Our server costs ~$56 per month to run. Please consider donating or becoming a Patron to help keep the site running. Help us gain new members by following us on Twitter and liking our page on Facebook!
Current time: March 29, 2024, 4:39 am

Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Mathematicians who are finitists.
#81
RE: Mathematicians who are finitists.
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)
I am the Infantry. I am my country’s strength in war, her deterrent in peace. I am the heart of the fight… wherever, whenever. I carry America’s faith and honor against her enemies. I am the Queen of Battle. I am what my country expects me to be, the best trained Soldier in the world. In the race for victory, I am swift, determined, and courageous, armed with a fierce will to win. Never will I fail my country’s trust. Always I fight on…through the foe, to the objective, to triumph overall. If necessary, I will fight to my death. By my steadfast courage, I have won more than 200 years of freedom. I yield not to weakness, to hunger, to cowardice, to fatigue, to superior odds, For I am mentally tough, physically strong, and morally straight. I forsake not, my country, my mission, my comrades, my sacred duty. I am relentless. I am always there, now and forever. I AM THE INFANTRY! FOLLOW ME!
Reply
#82
RE: Mathematicians who are finitists.
(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?
Reply
#83
RE: Mathematicians who are finitists.
(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.
Reply
#84
RE: Mathematicians who are finitists.
(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.
Reply
#85
RE: Mathematicians who are finitists.
(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.
Reply
#86
RE: Mathematicians who are finitists.
(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?
Reply
#87
RE: Mathematicians who are finitists.
(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.
Reply
#88
RE: Mathematicians who are finitists.
(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]

[Image: Rosen-2.jpg]

[Image: Rosen-3.jpg]
Reply
#89
RE: Mathematicians who are finitists.
(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?
Reply
#90
RE: Mathematicians who are finitists.
(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.
Reply



Possibly Related Threads...
Thread Author Replies Views Last Post
  Question for finitists -- 0.999... = 1? Jehanne 23 3041 November 26, 2022 at 8:40 pm
Last Post: Jehanne



Users browsing this thread: 1 Guest(s)