As the definition and Sarkus stated, the repeating decimal approximates the limit,but never equals it.
The Squared Circle
- Thread starter Motor Daddy
- Start date
Neddy Bate
Valued Senior Member
No, 33.333...% (repeating decimal percent) is exactly equal to 0.333... (repeating decimal) which is exactly equal to 1/3. Observe:
x = 0.333...
Express it so that everything after the decimal point is the same:
10x = 3.333...
Then subtract one from the other to remove everything after the decimal point (10x - 1x = 9x).
10x = 3.333...
-1x = -0.333...
__________
9x = 3
Solve for x:
x = 3/9
Divide numerator and denominator by 3:
x = 1/3
Last edited:
I never said it approximates, I said that it equals it. Not "almost exactly" but exactly.
Please stop misquoting people.
You remain wrong when considering the entire infinite sequence. You are arguing against the value of the infinite sequence by only considering a finite one. Until you start examining the actual case in question you will continue to be wrong.
This is all well and good, but the notion of a "limit" is very well-defined in mathematics. For example, consider the sequence:
$$y_n=1-1/n$$
Listing some values, we have:
y_1 = 0
y_2 = 0.5
y_3 = 0.666...
y_4 = 0.75
y_5 = 0.8
etc.
The following is a true and well-defined mathematical statement:
$$\lim\limits_{n\rightarrow \infty} y_n = 1$$
Mathematically, this means that for any specified $$\epsilon >0$$, there exists a corresponding n value such that $$1-y_n<\epsilon$$.
This notion of the limit of a sequence is vitally important in defining what is meant by the derivative of a function, for example. Without it, calculus would not exist as a well-defined part of mathematics.
If you insist on doing mathematics without using the concept of a limit, you will be severely limiting yourself.
Sarkus;
The red portion is what I referred to as equivalent to the definition, using 'approximates' in place of the string of words. I don't find it necessary to quote everything verbatim.
You can only compare finite sequences since there is no complete sequence. If there is, what is it? How does the sequence of 9's become 0's?
If it required an 'infinite' amount of time to build a pyramid, there wouldn't be any.
No different for any 'infinite' sequence. If it has no boundary for its length, then there is no end.
...
I used the phrase 'for all n', n being the set of integers, which is common knowledge.
The red portion is what I referred to as equivalent to the definition, using 'approximates' in place of the string of words. I don't find it necessary to quote everything verbatim.
You can only compare finite sequences since there is no complete sequence. If there is, what is it? How does the sequence of 9's become 0's?
If it required an 'infinite' amount of time to build a pyramid, there wouldn't be any.
No different for any 'infinite' sequence. If it has no boundary for its length, then there is no end.
...
I used the phrase 'for all n', n being the set of integers, which is common knowledge.
Last edited:
Yes, there is a complete sequence. It just happens to be infinitely long. There are numerous ways to express that, some of which have even been provided in this thread.
Let A = 1/9 = 0.111... (or are you denying that as well?)
9A = 9/9 = 1 = 9 x 0.111... = 0.999...
It's not rocket science.
Strawman. Noone is talking about things taking an infinite time. Maths doesn't require "time". One can sum an infinite sequence in seconds. You do it yourself, you just notate the sum as "0.9R" or "0.999...". The argument is whether or not that equals 1 (which it does), not how long it takes to sum the sequence (which even you do in seconds). So stop introducing nonsense.
Not if you want to write it out in long-form. Fortunately it also equals 1 in this case, so you can just write that.
I note you still haven't told me what the number is between 0.999... and 1.
If A = 1 and B = 0.999... (or 0.9R) then what is (A+B)/2
For you to be able to prove that A and B are not equal, you must surely be able to show that there is a number that falls between 0.999... and 1.
If A = 1 and B = 2, there exists a number (A+B)/2 that falls between them. In this case 1.5.
So what is that number that lies between 1 and 0.999...?
[/quote]I used the phrase 'for all n', n being the set of integers, which is common knowledge.[/QUOTE]Which is an infinitely large set. So "for all n" you have to consider the entire infinite sequence. You are refusing to do that. And so you arrive at an irrelevant conclusion.
James;
View attachment 4954
Why does he define a limit L for f(x), and not f(x)=L?
Could it be that he knows f(x) never =2, but he can set a limit for it? I.e., he is distinguishing two different things, limit and function, and assigning the value 2 to the L.
View attachment 4954
Why does he define a limit L for f(x), and not f(x)=L?
Could it be that he knows f(x) never =2, but he can set a limit for it? I.e., he is distinguishing two different things, limit and function, and assigning the value 2 to the L.
Attachments
Neddy Bate
Valued Senior Member
I think phyti is denying that. I think he will say that 0.111... is slightly less than 1/9, just as he has been saying that 0.333... is slightly less than 1/3!
He does not seem to realise that the process of dividing 1 by 9 in long division is the process that generates 0.111... in the first place. Or that 0.333... actually means, "When we divide 1 by 3, we get a decimal point followed by a 3, followed by another 3, followed by another 3, repeating forever." In other words, 0.333... is equal to 1/3 because it is result of that division.
In his mind, since the "repeating forever" part can't be completed as a sequential process, it must mean that he is free to stop after however many repeats he feels like using, and then say that's what 0.333... really means. He thinks it is an approximation. He and Motor Daddy must have had the same mathematics teacher!
Last edited:
Neddy Bate
Valued Senior Member
Correct, the limit L means the value that the function is approaching even if the function itself does not equal that value at that place. For example, a function which is not defined at a certain value of x can still have a defined limit as x approaches that value.
See here:
"So the limit of g at x=3 is equal to 5 but the value of g at x=3 is undefined! They are not the same!"
https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-2/a/limits-intro
What we are saying when we write 0.333... = 1/3 is that it is equal to 1/3 in the limit as the number of decimal places approaches infinity, even though it never equals 1/3 with a finite number of decimal places.
Last edited:
Neddy Bate
Valued Senior Member
In the limit as n --> infinity, your second term 1/10^n --> 0. So .9R = 1 - 1/10^n is not less than 1 in that case, it is equal to 1.
Yes. I agree with this. Nevertheless, the limit of the function is well defined and unambiguous.
Extending this idea, we can find the limits of infinite series, formally and unambiguously.
Also, the notion of a limit is also typically used to define the operations of differentiation and integration, which are fundamental to entire sub-fields of mathematics.
I'm sure you will agree. (?)
But it can be completed. We complete infinite things all the time, like moving from A to B and crossing all the infinite points inbetween.
I think he is confusing mathematics with practical matters, which is why he brings the issue of how long it might take into the equation (pun intended!). But even then he is wrong. Practically summing an infinite sequence element by element is impossible if and only if it takes a non-zero finite time each time , as even if we could do each sum in some small finite amount of time, it would take an infinite amount of time. But maths isn't about what is possible practically. That is left to physics etc. Mathematics is abstract.
The way it was explained to me, back at school:
Imagine a ball is shot up from ground level and reaches a height of X metres. It then comes back down and hits the ground after time Y seconds. It loses energy each time which means that the subsequent bounce takes only Y/2 seconds to go up and down (to a reduced height). And loses energy such that the next bounce takes Y/4 seconds, and then Y/8 etc.
In the realm of mathematics, this will result in an infinite sequence of bounces, to a reducing height each time, right?
So how long does this infinite sequence take to complete?
It sounds counter-intuitive (to those who have muddled thinking, at least) but an infinite series of events can conclude in a finite time - at least in the abstract world of mathematics. In this case the infinite number of bounces lasts precisely 2Y seconds (Y + 1/2 Y + 1/4 Y... etc).
So once 2Y seconds have elapsed, an infinite number of bounces have taken place.
Phyti, and MD, seem to think that it would be mathematically impossible to ever get to 2Y seconds. Which is pretty much Zeno all over again.
James;
An example from Wiki:
For example, consider f(x) = 2x/(x + 1).f(100) = 1.9802 f(1000) = 1.9980 f(10000) = 1.9998As x becomes extremely large, the value of f(x) approaches 2, and the value of f(x) can be made as close to 2 as one could wish just by picking x sufficiently large. In this case, we say that the limit of f(x) as x approaches infinity is 2.In words:
The limit of f(x)=2, as x →∞.
-----------------------------------------
There are two different things being considered, f(x) and limit. The value 2 is assigned to the limit, not f(x).
Whether sequences converge or diverge or become an equality, depends on the f(x).
Specifically to the subject of inverse integers 1/n, if n is not a factor of 10 (2 or 5), a repeating decimal occurs.
In the world of pure mathematics, the f(x) for these sequences do not equal their assigned limits.
In the world of applied mathematics, these sequences are truncated for practical/convenient reasons.
Here is an example of using a limit as it relates to calculus.
Given, a continuous curve containing the points p, q', q.
The secant is the straight line v from a fixed point p to a variable point q.
As v rotates clockwise, it intersects the curve at a point q' closer to p.
The slope of v is m=y/x, the coordinates of its end point.
As q moves closer to p, the limit of the angle corresponding to m is theta.
When q coincides with p, v becomes the tangent for the curve at p.
An example from Wiki:
For example, consider f(x) = 2x/(x + 1).f(100) = 1.9802 f(1000) = 1.9980 f(10000) = 1.9998As x becomes extremely large, the value of f(x) approaches 2, and the value of f(x) can be made as close to 2 as one could wish just by picking x sufficiently large. In this case, we say that the limit of f(x) as x approaches infinity is 2.In words:
The limit of f(x)=2, as x →∞.
-----------------------------------------
There are two different things being considered, f(x) and limit. The value 2 is assigned to the limit, not f(x).
Whether sequences converge or diverge or become an equality, depends on the f(x).
Specifically to the subject of inverse integers 1/n, if n is not a factor of 10 (2 or 5), a repeating decimal occurs.
In the world of pure mathematics, the f(x) for these sequences do not equal their assigned limits.
In the world of applied mathematics, these sequences are truncated for practical/convenient reasons.
Here is an example of using a limit as it relates to calculus.
Given, a continuous curve containing the points p, q', q.
The secant is the straight line v from a fixed point p to a variable point q.
As v rotates clockwise, it intersects the curve at a point q' closer to p.
The slope of v is m=y/x, the coordinates of its end point.
As q moves closer to p, the limit of the angle corresponding to m is theta.
When q coincides with p, v becomes the tangent for the curve at p.
phyti:
I can't tell from your most recent post whether you agree or disagree with what I wrote in my previous post.
Example: consider the series 0.1 + 0.01 + 0.001 + ...
One of the partial sums for this series is:
$$S_n=\sum \limits_{i=1}^{n} \frac{1}{10^i}$$
For instance, when n=3, the partial sum $$S_3=0.111$$.
Now, when we consider the infinite series, we consider the limit of the sequence of partial sums (i.e. $$S=\lim \limits_{n\rightarrow \infty} S_n$$ is the sum of the infinite series).
For this particular series, the sum is 1/9. Therefore, we are justified in writing
0.1 + 0.01 + 0.001 + 0.0001 + ... = 0.1111... = 1/9
It is a convention of mathematics that the notation "..." means "continue in the same way" (forever, in this context). In the case of the sum of an infinite series, this implies that we take the limit as $$n\rightarrow \infty$$ of the partial sums.
Do you agree?
Are we in agreement or not? Is there anything in what I've written here that you would dispute?
I can't tell from your most recent post whether you agree or disagree with what I wrote in my previous post.
Yes. This is correct.
Correct.
Yes.
Strictly speaking, when dealing with decimal expansions, we're talking about series, not sequences. Although, having said that, I'm sure that you're aware that the sum of a series is defined to be the limit of the sequence of partial sums.
Example: consider the series 0.1 + 0.01 + 0.001 + ...
One of the partial sums for this series is:
$$S_n=\sum \limits_{i=1}^{n} \frac{1}{10^i}$$
For instance, when n=3, the partial sum $$S_3=0.111$$.
Now, when we consider the infinite series, we consider the limit of the sequence of partial sums (i.e. $$S=\lim \limits_{n\rightarrow \infty} S_n$$ is the sum of the infinite series).
For this particular series, the sum is 1/9. Therefore, we are justified in writing
0.1 + 0.01 + 0.001 + 0.0001 + ... = 0.1111... = 1/9
It is a convention of mathematics that the notation "..." means "continue in the same way" (forever, in this context). In the case of the sum of an infinite series, this implies that we take the limit as $$n\rightarrow \infty$$ of the partial sums.
Do you agree?
No partial sum is equal to the sum of the infinite series, if that's what you mean.
It depends on the context. An 8-digit calculator might truncate 1/9 to 0.11111111, but that doesn't mean that an engineer will always write 1/9 as 0.11111111. Often, the exact value will be used, in practice.
Yes. And so...?
Are we in agreement or not? Is there anything in what I've written here that you would dispute?
Last edited:
Sarkus;
There is nothing 'infinite' in human experience.
There is no such thing as 'infinitely' long. Have you looked up the meaning of 'infinite'?
It also means not measurable.You can't in reality or fantasy, measure a thing/sequence without a length (boundary).
There is no number between .9r and 1. That's what you have to accept if you insist on using 'infinity' as a number, and making an abstraction real.
I can't speak for MD, but Zeno showed that the concept of continuous space and time led to illogical results.
There is nothing 'infinite' in human experience.
There is no such thing as 'infinitely' long. Have you looked up the meaning of 'infinite'?
It also means not measurable.You can't in reality or fantasy, measure a thing/sequence without a length (boundary).
There is no number between .9r and 1. That's what you have to accept if you insist on using 'infinity' as a number, and making an abstraction real.
You are contradicting yourself.
It's also moving a finite distance, and points being 'dimensionless', cannot be measured.
The finite number of oscillations is reduced by loss of energy to the ground until the ball reaches a state of equilibrium and it stops. The time will depend on the properties of the ball.
I can't speak for MD, but Zeno showed that the concept of continuous space and time led to illogical results.
That has nothing to do with it. Infinity is not a number but try to eliminate it and still do calculus. There are also no imaginary numbers in the 'human experience', but it is a great mathematical tool that helps solve real problems that are in the 'human experience'.
Trying to get through to you certainly feels like it.
But so what if there is nothing "infinite" in the human experience? Seriously. So what?
Maths isn't about human experience. It isn't even necessarily about reality. It is an abstraction. It is filled with concepts that have no "reality" but are meaningful tools.
Yes, there is such a thing as "infinitely" long. Not in physical reality, but in the abstract world of maths there clearly are: the sequence of real numbers is infinitely long etc. Why are you only focussing on what is possible practically?
Yes you can. An infinite sequence can sum to a finite number, as we have been trying to get through to you. The sum of the sequence 1 + 1/2 + 1/4 + 1/8 + ... etc sums to a finite number.
Therefore they are the same number, for all numbers that are not equal have a different number half-way between them.
I'm not making an abstraction real, and I am using infinity as a concept, not a number.
No I'm not. Maths requires no element of time. Humans do to perform mathematical functions, but maths itself requires no element of time. 2+2=4 requires no time to be correct, only for humans to do the sum. So no contradiction.
So what. No one's asking you to measure them. It's conceptual.
In the abstract world of maths, in this example there is a series of infinite oscillations, and they are completed in a finite time. Real world physics need not be applied. It is conceptual / abstract. It is maths.
Nope, I've given you all the information you need to come up with the answer I gave.There is no "equilibrium" in the maths I gave. Just an infinite series of bounces, all complete in a finite time. That's maths for you. Your insistence on only considering real-world applications is doing you no favours.
James;
No. You agreed the value is assigned to the limit L, and f(x) will become closer to L as x increases. It qualifies as convergent, but closer is not equal. The string/sequence of 1's remains a string of 1's. Multiply the string by 9 forming a string of 9's. It would be magic for the 9's to become 0's. If x in the f(x) approaches a specific value v within the interval, then it's possible for f(x) to equal the limit v.
No it isn't. The sample of the sequence is supposed to show the form of the terms, and typically an ellipsis (...) to indicate no last term. Using the convention of an index n implies the natural integers, or 'for all n'.
There are no infinite series, or infinite numbers, but there are processes (rules of formation) with the potential to form numbers of any size. An example is n'=n+1 which generates the set of integers. It can form greater numbers, but not a greatest number.
The common problem to these disputes is the abstract concept of the continuum with its unlimited division, and using 'infinity' as a number.
Chemistry and now quantum physics reveals the world as composed of discrete quantities of matter and discrete amounts of energy.
No. You agreed the value is assigned to the limit L, and f(x) will become closer to L as x increases. It qualifies as convergent, but closer is not equal. The string/sequence of 1's remains a string of 1's. Multiply the string by 9 forming a string of 9's. It would be magic for the 9's to become 0's. If x in the f(x) approaches a specific value v within the interval, then it's possible for f(x) to equal the limit v.
No it isn't. The sample of the sequence is supposed to show the form of the terms, and typically an ellipsis (...) to indicate no last term. Using the convention of an index n implies the natural integers, or 'for all n'.
There are no infinite series, or infinite numbers, but there are processes (rules of formation) with the potential to form numbers of any size. An example is n'=n+1 which generates the set of integers. It can form greater numbers, but not a greatest number.
The common problem to these disputes is the abstract concept of the continuum with its unlimited division, and using 'infinity' as a number.
Chemistry and now quantum physics reveals the world as composed of discrete quantities of matter and discrete amounts of energy.