Author Topic: Arithmetic for Cavebear  (Read 182 times)

Offline Cavebear

Re: Arithmetic for Cavebear
« Reply #15 on: November 07, 2017, 01:47:23 PM »
Excellent ... such objects are the grist of civilization.  Utopian politics, less so ... all failures.

Some maths isn't failed, it is impossible.  But the gullible are taken in by math and science charlatans.

Thank you.  So may I be forgiven that at age 15 I did not provide a 600 page proof that "1+1=2"?
Atheist born, atheist bred.  And when I die, atheist dead!  b 1950

Offline Baruch (OP)

Re: Arithmetic for Cavebear
« Reply #16 on: November 07, 2017, 08:08:49 PM »
Thank you.  So may I be forgiven that at age 15 I did not provide a 600 page proof that "1+1=2"?

Proof is over-rated.  Pythagoras went around asking girls, you want to see my equations?
שלום

Offline Baruch (OP)

Re: Arithmetic for Cavebear
« Reply #17 on: November 07, 2017, 08:16:12 PM »
There are numbers that are inaccessible to any algorithm?   Really?

Yes, an algorithm just gives a name to a point in the number line etc.  it is not possible to name all the points, because the number of points is an infinity greater than the infinity of rational numbers 1432.234234 (terminates or repeats at some point).  The continuum is the next larger, incommensurable infinity.  See Georg Cantor.  A non-terminating digital number isn't a proper name (say Pi) ... you can give a name (Pi) to such a number, but there aren't an infinite number of them, and aside from the digital representation of the rational numbers ... there is no other system of naming an infinite collection.  You can't simply reuse the rational numbers you already used ... and in fact, you would need to cheat that way, not just once, but an infinite number of times.  The number of rational numbers is a really big number, but the continuum is infinitely bigger than that.

Now it doesn't matter which number system you use.  The simplest is binary .. but using decimal is simply an alternative naming scheme for the same points.
שלום

Offline Baruch (OP)

Re: Arithmetic for Cavebear
« Reply #18 on: November 07, 2017, 08:23:55 PM »
The last digit of pi?

All transcendental numbers are like that.  And in a practical way, you can say that there is an algorithm for calculating the next digit of Pi ... but not the last one.  But there are transcendental numbers where you can't even calculate the first digit (the normal transcendental).  Not very practical, but still there.  Here is how that works ... in an algorithm to calculate a number, the answer either converges or diverges.

A good algorithm not only converges but does so rapidly.  Pi is poorly behaved, it converges, but slowly.  Consider if there is number, such that there is not only no converging algorithm that converges rapidly, but that there is no possible algorithm that converges at all (this is at the extreme limit).  That is a normal transcendental.  A ordinary transcendental like Pi is fortunately useful.

This also appears in computer science.  There is the halting problem ... can I tell if a given computer, set up and running, will terminate?  In general you can't say it does, just in particular cases.  So consider an algorithm that if it halts, generates the number 2 ... but it is of the type that you can't tell it will ever terminate.  In that case, the time to the generation of the first digit is indefinite.
שלום

Re: Arithmetic for Cavebear
« Reply #19 on: November 08, 2017, 03:21:27 PM »
Transcendental numbers are non-algebraic, that can't ever be the solution to an algebraic equation. But how can they be identified otherwise? Is there a method of determining whether a given irrational number is transcendental or not?
God Not Found
“Money supplants skill; it's possession allows us to become happily stupid.”
Bill McKibben, The Age of Missing Information

Offline Baruch (OP)

Re: Arithmetic for Cavebear
« Reply #20 on: November 08, 2017, 07:04:57 PM »
Transcendental numbers are non-algebraic, that can't ever be the solution to an algebraic equation. But how can they be identified otherwise? Is there a method of determining whether a given irrational number is transcendental or not?

No.  All algebraic numbers are either irrational or rational.  But some transcendental numbers (Pi) are not even algebraic.  Of course all transcendental numbers are irrational, because all rational numbers are algebraic.  Being algebraic is a proper subset of algorithmic derived numbers ... interestingly the cardinal number of algebraic numbers is the same as the cardinal number of rational numbers (dealing with infinity allows certain paradoxes as legitimate ... and this is why finitists reject infinity in maths).  So an algorithm can be devised to derive all rational numbers (they are all algebraic).  But no algorithm can be devised to derive all irrational numbers.  Such an algorithm can be devised for all algebraic numbers (so all irrational numbers that are also algebraic).  But there is a set of non-algebraic numbers, that are irrational ... that can be computed by an algorithm(Pi or e).  There are plenty of well defined irrational numbers, that have not been proven to be transcendental (non-algebraic).

Proof falls under the same bug-bear as any other algorithm.  There are constants, which haven't even been proven to be rational vs irrational, let alone algebraic vs transcendental.

https://en.wikipedia.org/wiki/Transcendental_number

Simple examples are not proven as to what class they are (Pi plus e).

One could try to escape this by allowing proofs that are not covered by an algorithm ... a proof with an infinite number of steps ... but that would be pulling a rabbit out of your hat.  These are called oracles in Turing machine theory ... a way of formalizing cheating.  Though in fact, there is no way to cheat.  Basically it amounts to the proctor (oracle) handing you the answer to the question.

The question of what is a true random number vs a pseudo-random number, is critical in cryptology.  A truly random number could be a normal transcendental for example, because by definition there would be no possible converging algorithm to calculate it.  There may be degrees of true randomness, as the normal transcendentals aren't the only non-algorithmic transcendentals, but others nearly as bad.  This comes up in computational complexity theory (which again plays into the subject of cryptology).  I studied all this 2 summers ago, when studying cryptology.
« Last Edit: November 08, 2017, 07:20:18 PM by Baruch »
שלום

Offline Hakurei Reimu

Re: Arithmetic for Cavebear
« Reply #21 on: November 09, 2017, 12:55:49 PM »
All rational numbers are algebraic and therefore not transcendental, because a rational number a/b is the solution to the polynomial bx - a = 0. Ergo, all transcendental numbers are irrational as well.

"Transcendental number" is a catch-all for "any number not the root of a polynomial equation with integer coefficients." It is necessarily exclusionary. We know enough about polynomials to know that some numbers are necessarily transcendental (e is an example) and others are necessarily algebraic (square roots of rational numbers), but outside those cases there's no finite number of tests you can perform on a number to show that it is not the solution to some polynomial of some finite degree.
Warning: Don't Tease The Miko!
(she bites!)
Spinny Miko Avatar shamelessly ripped off from Iosys' Neko Miko Reimu

Offline Baruch (OP)

Re: Arithmetic for Cavebear
« Reply #22 on: November 09, 2017, 01:00:57 PM »
All rational numbers are algebraic and therefore not transcendental, because a rational number a/b is the solution to the polynomial bx - a = 0. Ergo, all transcendental numbers are irrational as well.

"Transcendental number" is a catch-all for "any number not the root of a polynomial equation with integer coefficients." It is necessarily exclusionary. We know enough about polynomials to know that some numbers are necessarily transcendental (e is an example) and others are necessarily algebraic (square roots of rational numbers), but outside those cases there's no finite number of tests you can perform on a number to show that it is not the solution to some polynomial of some finite degree.

Correct and clear.
שלום