)

I claim that |x_n^2 - 2| <= 1/2^n. By induction:

Base case: |x_0^2 - 2| = 1 <= 1/1
Step case:

|x_{n+1}^2 - 2| = |(x_n^2 - 2) / (1 + x_n)^2| <= (1/2^n) / (1 + x_n)^2 < 1/2^{n+1}.

All the rational algebra here is finitistic.

Next, for any rational epsilon ε > 0, find the smallest power p of 2 such that 2^p > 1/ε. This can be found with another primitive recursive function. We then have

x_p^2 - 2 <= 1/2^p < ε.

Jehanne · July 8, 2019 at 3:38 pm

(July 8, 2019 at 8:19 am)A Toy Windmill Wrote:
(July 8, 2019 at 7:05 am)Jehanne Wrote: Your ticket to fame, perhaps? On the other hand, Mike Pence once said on the floor of the United States Senate that he believed that Science would someday vindicate ID. "Someday" is, of course, a long, long time.
You make me sad.

Take the following primitive recursive function

x_0 = 1
x_{n+1} = 1 + 1 / (1 + x_n)

I claim that |x_n^2 - 2| <= 1/2^n. By induction:

Base case: |x_0^2 - 2| = 1 <= 1/1
Step case:

|x_{n+1}^2 - 2| = |(x_n^2 - 2) / (1 + x_n)^2| <= (1/2^n) / (1 + x_n)^2 < 1/2^{n+1}.

All the rational algebra here is finitistic.

Next, for any rational epsilon ε > 0, find the smallest power p of 2 such that 2^p > 1/ε. This can be found with another primitive recursive function. We then have

x_p^2 - 2 <= 1/2^p < ε.

As p goes to infinity? Define your universe for p?

A Toy Windmill · (This post was last modified: July 9, 2019 at 3:50 am by A Toy Windmill.)

(July 8, 2019 at 3:38 pm)Jehanne Wrote: As p goes to infinity? Define your universe for p?

p does not go anywhere. It would not even appear in the formalized theorem.

The only variables in the formalized theorem are the numerator and denominator of ε. p = f(ε) where f is the primitive recursive function which sends n to the smallest p such that 2^p > n. We could, however, just take f to be the successor function, and apply it to the numerator of ε. The proof would be fine, since 2^{m+1} >= 2^{(m + 1) / n} > n.

Jehanne · July 9, 2019 at 7:51 am

(July 9, 2019 at 2:54 am)A Toy Windmill Wrote:
(July 8, 2019 at 3:38 pm)Jehanne Wrote: As p goes to infinity? Define your universe for p?
p does not go anywhere. It would not even appear in the formalized theorem.

The only variables in the formalized theorem are the numerator and denominator of ε. p = f(ε) where f is the primitive recursive function which sends n to the smallest p such that 2^p > n. We could, however, just take f to be the successor function, and apply it to the numerator of ε. The proof would be fine, since 2^{m+1} >= 2^{(m + 1) / n} > n.

Here's what Professor Kenneth Rosen of Columbia University has to say about the (weak) principle of mathematical induction in his book, Discrete Mathematics, 8th edition:

[Image: Rosen-1.jpg]

polymath257 · July 9, 2019 at 9:56 am

(July 8, 2019 at 8:19 am)A Toy Windmill Wrote:
(July 8, 2019 at 7:05 am)Jehanne Wrote: Your ticket to fame, perhaps? On the other hand, Mike Pence once said on the floor of the United States Senate that he believed that Science would someday vindicate ID. "Someday" is, of course, a long, long time.
You make me sad.

Take the following primitive recursive function

x_0 = 1
x_{n+1} = 1 + 1 / (1 + x_n)

I claim that |x_n^2 - 2| <= 1/2^n. By induction:

Base case: |x_0^2 - 2| = 1 <= 1/1
Step case:

|x_{n+1}^2 - 2| = |(x_n^2 - 2) / (1 + x_n)^2| <= (1/2^n) / (1 + x_n)^2 < 1/2^{n+1}.

All the rational algebra here is finitistic.

Next, for any rational epsilon ε > 0, find the smallest power p of 2 such that 2^p > 1/ε. This can be found with another primitive recursive function. We then have

x_p^2 - 2 <= 1/2^p < ε.

After thinking a bit, this does not quite prove the convergence since you need to show that
|x_n^2 -2|<eps for ALL n>=p where p is that smallest exponent.

Is there a finitistic definition that gets around this issue?

A Toy Windmill · July 9, 2019 at 1:34 pm

(July 9, 2019 at 9:56 am)polymath257 Wrote: After thinking a bit, this does not quite prove the convergence since you need to show that
|x_n^2 -2|<eps for ALL n>=p where p is that smallest exponent.

Is there a finitistic definition that gets around this issue?

Yes, you're right, but that use of a quantifier is still legal.

n >= p implies that |x_n^2 - 2| <= 1/2^n <= 1/2^p < ε.

Login
Username:
Password:	Lost Password?
	Remember me

Possibly Related Threads...
Thread		Author	Replies	Views	Last Post
	Question for finitists -- 0.999... = 1?	Jehanne	23	4536	November 26, 2022 at 8:40 pm Last Post: Jehanne