My Math Forum Is an Infinite Binary Sequence a Natural Number?

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July 17th, 2017, 09:31 AM   #31
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Quote:
 Originally Posted by zylo Both infinite sequences can be uniquely associated with a natural number.
Sure, that's what you SAY! I want you to actually PROVE it.

July 17th, 2017, 01:46 PM   #32
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Quote:
 Originally Posted by Micrm@ss Sure, that's what you SAY! I want you to actually PROVE it.
Definition of binary (decimal) representation of a natural number.

It seems you believe there is a largest natural number.

July 17th, 2017, 01:54 PM   #33
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Quote:
 Originally Posted by zylo Definition of binary (decimal) representation of a natural number.
OK, please give me that definition. Then prove me that every decimal representation corresponds to a natural number in a unique way and vice versa.

Come on, don't just state things. Actually prove them too! I'd prefer if you'd rely on the Peano axioms alone, but hey, it's up to you.

July 17th, 2017, 02:39 PM   #34
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Quote:
 Originally Posted by Micrm@ss OK, please give me that definition. Then prove me that every decimal representation corresponds to a natural number in a unique way and vice versa. Come on, don't just state things. Actually prove them too! I'd prefer if you'd rely on the Peano axioms alone, but hey, it's up to you.
As I said before, it's a definition.

Quote:
 Originally Posted by zylo The infinite binary (or decimal) sequence $\displaystyle a_{1}a_{2}a_{3}.......$ defines a unique natural number or unique binary fraction, depending on interpretation: Natural Number: $\displaystyle \sum a_{n}2^{n}$ which approaches a unique natural number as n -> infnity. Binary Fraction: $\displaystyle \sum a_{n}2^{-n}$ which approaches a unique fraction in [0,1) as n -> infinity. Note the unique association between natural number and binary fraction.

If any a$\displaystyle _{n}$'s are different, the N's are different.

If you can conceive of an arbitrary small $\displaystyle \delta$, then you should be able to conceive of an arbitrarily large 1/$\displaystyle \delta$.

What is pi? 3.141592654......, by definition. It's the unique infinite sequence(s) that define pi.

EDIT:

If N=$\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}$ then N+1 is of the same form.
If N=$\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}$ is a natural number, then so is it for n=n+1.
By elementary arithmetic.

Last edited by zylo; July 17th, 2017 at 03:05 PM.

July 17th, 2017, 03:12 PM   #35
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Quote:
 Originally Posted by zylo What is pi? 3.141592654......, by definition.
That's a rather bad definition since it doesn't define pi uniquely. The problem is that we don't know all the digits of pi, so we shouldn't define pi by its decimal expansion. Not saying that pi doesn't have a decimal expansion, it's just not possible to define pi this way.

So perhaps let us start this way. What is your definition of the natural numbers and of the real numbers?

Last edited by skipjack; July 17th, 2017 at 03:17 PM.

 July 17th, 2017, 03:18 PM #36 Global Moderator   Joined: Dec 2006 Posts: 19,718 Thanks: 1806 As you, zylo, referred to a definition to support your assertion, the onus is on you to provide that definition, not Micrm@ss. I don't know of any accepted definition that supports your assertion that any infinite binary sequence defines a unique natural number. Your mention of pi isn't relevant in this matter, as pi isn't a natural number (it's a real number, though).
July 17th, 2017, 05:27 PM   #37
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Quote:
 Originally Posted by skipjack I don't know of any accepted definition that supports your assertion that any infinite binary sequence defines a unique natural number.
Due Cantor's Diagonal Argument, at best zylo you can only have a surjection from the infinite binary sequences onto the natural numbers that is not 1-to-1.

Naturally, there will be an infinite number of binary sequences for each natural number. There are infinitely many surjective mappings from the set of infinite binary sequences onto the natural numbers.

Instead of trying to contradict Cantor's Diagonal Argument (which is still your underlying focus here...), I challenge you to try and do the opposite. Try to prove to yourself that it is true. And by that, I do not simply mean for you to try and believe in the theorem, I mean for you to try and prove to your own satisfaction that it is true.

In the end, you will never disprove Cantor's Theorem. It's not possible to disprove. Your best shot is to wage war on any use of the axiom of infinity.

Set theory's current state of affairs is one of "assume a certain axiom or conjecture, and then X is (not) provable, but don't, and then Y is (not) provable." The continuum hypothesis is one example. Given the standard axioms, we can neither prove nor disprove it. We can conjecture the hypothesis is true and derive certain results, but we can likewise conjecture it is false and derive other results. In a sense, it isn't math any more. Hugh Woodin makes a similar comment in his lecture here (he is an extremely respected set theorist).

My personal intuition is that the axiom of infinity, once accepted, will always lead to uncertainty. Compare that intuition with Gödel's incompleteness theorems showing that a complete and consistent set of axioms for all of mathematics is not possible.

If you want to waste your life slaving away trying to disprove a theorem you simply cannot disprove, then I'm sorry to hear that. Your motivation for doing so cannot be sane, quite frankly. It's like flapping your arms because you're convinced you can fly.

But, if you want to try and get onto the forefront of set theory, perhaps there is something more left to prove. Find a resolution to Scott's theorem by showing whether or not a measurable cardinal exists (ie, if there exists a measurable cardinal, then $V \neq L$). Perhaps that would net you a Nobel Prize. In the very least, your name would go down in history (that appears to be your delusional goal with respect to disproving Cantor's Theorem...).

July 17th, 2017, 07:16 PM   #38
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Quote:
 Originally Posted by AplanisTophet Perhaps that would net you a Nobel Prize.
Somewhat unlikely, since there is no Nobel Prize for mathematics.

July 17th, 2017, 07:18 PM   #39
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Quote:
 Originally Posted by zylo As I said before, it's a definition.
It's not anybody's definition but your own. It doesn't agree with anybody else's definition either, so whatever results you get from it are nothing to do with anybody else's mathematics: specifically Cantor's.

July 17th, 2017, 07:19 PM   #40
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Quote:
 Originally Posted by v8archie Somewhat unlikely, since there is no Nobel Prize for mathematics.
Fields Medal perhaps then.

There have been a number of prizes awarded to mathematicians, but these were for their contributions to Physics. Wooden also discusses set theory's potential application to Physics in the above lecture, but I have my doubts...

http://www-history.mcs.st-andrews.ac...urs/Nobel.html

Last edited by AplanisTophet; July 17th, 2017 at 07:22 PM.

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