The meaning of infinity

[bilby]

No. I accept that given 0.333... 1/3 is an integer approximation. Given 0.555... 5/9 is an integer approximation. Given 0.999... 9/9 is an integer approximation. The method does not serve as a proof, applied to repeating decimals it provides integer approximations.

0.999.. as an infinite defined infinite entity and a finite quantification are mutually exclusive.

That you are too shallow and inexperienced in applying math to follow my counter arguments is not my problem. Read, consider, and reply to my objections. What I have found is personal attack rises in proportion to lack of competence.

I new when you put me on ignore it would not last, a moth to a flame. If I an stupid put me on ignore and forget about me.

That's the first sensible thing you have said in this thread.

 

[Kharakov]

Hint (for dealing with .9999....): What is the number that signifies something smaller than any real amount?

Hint: you can use pi to help approximate the circumference of a symbol that resembles it.

 

[Juma]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An <= 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Do you agree on this?

I do not agree that 0.999... and the series representaion are the same thing. By definition 0.999... is infinite and can not be reduced to a finite number. The only comclusion is that the method does not infer a finite 1 equality, it infers a fractional 9/9 approximation.

Beyond that we will be going in circles. For me the conflict between an infinite unchanging sequence 0.999.. and and a finite number can not be resolved. Therefore while the result of 09/9 is correct within the bounds of what the method provides, integer approximations to repeating decimals, conclusion is wrong.

do you agree with the definition or not?

 

[steve_bank]

I do not agree that 0.999... and the series representaion are the same thing. By definition 0.999... is infinite and can not be reduced to a finite number. The only comclusion is that the method does not infer a finite 1 equality, it infers a fractional 9/9 approximation.

Beyond that we will be going in circles. For me the conflict between an infinite unchanging sequence 0.999.. and and a finite number can not be resolved. Therefore while the result of 09/9 is correct within the bounds of what the method provides, integer approximations to repeating decimals, conclusion is wrong.

do you agree with the definition or not?

My position is clear. You and I will not agree. There is nothing more for me to get out of the problem.

 

[Kharakov]

All you have to do is accept the fact that anything larger in magnitude than zero has actual length, even if it is transcendental or irrational.

The question you should ask:

is there a transcendental .999... that does not equal 1?

 

[Artemus]

How so? A Reimann integration is a limut as dx goes to zero. It is an approximation.

No it isn't. It is an exact operation. Please cite any mathematical text or peer reviewed article that says it is an approximation.

Riemann integration is a limit as the interval of integration is divided into infinitesimal divisions, or dx. A limit as dx goes to zero. I went through this when I studied Numerical Methods in the 80s. The reason I got my first PC.

Calculate integral cos(x) over 0 pi/4. First use the exact solution int cos() = sin(). Then calculate using Rei,amm integration, or sums. As the dx or the number of intervals get small the numerical solution gets close to to the first integration. Get up to a few thousand intervalues and the result will equal the accuracy of the truncated series on a calculator.

https://en.wikipedia.org/wiki/Riemann_sum
https://en.wikipedia.org/wiki/Riemann_integral

"The basic idea of the Riemann integral is to use very simple approximations for the area of S. By taking better and better approximations, we can say that "in the limit" we get exactly the area of S under the curve."

We say it is exact because the error becomes infinitesimally small. Not because it is really exact.


The Fundamental Theorem results from a limit to infinity


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

We say integral x^n = (x/n)^n+1 is an exact integral, but it is really an approximation. We say it is exact because by the fundamental theorem it reduces to an algebraic equation.,albeit in the limit. A limit that s never reached. An exact integral means not having to resort to numerical solutions.

To show that the integral isn't exact you quote a text that says it is exact but then claim that exact doesn't really mean exact? Seriously? And you again refer to the "the exact solution" of an equation you say is approximate? And once again say that the limit operation is not exact because there is an infinitesimal error when it has been pointed out ad nauseum that only infinitesimal in the real number system is zero? At this point I can only conclude that you are either willingly or pathologically unable to see the self-contradictions of your own arguments.

 

[steve_bank]

Riemann integration is a limit as the interval of integration is divided into infinitesimal divisions, or dx. A limit as dx goes to zero. I went through this when I studied Numerical Methods in the 80s. The reason I got my first PC.

Calculate integral cos(x) over 0 pi/4. First use the exact solution int cos() = sin(). Then calculate using Rei,amm integration, or sums. As the dx or the number of intervals get small the numerical solution gets close to to the first integration. Get up to a few thousand intervalues and the result will equal the accuracy of the truncated series on a calculator.

https://en.wikipedia.org/wiki/Riemann_sum
https://en.wikipedia.org/wiki/Riemann_integral

"The basic idea of the Riemann integral is to use very simple approximations for the area of S. By taking better and better approximations, we can say that "in the limit" we get exactly the area of S under the curve."

We say it is exact because the error becomes infinitesimally small. Not because it is really exact.


The Fundamental Theorem results from a limit to infinity


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

We say integral x^n = (x/n)^n+1 is an exact integral, but it is really an approximation. We say it is exact because by the fundamental theorem it reduces to an algebraic equation.,albeit in the limit. A limit that s never reached. An exact integral means not having to resort to numerical solutions.

To show that the integral isn't exact you quote a text that says it is exact but then claim that exact doesn't really mean exact? Seriously? And you again refer to the "the exact solution" of an equation you say is approximate? And once again say that the limit operation is not exact because there is an infinitesimal error when it has been pointed out ad nauseum that only infinitesimal in the real number system is zero? At this point I can only conclude that you are either willingly or pathologically unable to see the self-contradictions of your own arguments.

Nice try. I maintain any function divide with an infinite limit is approximate, albeit small error.t. I believe the link stes the Riemann Integral as an approximation taken as exact in the limit. . As an x goes to infinity at what x does it it become infinite?

A Riemann Sum can never have an infinite partition. As such it is an approximation.

How would you integrate across a a discontinuity?

What is your answer to my last post on limits? It is the same question as with Riemann Integrals.

 

[Juma]

I do not agree that 0.999... and the series representaion are the same thing. By definition 0.999... is infinite and can not be reduced to a finite number. The only comclusion is that the method does not infer a finite 1 equality, it infers a fractional 9/9 approximation.

Beyond that we will be going in circles. For me the conflict between an infinite unchanging sequence 0.999.. and and a finite number can not be resolved. Therefore while the result of 09/9 is correct within the bounds of what the method provides, integer approximations to repeating decimals, conclusion is wrong.

do you agree with the definition or not?

My position is clear. You and I will not agree. There is nothing more for me to get out of the problem.

Your position is far from clear. Do you accept the definition of decimal numbers above, or not? If not, specify what is wrong with the actual definition as it stands, do not bring up any other issues we may have, just concentrate on the definition at it stands.

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An <= 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

 

[steve_bank]

My position is clear. You and I will not agree. There is nothing more for me to get out of the problem.

Your position is far from clear. Do you accept the definition of decimal numbers above, or not? If not, specify what is wrong with the actual definition as it stands, do not bring up any other issues we may have, just concentrate on the definition at it stands.

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An <= 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Some around here seem to think they are locked into a life or death struggle. For me it about exploring and staying active. I learned a few things in the debate. It is not a peer competition.

I can not continually restate.

1. Infinite and finite are mutually exclusive. If you argue that forget the rest. If you accept that then 0.999.. can never = a finite 1.

2. geometric series applied to repeating decimals yields fractional approximations. 0.333... goes to 3/9, a fractional approximation. 0.999... goes to 9/9, a fractional approximation. It says use 1 as an approximation for 0.999.., it does not say 0.999.. equals a finite 1.

3. if 0.333.. goes to 1/3 goes to 0.333... does not round up but 0.999.. seems to do, then 0.999.. as a proof it is 1 is questionable, the method is inconsistent on that point. The method works as intended but can not be relied on for the assumed proof. I could cite examples when I fell into that kind of a trap but it would not mean anything to you.

That is about it for me. If it is not clear I can not help you. I'd suggest letting go of hanging onto a web link and try some independent thought. Learn to criticize yourself as you would me, introspection is important. Stretch and take some risks, it is what the forum affords.

 

[Juma]

My position is clear. You and I will not agree. There is nothing more for me to get out of the problem.

Your position is far from clear. Do you accept the definition of decimal numbers above, or not? If not, specify what is wrong with the actual definition as it stands, do not bring up any other issues we may have, just concentrate on the definition at it stands.

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An <= 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Some around here seem to think they are locked into a life or death struggle. For me it about exploring and staying active. I learned a few things in the debate. It is not a peer competition.

I can not continually restate.

1. Infinite and finite are mutually exclusive. If you argue that forget the rest. If you accept that then 0.999.. can never = a finite 1.

2. geometric series applied to repeating decimals yields fractional approximations. 0.333... goes to 3/9, a fractional approximation. 0.999... goes to 9/9, a fractional approximation. It says use 1 as an approximation for 0.999.., it does not say 0.999.. equals a finite 1.

3. if 0.333.. goes to 1/3 goes to 0.333... does not round up but 0.999.. seems to do, then 0.999.. as a proof it is 1 is questionable, the method is inconsistent on that point. The method works as intended but can not be relied on for the assumed proof. I could cite examples when I fell into that kind of a trap but it would not mean anything to you.

That is about it for me. If it is not clear I can not help you. I'd suggest letting go of hanging onto a web link and try some independent thought. Learn to criticize yourself as you would me, introspection is important. Stretch and take some risks, it is what the forum affords.

Ponder this: why cant you answer my question? There must be some sort of psychological effect here.
The question is easy. Do you accept the definition of decimal numbers as I described it or not?

 

[Jokodo]

My position is clear. You and I will not agree. There is nothing more for me to get out of the problem.

Your position is far from clear. Do you accept the definition of decimal numbers above, or not? If not, specify what is wrong with the actual definition as it stands, do not bring up any other issues we may have, just concentrate on the definition at it stands.

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An <= 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Some around here seem to think they are locked into a life or death struggle. For me it about exploring and staying active. I learned a few things in the debate. It is not a peer competition.

I can not continually restate.

1. Infinite and finite are mutually exclusive. If you argue that forget the rest. If you accept that then 0.999.. can never = a finite 1.

Then tell us, how can an infinite 1.000... ever be equal to 1?

The solution is simple: It's the string 0.999... (or rather, an alternative string defined to be equivalent within the symbolic system we're using) that's infinite. That doesn't allow us any conclusions about the nature of the number. Indeed, just by looking at the first too symbols, "0" and ".", we can definitely conclude that the label must be referring to a finite real number between 0 and 1 inclusive. And every algorithm to transform decimal representations to fractional representations tells us this number is 1/1.

2. geometric series applied to repeating decimals yields fractional approximations. 0.333... goes to 3/9, a fractional approximation. 0.999... goes to 9/9, a fractional approximation. It says use 1 as an approximation for 0.999.., it does not say 0.999.. equals a finite 1.

Not an approximation, an exact result.

3. if 0.333.. goes to 1/3 goes to 0.333... does not round up but 0.999.. seems to do, then 0.999.. as a proof it is 1 is questionable, the method is inconsistent on that point. The method works as intended but can not be relied on for the assumed proof. I could cite examples when I fell into that kind of a trap but it would not mean anything to you.

0.999... does not "round up". Saying it does is assuming your conclusion: Only when there's a difference between the number represented by 0.999... and the number represented by 1.000... to start with does equating them amount to rounding.

That is about it for me. If it is not clear I can not help you. I'd suggest letting go of hanging onto a web link and try some independent thought. Learn to criticize yourself as you would me, introspection is important. Stretch and take some risks, it is what the forum affords.

This is some heavy irony right there. How about you try to criticize yourself.

Here's another proof that 0.999... = 1.000...: What is 1 - 0.999...?

We know that subtracting a digit 9 from a digit 0 yields a 1 in the corresponding position of the result if no non-zero digits follow, a 0 otherwise. for the expression "1 - 0.999..." this means a 0 for every position except that of the last 9. Since by definition there is no last 9, the result has to be 0.000..., or 0. In other words, the terms terms are equal.

 

[Artemus]

Nice try. I maintain any function divide with an infinite limit is approximate, albeit small error.t. I believe the link stes the Riemann Integral as an approximation taken as exact in the limit. . As an x goes to infinity at what x does it it become infinite?

A Riemann Sum can never have an infinite partition. As such it is an approximation.

How would you integrate across a a discontinuity?

What is your answer to my last post on limits? It is the same question as with Riemann Integrals.

The limit is a precisely defined, exact mathematical operation. Did you even read the Wiki article on  (ε,_δ)-definition_of_limit that clearly states that need for an infinitesimal is removed? If you did not then your problem is willful. If you did and still claim it is inexact then it is pathological. But tell you what, you clearly believe that your intuition has demonstrated that the Wiki article, and therefore reference [11] that it cites, is wrong. So put your money where your mouth is. Write up your proof and submit it to a mathematical journal. Proving that the fundamental theorem of calculus is only approximate will surely make you famous and textbooks will refer to the "Bank proof" for hundreds of years.

Let us know how the peer review turns out.

(And for the record, the Reimann integral of (some) discontinuous functions exists precisely because when the limiting operation is applied the zero-width intervals never cross the discontinuity. And direct Reimann sums will almost certainly be the worst way possible to efficiently compute a  numerical quadrature in any practical problem. See QUADPACK.)

 

[steve_bank]

As to numerical integration

The limit as x goes to infinity 1/ [1 + (1/x)] exactly equals 1.

The limit as x goes to infinity 1/ [1 + (1/x)] approaches 1 but never gets there.. As x gets large the error is infinitesimally small but not zero, so we take the limit as 1.

I take the second statement as correct.

When it comes to partitions for Riemann Sums, as the number of partitions get large over an integration interval the area is not exact, but the error is infinitesimally small. That is my view. Taken something to infinity is conceptual not actual. It is a theoretical concept. There is no finite number of sums that will yield an exact area.

It goes to the OP, the meaning of infinity. Infinity is not reachable.

 

[Juma]

As to numerical integration

The limit as x goes to infinity 1/ [1 + (1/x)] exactly equals 1.

The limit as x goes to infinity 1/ [1 + (1/x)] approaches 1 but never gets there.. As x gets large the error is infinitesimally small but not zero, so we take the limit as 1.

I take the second statement as correct.

When it comes to partitions for Riemann Sums, as the number of partitions get large over an integration interval the area is not exact, but the error is infinitesimally small. That is my view. Taken something to infinity is conceptual not actual. It is a theoretical concept. There is no finite number of sums that will yield an exact area.

It goes to the OP, the meaning of infinity. Infinity is not reachable.

You talk about a difference between actual and theoretical concepts. There is no such difference, even a sum of three terms is a theoretical concept. Math is a theoretical concept altogether.
And a finite sum of 200^(200^200) different terms is as ”unreachable” as an infinite sum.

 

[Artemus]

While I don't expect to convince certain posters, in case anyone wants to learn more here is another description of the epsilon-delta definition of limits that may be easier to follow than the wikipedia page. It includes limits to infinity with a relevant example. The limit is the number L, not an evaluation of a function when x "gets to" infinity. And no matter how "infinitesimal" you make \(\epsilon\) you can always make it smaller still in the real number system. L is therefore a rigorously defined exact number. The argument that you "can't get to infinity" is meaningless. The same argument applies to the definition of the Reimann integral. The integral is S and exists and has an exact value if the epsilon-delta criterion is met for all \(\epsilon > 0\). The goal is to mathematically prove that the limit exists, not to directly add an infinite number of terms.

 

[Speakpigeon]

While I don't expect to convince certain posters, in case anyone wants to learn more here is another description of the epsilon-delta definition of limits that may be easier to follow than the wikipedia page. It includes limits to infinity with a relevant example. The limit is the number L, not an evaluation of a function when x "gets to" infinity. And no matter how "infinitesimal" you make \(\epsilon\) you can always make it smaller still in the real number system. L is therefore a rigorously defined exact number. The argument that you "can't get to infinity" is meaningless. The same argument applies to the definition of the Reimann integral. The integral is S and exists and has an exact value if the epsilon-delta criterion is met for all \(\epsilon > 0\). The goal is to mathematically prove that the limit exists, not to directly add an infinite number of terms.

Brilliant!

As I recall, it's how I learnt about limits in my student days.

Still, this assumes that epsilon can get infinitely small for the limit to be what it is said to be so people objecting somehow to infinity will keep objecting. This just removes the offending infinite sum for an obnoxious approach infinitely close to x₀.
EB

 

[Speakpigeon]

As to numerical integration

The limit as x goes to infinity 1/ [1 + (1/x)] exactly equals 1.

The limit as x goes to infinity 1/ [1 + (1/x)] approaches 1 but never gets there.. As x gets large the error is infinitesimally small but not zero, so we take the limit as 1.

I take the second statement as correct.

When it comes to partitions for Riemann Sums, as the number of partitions get large over an integration interval the area is not exact, but the error is infinitesimally small. That is my view. Taken something to infinity is conceptual not actual. It is a theoretical concept. There is no finite number of sums that will yield an exact area.

It goes to the OP, the meaning of infinity. Infinity is not reachable.

The question of "reaching" infinity is irrelevant. The question is whether the result of the mathematical expression assuming infinity is exact. You're not capable of proving it's not. You're not even capable of showing there's any kind of difference.

And I'm not going to be contradicted here.
EB