|
Joined: Mar 2000
Posts: 315
enthusiast
|
enthusiast
Joined: Mar 2000
Posts: 315 |
I was expressing (4-0.99999...) as 3.000...0001 It is not correct to use "usual method" for computation in the infinite case. Before computation, you have to give a precise meaning to symbols! Infinity has been very controversial in the history of maths. The fact is that a real number (that can be thought as a point in the line, measured respect a given unit segment" )has just one decimal representation, or two, but it can happen just in the cases similar to that one, for example 2,34 = 2,33999999999999.......
The only time I have seen this fact used was to prove that "real numbers" are "more" than "integers numbers" I know that it is hard to imagine that there are several steps of infinities... (infinite,indeed) and in fact it has been understood at the end of XIX century . The meaning - more or less - is that it is possible to imagine the integers numbers as a subset of the real numbers, but not viceversa.
I suppose that this post will start another discussion: we would say that I am throwing a stone in the pond ( gettare un sasso nello stagno), in order to excite something quiet.
|
|
|
|
Joined: Jan 2001
Posts: 1,819
Pooh-Bah
|
Pooh-Bah
Joined: Jan 2001
Posts: 1,819 |
Thank you but I am not having trouble with the concept of infinity or infinities, and I understand that the integers are a subset of the real numbers. In fact, the relationship of integers to real numbers can be used to illustrate that there is more than one infinity. We all agree that there are an infinite number of integers, right? And yet for any two consecutive integers such 2 and 3 (or any two integers for that matter), there is an infinite number of real numbers that can be represented in decimal form between them, such as 2.9, 2.99, 2.998, 2.999... So the infinite number of real numbers must be greater than the infinite number of integers.
All I am arguing is that there are subtle mathematical differences between two numbers if one of them a tiny bit larger than 3 and the other is an equally tiny bit smaller than three. I can easily imagine 2.99999... I can just as easily imagine a number which is greater than three by the same degree. They are all three points on a number line, with the integer 3 in the middle and the two repeating-decimal numbers flanking it, equidistant on either side. The fact that the difference between them is essentially unmeasurably small doesn't change the fact that they are different.
To put it another way, take the limit of x as x approaches 3 from a number greater than three, and then take the limit of x as x approaches 3 from a value less than three. You plug those two values into f(x)=3/(x-3) and your answers will converge on to two different values. In fact, the answers converge on infinity, the first one with a positive sign and the other with a negative sign. That's because in the first case the denominator remains positive, because you are subtracting a number less than 3 from 3. As x converges on its limit of 3 from a number greater than three, (x-3) converges on zero, but remains greater than zero. The casual observer can see that for a number converging on 3 from say, 2, (x-3) remains a negative number that converges on zero "from below."
So my argument is that if 2.9999... is the same as 3, then (4-0.999999999...) is the same as three. But they are not as f(x)=3/(x-3) shows. What I am arguing more or less is that 2.9999... converges on 3 but is not quite the same as 3. Now maybe there is some flaw in my reasoning, but nobody has pointed it out yet to my satisfaction.
|
|
|
|
Joined: Jan 2001
Posts: 1,156
old hand
|
old hand
Joined: Jan 2001
Posts: 1,156 |
I think that by putting some of your above statements together, we've got all the bits we need to prove that 2.999999... = 3. The key one was where you said that you were writing 0.0000....000001 as 1/(10^n) (or 10^(-n)) with n approaching infinity. And as Faldage correctly stated, that's equal to zero. So now put it all together:
2.99999999... = lim (n->inf) (3 - 10^(-n)) - this is more or less what was stated above, right? That it differs from 3 by a small amount, namely 10^(-n).
But using the rules of limits, which have been implicitly used above, so we must be in agreement on them, re-write the RHS, so now we have
2.99999999... = lim (n->inf) (3) - lim (n->inf) (10^(-n))
Again using the rules of limits, the limit of a constant (the first term) is a constant itself, and the limit of the second term is, as Faldage said, zero, so we have
2.99999999... = 3 - 0 2.99999999... = 3
Making up an entirely different function from the one being discussed that shows that the limit OF THE PARTICULAR FUNCTION is different when approaching from above or below has no relevance to this argument, which is "how do we write the number 3?", not "can we think of a function that has a limit that is undefined at 3?". Certainly f(x) = 3/(x-3) is undefined at x = 3 but that doesn't tell you a lot about the point x = 3 on the number line.
|
|
|
|
Joined: Dec 2000
Posts: 13,803
Carpal Tunnel
|
Carpal Tunnel
Joined: Dec 2000
Posts: 13,803 |
In emanuela's post above, about the real numbers being of a higher order of infinity than the integers, it should be noted that the numbers represented by, e.g., 2.999999999... do not exhaustively represent the real numbers. A rational number is one that can be represented as the ratio of two integers. Thus, 0.3333333... can be represented as the ratio of 1 to 3. A feature of the rational numbers is that they can be represented as a repeating decimal, i.e., one in which a pattern of some finite number of digits repeats indefinitely at the end. 1/11=0.090909..., for example and 1/13=0.076923076923076923.... In some cases the repeating digit(s) will be 0, e.g., 1/8=0.12500000... In this case there is a non-repeating string of digits before we get to the repeating part. The repeating part of one of this type doesn't have to be zero; 2/15=0.13333333...
There is another type of real number, the irrational. Examples are pi, the square root of two, and e, the base of the natural logs. These cannot be represented as repeating decimals.
The the order of infinity of the rational numbers is the same as the order of infinity of the integers. Bean or emanuela could prove this for us. Add in the irrational numbers and *that's when you get the higher order of infinity.
This notwithstanding the fact that, as Alex pointed out, between any two rational numbers you can put more rational numbers, even an infinite number of rational numbers; there are no adjacent pairs of rational numbers.
|
|
|
|
Joined: Jan 2001
Posts: 1,819
Pooh-Bah
|
Pooh-Bah
Joined: Jan 2001
Posts: 1,819 |
From M-W:... In reply to:
LIMIT: 6 a : a number whose numerical difference from a mathematical function is arbitrarily small for all values of the independent variables that are sufficiently close to but not equal to given prescribed numbers or that are sufficiently large positively or negatively b : a number that for an infinite sequence of numbers is such that ultimately each of the remaining terms of the sequence differs from this number by less than any given positive amount
The above definition explains why I think that 2.9999... illustrates the concept of a limit of 3. The limit is 3, but as x approaches the limit it becomes infinitely close to 3 without quite becoming three. Sounds like 2.99999... to me.
Bean wrote: In reply to:
Making up an entirely different function from the one being discussed that shows that the limit OF THE PARTICULAR FUNCTION is different when approaching from above or below has no relevance to this argument, which is "how do we write the number 3?", not "can we think of a function that has a limit that is undefined at 3?". Certainly f(x) = 3/(x-3) is undefined at x = 3 but that doesn't tell you a lot about the point x = 3 on the number line..
Yes but my point has to do with those two values that are not quite three. The function serves to illustrate the difference between 2.999999... and (4-0.99999...). The point at which x=3 itself has no bearing whatsoever on my argument.
Of course the big problem here is that I seem to be picking one nit and you seem to be picking another.
|
|
|
|
Joined: Mar 2000
Posts: 315
enthusiast
|
enthusiast
Joined: Mar 2000
Posts: 315 |
My question is : what do you say to be a number? You cannot treat numbers before knowing what they are, and giving a meaning to such infinite way of writing.
My answer would be " a point in the real line, in which I fixed a point 0, and a point 1, in order to have the unit to measure segments with" In this context, which number is 2,99......? BY DEFINITION, it is the limit of the points given by the succession 2 2,9 2,99 ... so that your arguments prove indeed that 2.99999...is 1.
Interesting to me, the discussion here is similar to that historically happened about infinity "in fieri" = latin,for "becoming" and "in acto" (not sure) = "already existing".
|
|
|
|
Joined: Jan 2001
Posts: 1,819
Pooh-Bah
|
Pooh-Bah
Joined: Jan 2001
Posts: 1,819 |
I feel like emanuela and I are arguing semantics at this point rather than mathematics, and it looks like the kind of thing that can just go around and around in circles (thus leading to a discussion of pi, perhaps).
I may be wrong or I may be right; I don't know. I don't claim to be a professional mathematician, and it has been 8 years since my bachelor's degree in mathematics, so the odds are that I'm wrong and emanuela is right. But my intellectual curiosity is not really satisfied. Maybe someone else can set me straight on this. There must be some demonstrable flaw in my logic.
|
|
|
|
Joined: Dec 2000
Posts: 13,803
Carpal Tunnel
|
Carpal Tunnel
Joined: Dec 2000
Posts: 13,803 |
I think the point here, Alex, is that all values of the independent variables that are sufficiently close to but not equal to given prescribed numbers holds as long as you have a finite number of 9s trailing after your 2. You can add 9s arbitrarily and you will get closer and closer to the limit of 3 as long as you have a finite number of 9s. Once you reach an infinite number of 9s (not possible in the real world, but what the notation 2.99999... means) you are outside the realm of the defintion you quoted. You are no longer approaching the limit, you have reached it.
|
|
|
|
Joined: Aug 2001
Posts: 10,713 Likes: 2
Carpal Tunnel
|
Carpal Tunnel
Joined: Aug 2001
Posts: 10,713 Likes: 2 |
One problem repeatedy appearing in this thread is that the intuitive meaning of some of our words is not the same as the mathematical one. In particular (not that I would ever try to get us into a political discussion) it all comes down to precisely what the meaning of "is" is. Or "equals," more specifically.
In mathematics, "equality" means that no matter how small a difference you specify, the difference between the two objects under consideration is less than that. For the matter at hand, it translates as: no matter how close you want to require, the series 0.999999... is closer still to 1.000000... . And there is no "exactly 1"; it's just a convenient shorthand for the endless decimal 1.000000..., which is the same as 0.999999..., just written in a different way.
Mathematics has some very precise meanings for its words, if for no other reason than to resolve this kind of quandary. True, if you reject the mathematical definition of the word, the discrepancy won't go away. There is an underlying assumption of "under the rules of mathematics..." that some are invoking and others are ignoring.
"Infinite" and "infinity" are two other words that cause confusion, because the mathematical and the common meanings aren't quite the same.
See - it all comes back to words, after all!
|
|
|
Forums16
Topics13,913
Posts229,580
Members9,187
|
Most Online3,341 Dec 9th, 2011
|
|
0 members (),
332
guests, and
0
robots. |
Key:
Admin,
Global Mod,
Mod
|
|
|
|