Mtgman, believe it or not, I was about to bring up *a priori * and a posteriori many posts back, but I had the intuition that it wasn’t the right time!
Why did you decide to bring up the a priori and a posteriori? Why is now the right time?
Believe it or not, I honestly don’t know where I’m going with this. I was hesatant to bring up Pi. Pi is indeed the “puzzle” I intended to reveal. But I don’t know where I’m going with this. I begin with ignorance and hope to end with knowledge.
Here we go: a[sub]k[/sub] = 2(k!)/((2k + 1)2[sup]2k/sup[sup]2[/sup])
:smack: This is what I get for rushing a post. Yes, of course. Gaudere’s law in mathematical form strikes again. 
And so, what do you think of the Gettier problem?
Kosmik, I believe you are discussing territory covered by mathematician Roger Penrose in his book The Emperor’s New Mind.
A problem that is solvable by a computer is solvable by an algorithmic process, a “mechanical” process if you like. You can set the problem up with digital electronics, or with gearwheels and linkages, and crank through it and the solution pops out. There are plenty of problems which are not solvable by algorithmic processes - the classic is the Turing Halting Problem.
If I’m reading you correctly, you are regarding truths that can be proven algorithmically, i.e. a proof that a computer can crank out, as “knowledge with proof”. Whereas a truth that is arrrived at without being able to use an algorithmic method, i.e. a computer cannot crank out a proof, is “knowledge without proof”. And you are giving the subtraction of an irrational from the same irrational being zero as an example of a non-computable truth.
However, people are getting impatient with you because the subtraction of irrationals are NOT examples of non-algorithmic proofs. Irrationals can’t be represented as ratios of whole numbers, true, but that doesn’t mean they are intractable by algorithmic processes. You can let an abacus bead stand for a whole number, an irrational number, an imaginary number, etc. It’s simple manipulation of symbols.
Penrose on the other hand did give some examples that he regarded as non-algorithmic - tiling problems. Some examples are given here (scroll down a bit)
Penrose argued that although there was definitely no algorithmic way of deciding whether a particular tile shape can tile an infinite plane, a human could see whether it could or couldn’t, and therefore was capable of non-algorithmic operations - “knowledge without proof”. This led him to argue that there were quantum-mechanical elements to the operation of the brain that allowed the limits of algorithmic processes to be exeeded. I’m in no position to critique his views, in the first place I don’t understand them.
Now, one view of the human brain is that it is purely a material device, no different in principle from a very complicated arrangement of gearwheels and linkages. In which case, human thought would be purely algorithmic in nature. This point of view is called physicalism and has some strong proponents and opponents on this board.
More discussion here:
I’m pretty sure that there are group-theoretic methods for determining whether a set of polytopes will tessellate n-space. That aside, any proof (in the formal sense) that can be discovered by a human can also be discovered by a computer (in the formal sense).
Why is this a problem?
Because π - π = π / π - 1 is intractable.
You don’t understand what intractable means. In any computation model that’s equipped to handle real computation, simple arithmetic like this is one of the easiest problems to solve.
I think you’re confused about the definition of “nonterminating algorithm.” And also “infinite precision.”
[symbol]p[/symbol] - [symbol]p[/symbol] = [symbol]p[/symbol] / [symbol]p[/symbol] - 1 is solvable. I’ll solve it right now.
[symbol]p[/symbol] - [symbol]p[/symbol] = [symbol]p[/symbol] / [symbol]p[/symbol] - 1 = 0
You don’t seem to accept the method I used because it depends on a priori justification, based on definitions and assumptions about the nature of the entities involved. That’s your perogative, but it isn’t binding on anyone else. The vast majority of the rest of the world has no problem accepting [symbol]p[/symbol] - [symbol]p[/symbol] = 0 or [symbol]p[/symbol] / [symbol]p[/symbol] - 1 = 0
Enjoy,
Steven
π must be determined via an infinite process; therefore, the computation of π - π = π / π - 1 is also an infinite process;
therefore, π - π = π / π - 1 can never be computed.
This algorithmic equation is non-computable because it can never terminate. Yet the paradox is that we know that it does terminate.
π - π = π / π - 1 can never be computed
π - π = π / π - 1 can instantly be solved
What you don’t seem to understand is that there’s a difference between [symbol]p[/symbol] and the decimal representation of [symbol]p[/symbol]. [symbol]p[/symbol] is not determined by an infinite process any more than 1/2 or -1 is, and the decimal representations of all three are exactly the same length.
Maybe it’s time we fessed up. It’s obvious he isn’t going to fall for any more of our coy tricks. We may as well admit what he alreadys knows we know — numbers are magical.
Ok, plug in [symbol]p[/symbol] - [symbol]p[/symbol] into this online graphing calculator, hit “eval” and see what you get. Since it is online it must have been implemented algorithmically, and yet it still shows 0 as the answer and it comes up with the answer immediately. It doesn’t process for infinite time to come up with the answer.
Enjoy,
Steven
The pages you link to talk about computability, as in the sorts of results that are computable on a Turing machine — which defines a general model of all computation that can be done “by machine”. By definition, algorithms must indeed be guaranteed to terminate after a finite number of steps.
But this doesn’t have much to do with your claim that pi - pi = pi/pi - 1 is intractable or unproveable for a couple reasons: (A) the set of problems that can be solved mechanically is only a subset of those problems solvable by humans, and (B) the fact that most real numbers require an infinite number of symbols to express in any particular notation system (such as standard decimal) does not mean that they are fundementally “infinite” or “unattainable” in some way, and that you therefore can’t do arithmetic on them, or have confident knowledge about some of their properties.
As another example: the cube root of 12 is an irrational number, like e and pi, and cannot be expressed in decimal with a finite number of digits. The same goes for the cube root of 18. But the product of these two numbers is the cube root of 216, which is exactly 6. You can compute this result, and know it to be true, without making any approximations of either number, and without filling up notebook paper with thousands of digits. Moreover, you can know it’s exactly 6 even if your calculator tells you it’s 5.9999998 instead. That’s a limitation of calculators, not mathematics or the real number system.
What’s correct to say is that the expression of pi in base 10 (or any other integer base) is incomputable. No computer program that attempts to output it will ever terminate, obviously. (I think that was a Star Trek episode in fact.) But pi - pi, pi/pi, pi[sup]0[/sup], and many other such expressions certainly are computable in a finite number of steps. You might, however, need to use a more sophisticated representation of numbers than standard decimal.
Yup, I agree.
I’m shaking the dust, folks. Enjoy.
I think this is your fundamental problem. Can
x - x = 0
ever be computed. In your view, no, since we don’t even know the value of x. However it can easily be shown to be true since we don’t need to know the value of variables to manipulate them. π isn’t a variable, of course, but it is a symbol ,standing for a number, which can be manipulated just like x.
The issue with the Halting Problem is not to show all programs halt (you can easily prove that some don’t) but that there are cases in which you can’t tell if the program halts or not. I’m not sure that this has been made clear.
Nope. Ever since mankind first figured out that anything minus itself is 0, it has no longer been necessary to determine the value of π before evaluating “π - π”. That’s the point of using algebra - we don’t need to find a decimal value immediately, because we can manipulate the symbols instead, and in this case it turns out that we can eliminate π from the equation without having to look for its decimal value at all.
What do you think the value of π - π is? Even if you’re incapable of thinking of π as anything but an infinitely long decimal number, a simple thought experiment should show that if you subtract an infinite series of digits from itself, you’ll end up with nothing but zeros.
3.1415926...
- 3.1415926...
==============
0.0000000...
You may not know exactly what digits of pi come next, but you do know that they’ll be the same on the first and second lines, and therefore the digits of the result will continue to be 0 forever.
I don’t know if you’re seriousl, but I would imagine that calculator is only using an approximation of pi, stored internally to a few dozen (or a few hundred) decimal places.
I think that’s the point that *kozmik is labouring to make; that the process you describe, of subtracting the digits one from the other, cannot be completed.
That’s not the same as saying it cannot be proven