The meaning of infinity

[untermensche]

Your position is a joke.

You have some delusion that calling an infinite series a set or calling anything with infinite elements a set you have reduced the number of elements or somehow allowed infinite elements to be expressed.

It is so stupid you should be ashamed to even say it.

That you don't know it is such a worthless position is a little strange. I often wonder what is at the other end of my comments.

We're not discussing series...
EB

Yes we are.

We are discussing an infinite series that can abstractly be thought of as existing within a set.

The infinite series is still there.

The elements without end are still there.

And to translate the set to something real would require every element of the infinite series being expressed.

That this is news to you just proves what a waste of my time you are.

 

[Bomb#20]

Note: just to be clear: the non-computability of such numbers is a property of the problem stated (that states which value we want to compute) not a property of numbers (as in the set of reals).

As far as I understand it, Chaitin is a way of identifying numbers that are not computable. So, non-computability is a property of those numbers unless you can explain why Chaitin doesn't work.
EB

I suspected that you should believe something like that. Thats why i wrote the post.
There is nothing special with the numbers themselves, its just a problem of identifying which number.
It has nothing incommon with incommensrable numbers, transcendet numbers etc.

That turns out not to be correct -- it's a property of the numbers themselves. I've read there's a proof that Chaitin's number is non-computable per se, irrespective of how we state the problem, but I admit I didn't follow the tricky construction; so let's switch to a different non-computable number.

Consider the number L, specified thusly: treat every possible sequence of ASCII characters as a number N written in base 128. Let the real number L be a number between 0 and 1 whose binary expansion has its Nth binary digit determined by the properties of ASCII sequence N, as follows. If sequence N is not a legal C program, bit N of number L is 0. If sequence N is a legal C program that contains a command to read any input other than a file the program wrote itself, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it exits, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it goes into an infinite loop, bit N is 1.

The number L itself has the special property of being non-computable. This has nothing to do with how we state the problem of identifying L. If there existed a program P that could dependably compute the Nth bit of L in a finite amount of time, which is what it means for a number to be computable, then if you wanted to know whether some other program Q ever goes into an infinite loop, all you would need to do is read the program, interpret its ASCII characters as a number in base 128, feed that number into program P, and when program P finishes, see whether it says the corresponding digit of L is 0 or 1. If P says 1, L goes into an infinite loop. If P says 0, L halts. That means P solves the Halting Problem. But Alan Turing already proved it's impossible for any program to solve the Halting Problem.

 

[Bomb#20]

So, it's not really non-computable. Rather, it's non-computable in all probability. So, logically computable but probably not in practice, somewhat like the non-zero probability for a particle to be anywhere in the universe. Non-measurable in any place, sort of. And yet it has to be somewhere when we do measure it.

See my reply to Juma, post #182. There is allegedly a similar construction for Chaitin's Constant, showing how a program for computing it could be used to make a program that solves the Halting Problem.

 

[Bomb#20]

So let x = 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 90/10 + 90/100 + 90/1000 + ...

Therefore, 10x = 9/1 + 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 9/1 + x

Therefore, 9x = 9

Therefore, x = 1

So 0.999... = 0 + 1

bomb

Make 0.999... 0.777...

So let x =77/10 + 7/100 + 7/1000 + ...

I take it you meant "let x =7/10 + 7/100 + 7/1000 + ...". Carry on...

Therefore, 10x = 70/10 + 70/100 + 70/1000 + ...

Yep.

Therefore, 10x = 7/1 + 7/10 + 7/100 + 7/1000 + ...

Yep.

Therefore, 10x = 7/1 + x

Yep.

Therefore, 7x = 7

Nope. When I subtract x from both sides of "10x = 9/1 + x", I get "9x = 9", but that doesn't mean when you subtract x from both sides of "10x = 7/1 + x" you get "7x = 7". 10-1 does not equal 7, so 10x - 1x does not equal 7x. When you subtract x from both sides of "10x = 7/1 + x" you get...

Therefore, 9x = 7

Therefore, x = 1

Continuing from the corrected previous step, you get

Therefore, x = 7/9

So 0.777... = 0 + 1 ?

Continuing from the last step, you get

So 0.777... = 0 + 7/9

(Which is correct.)

What you did is show the distributive property.

x = a + b
k*x = k*a + k*b
k*x = k*[a + b]
k/k = [a + b]/x
1 = [a + b]/x

You took a stab at it and it did not work, happens to everybody. Improving math skills in large part is trial and error. Failure leads to insight and new doors to walk through.

Looks to me like what I did worked. As you can see from my examination of your derivation, when a derivation is wrong it's because one specific step goes from a true premise to a false conclusion; in your case it was your step from "10x = 7/1 + x" to "7x = 7". There's always a first incorrect statement, from which the rest of the derivation takes off on a wild goose chase.

If you still think my derivation is incorrect, which line in it do you think is the first incorrect statement?

What is 0.555... ? I might round to .6, .56 or .556 depending on calculations.

5/9. The proof is analogous to the 0.777 derivation.

How about 0. 123123123...? what does that equal when you take it to infinity?

x = .123123123...

1000x = 123.123123123...

1000x = 123 + x

999x = 123

x = 123/999

 

[Bomb#20]

If that can help, I understand that the probability is overwhelmingly for a non-computable number to be transcendental. But it's not clear to me that the "definition" we have allows to infer that non-computable number are all necessarily transcendental. However, the definition I think leaves some room for interpretation between numbers actually non-computable versus numbers we don't know how to compute. In the latter case, we could have non-transcendental numbers that would be non-computable based on some identification. In the former case, all non-computable numbers would be all transcental, I think.
EB

Keep in mind that "transcendental" simply means "not algebraic". We have a known algorithm for computing any desired digit of any algebraic number. So yes, the definition of non-computable lets us infer that non-computable numbers are all necessarily transcendental.

 

[DBT]

I'm not sure how something can be both ''infinite in some way and yet finite''

How about the infinity of rationals between 0 and 1, being bounded by both 0 and 1?

Oh, right, brilliant! If you're serving all the answers on a plate the guy will never learn anything!

That was a baby step. He could have done it on his own. You would need to restrain your impulse.
EB

Coming from Noddy the All Knowing, infinitely arrogant, smug and snide in each and every post, yet at the same time, an ignorant Prat.

 

[DBT]

Depends on the terminology and definitions being used. Some things may be defined into existence, mathematical constructs, etc. It's not something that I would consider to be actually infinite. Not like an actual Infinite Universe or an Infinite Multiverse where the 'infinity of rationals' between 0 and 1, being bounded by both 0 and 1 does not compare.

How does it not "compare"? The rationals is a pretty good conceptual representation of a space that would be infinitely divisible, which ordinary space might well be for all we know. How does it not "compare"?
EB

Rationals as a conceptual representation of a space that would be infinitely divisible being a mathematical representation, a mathematical construct, a concept of infinite divisibility that does not necessarily relate to the physical world, physical infinity, an actual Physically Infinite Universe.

 

[steve_bank]

I take it you meant "let x =7/10 + 7/100 + 7/1000 + ...". Carry on...

Therefore, 10x = 70/10 + 70/100 + 70/1000 + ...

Yep.

Therefore, 10x = 7/1 + 7/10 + 7/100 + 7/1000 + ...

Yep.

Therefore, 10x = 7/1 + x

Yep.

Therefore, 7x = 7

Nope. When I subtract x from both sides of "10x = 9/1 + x", I get "9x = 9", but that doesn't mean when you subtract x from both sides of "10x = 7/1 + x" you get "7x = 7". 10-1 does not equal 7, so 10x - 1x does not equal 7x. When you subtract x from both sides of "10x = 7/1 + x" you get...

Therefore, 9x = 7

Therefore, x = 1

Continuing from the corrected previous step, you get

Therefore, x = 7/9

So 0.777... = 0 + 1 ?

Continuing from the last step, you get

So 0.777... = 0 + 7/9

(Which is correct.)

What you did is show the distributive property.

x = a + b
k*x = k*a + k*b
k*x = k*[a + b]
k/k = [a + b]/x
1 = [a + b]/x

You took a stab at it and it did not work, happens to everybody. Improving math skills in large part is trial and error. Failure leads to insight and new doors to walk through.

Looks to me like what I did worked. As you can see from my examination of your derivation, when a derivation is wrong it's because one specific step goes from a true premise to a false conclusion; in your case it was your step from "10x = 7/1 + x" to "7x = 7". There's always a first incorrect statement, from which the rest of the derivation takes off on a wild goose chase.

If you still think my derivation is incorrect, which line in it do you think is the first incorrect statement?

What is 0.555... ? I might round to .6, .56 or .556 depending on calculations.

5/9. The proof is analogous to the 0.777 derivation.

How about 0. 123123123...? what does that equal when you take it to infinity?

x = .123123123...

1000x = 123.123123123...

1000x = 123 + x

999x = 123

x = 123/999

From the disttributive property no matter what you multiply both sides by you will end up with unity. Basic algebra.

 

[steve_bank]

10x = 7/1 + 7/10 + 7/100 + 7/1000 + ...
10x = 7 + .7 + .07 +.007 = 7.777
x = .7777

When you apply the distributive property the balance implied by the = sign is not altered. Applying the distrubitive property as above you can not get a change in magnitude of reults, that be like violating conservation something from nothing.

x = 4
10x - 40
x is still 4.

You can not get 1 from .999...

 

[steve_bank]

x = .9
n = 100

while( forever){

x = x + [x/n]
n = n*10

}

As the algorith chugs through eternity when does x become 1?

 

[steve_bank]

So let x = 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 90/10 + 90/100 + 90/1000 + ...

Therefore, 10x = 9/1 + 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 9/1 + x

No.
10x = 90/10 + 90/100 + 90/1000 + ...
10 x = 9 + .9 + .09 + .009
10x= 9.999
x = .9999

No matter ho many terms you use applying the distributive principle changes nothing.

I did not pick up your mistake and that led to mine.

 

[DBT]

Depends on the terminology and definitions being used. Some things may be defined into existence, mathematical constructs, etc. It's not something that I would consider to be actually infinite. Not like an actual Infinite Universe or an Infinite Multiverse where the 'infinity of rationals' between 0 and 1, being bounded by both 0 and 1 does not compare.

How does it not "compare"? The rationals is a pretty good conceptual representation of a space that would be infinitely divisible, which ordinary space might well be for all we know. How does it not "compare"?
EB

Rationals as a conceptual representation of a space that would be infinitely divisible being a mathematical representation, a mathematical construct, a concept of infinite divisibility that does not necessarily relate to the physical world, physical infinity, an actual Physically Infinite Universe.

In other words, take any measurement, say one inch, and apply Rationals, divide that space infinitesimally - which I do not deny or ague against, however the inch of space remains a finite distance between 0 and 1, what we call an inch, in the real world. You can play with Rationals as an exercise in Mathematics, do it with the diameter of the Moon, dividing 3 474.8 kilometres till the end of time, yet the Moon remains a Finite Object orbiting the Earth, itself a finite object orbiting a finite Sun.......

 

[Juma]

I suspected that you should believe something like that. Thats why i wrote the post.
There is nothing special with the numbers themselves, its just a problem of identifying which number.
It has nothing incommon with incommensrable numbers, transcendet numbers etc.

That turns out not to be correct -- it's a property of the numbers themselves. I've read there's a proof that Chaitin's number is non-computable per se, irrespective of how we state the problem, but I admit I didn't follow the tricky construction; so let's switch to a different non-computable number.

Consider the number L, specified thusly: treat every possible sequence of ASCII characters as a number N written in base 128. Let the real number L be a number between 0 and 1 whose binary expansion has its Nth binary digit determined by the properties of ASCII sequence N, as follows. If sequence N is not a legal C program, bit N of number L is 0. If sequence N is a legal C program that contains a command to read any input other than a file the program wrote itself, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it exits, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it goes into an infinite loop, bit N is 1.

The number L itself has the special property of being non-computable. This has nothing to do with how we state the problem of identifying L. If there existed a program P that could dependably compute the Nth bit of L in a finite amount of time, which is what it means for a number to be computable, then if you wanted to know whether some other program Q ever goes into an infinite loop, all you would need to do is read the program, interpret its ASCII characters as a number in base 128, feed that number into program P, and when program P finishes, see whether it says the corresponding digit of L is 0 or 1. If P says 1, L goes into an infinite loop. If P says 0, L halts. That means P solves the Halting Problem. But Alan Turing already proved it's impossible for any program to solve the Halting Problem.

? This problem is a problem of turing machine, nor reals.
It has nothing to do with any mathematical property of L.
Only a result of the halting problem.

 

[Juma]

So let x = 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 90/10 + 90/100 + 90/1000 + ...

Therefore, 10x = 9/1 + 9/10 + 9/100 + 9/1000 + ...

Therefore, 10x = 9/1 + x

No.
10x = 90/10 + 90/100 + 90/1000 + ...
10 x = 9 + .9 + .09 + .009
10x= 9.999
x = .9999

No matter ho many terms you use applying the distributive principle changes nothing.

I did not pick up your mistake and that led to mine.

If 10x = 90/10 + 90/100 + 90/100 + ...
then it does not follow that
10x = 9 + 0.9 + 0.09 + 0.009
You forgot the ”...”

 

[Juma]

x = .9
n = 100

while( forever){

x = x + [x/n]
n = n*10

}

As the algorith chugs through eternity when does x become 1?

When it finished. The clockcycles take no time since this an abstract machine.

 

[steve_bank]

x = .9
n = 100

while( forever){

x = x + [x/n]
n = n*10

}

As the algorith chugs through eternity when does x become 1?

When it finished. The clockcycles take no time since this an abstract machine.

Yes it is abstract, the point is to use your imagination to approach the question of 0.999.. equaling 1.

0.999... is an abstraction.

 

[Juma]

x = .9
n = 100

while( forever){

x = x + [x/n]
n = n*10

}

As the algorith chugs through eternity when does x become 1?

When it finished. The clockcycles take no time since this an abstract machine.

Yes it is abstract, the point is to use your imagination to approach the question of 0.999.. equaling 1.

Yes. And I did that. Takes no time.

0.999... is an abstraction.

So what?
1 is an abstraction.

 

[Phil Scott]

Consider the number L, specified thusly: treat every possible sequence of ASCII characters as a number N written in base 128. Let the real number L be a number between 0 and 1 whose binary expansion has its Nth binary digit determined by the properties of ASCII sequence N, as follows. If sequence N is not a legal C program, bit N of number L is 0. If sequence N is a legal C program that contains a command to read any input other than a file the program wrote itself, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it exits, bit N is 0. If sequence N is legal C and doesn't contain such an input command, and if you run the program it goes into an infinite loop, bit N is 1.

The number L itself has the special property of being non-computable. This has nothing to do with how we state the problem of identifying L. If there existed a program P that could dependably compute the Nth bit of L in a finite amount of time, which is what it means for a number to be computable, then if you wanted to know whether some other program Q ever goes into an infinite loop, all you would need to do is read the program, interpret its ASCII characters as a number in base 128, feed that number into program P, and when program P finishes, see whether it says the corresponding digit of L is 0 or 1. If P says 1, L goes into an infinite loop. If P says 0, L halts. That means P solves the Halting Problem. But Alan Turing already proved it's impossible for any program to solve the Halting Problem.

But as I've pointed out elsewhere, C isn't Turing complete You can determine halting of all those C programs with another program of sufficient power, since each C program is effectively a very very large state machine.

Plenty of languages are Turing complete though, and any of those works for your argument.

 

[Phil Scott]

You took a stab at it and it did not work, happens to everybody. Improving math skills in large part is trial and error. Failure leads to insight and new doors to walk through.

Can you please respond to my post here steve.

 

[untermensche]

If you imagine a string of repeating decimals completes it does.

In the imagination.

In the imagination 1.000 = 0.999...

Because in the imagination you can pretend the repeating decimals somehow complete.

In the real world there is no such thing as a repeating decimal.

It is all pretend knowledge.

Like knowing about Archie and Jughead.