Actually, I'm saying exactly the opposite
Infinite should not have a finite value attached to it ever. It should not be used as an adjective to exemplify the finite and finite should not be use to exemplify infinite
Infinitesimally finite, infinitely finite, finite infinity, etc. These are plain english examples of mathematics and they are not accurate
I'm not objecting to "inventing something outside the reals". I'm objecting to something (infinity) that IS outside the reals, being allowed to be equal to or lesser than the reals.
I'm objecting to "something so small it behaves like zero, even if it isn't".
I'm also objecting to something being so large it behaves like 1, even if it isn't.
I'm also objecting to "the idea that the infinitely small and large has served us well and likely doesn't create any serious problems".
And lastly, "I'm of the mind that YOUR "finite infinite" creates more problems than it solves and results in more inconsistancies and illogical conclusions" is NOT MINE, but belongs to mathematics.
"It would be interesting to see you show that the world would be better without infinity", is again, the very opposite of what I'm actually saying.
The world would be better if it refused to accept, that infinity can be "toyed with" like mathematics has done.
Infinite is always greater than finite, not in numbers however, but in principle. Infinity doesn't have anything whatsoever, to do with finite numbers or geometry.
It is the field upon which finite numbers and geometry "occur", but it is never affected by nor should it be induded with, the finite in any way.
The only thing the "infinte field" is made up of, is a dimensionless "center point" that can't be counted, grow in size, rotate, or have its infinite spacing altered by the finite.
There is something greater than infinity, but it is not the finite.
I think you can trace the logic and see the equivalent of a cyclic loop and people intuitively understand that we can't continually reevaluate it it over and over again, so a point to break out is selected arbitrarily and one person can get a different result than someone else (it's like tossing something in either the infinite or infinitesimal bins).
So we have the cycle of:
1) .9 is <1
2) .9<.99
then back to 1:
1) .99<1
2) .99<.999<1
etc.
So it just cycles back and forth and becomes an infinite (not in any numerical sense, but in a logic one). There is no specific value for .9r because the number of digits isn't specified though we know it can't be 1 because no iteration of this loop ever makes it equal to 1, but it could be as close as we'd generally ever need it to be to 1, so it's generally assumed to be 1, but the details of how that relationship were derived is overlooked.
The claim that .9r=1, isn't true in a simple numeric sense. Instead .9r is swapped, using a priorly known or defined limit of its convergence, 1. 1 wasn't derived from .9r but instead .9r is derived from 1. So if there's a conflict in the logic, you break out before the infinite loop and backtrace.
Here's a good question - how was .9r derived? Likely the vast majority of its derivations are from an uncompleted division. 1/3 of course can't be precisely written as a decimal value alone. At each iteration of the division there's a constant remainder that never disappears, though on a decimal scale it could be become arbitrarily small, so the logic is that if we could keep extending it out forever, then we could reach a point at which the value becomes insignificant. Truly the value was initially just 1/3 and the remainder is the 1 in 1/3, so it's explicit and has no error .. .1=1, by definition. We don't have to compute whether or not 1/3=1/3 (and we can prove 2/6=3/9=4/12=...=n/3n=1/3 without creating any logic conflicts).
But what's overlooked is that multiple infinities or multipled infinitesimals can interact and create a new value.
So for example, if we have the sequence 1/2+1/4+1/8+..., we might try to assume this sequence sums to exactly 1 when we have an infinite number of terms, but truly it only approaches 1 and never equals 1 (you could rewrite this as 1/2^1+1/2^2+1/2^3+...+1/2^n+1/2^n and double up the last term and then this could be proven to be equal to one without any infinitesimal remainders, even as n goes off to infinity).
But we can see the difference here:
(1/2)^2=1/4
(1/2+1/4)^4~=.3164
(1/2+1/4+1/8)^8~=.3436
...
(1/2+1/4+1/8+...+1/2^n)^2^n~=1/e~=.3679 (n>>0)
Notice these values are actually approximate values that depend upon n, even as n goes toward infinity. Though in many cases we can substitute them with a known limit that has a non-infinite derivation and can be proven to be arbitrarily close to the value as n goes toward infinity, but the proof is never a result of an infinite loop, but instead a backtracing that avoids the infinite loop and makes substitutions for the recursion instead that provide a straight, "feed forward" computation instead. In the case of e, there is no such way to compute e and e only stands for an infinite process. It's truly symbolic for an infinite process and not a number, though we can compute numbers very close to e.
One the other hand if we take this process and evaluate it:
(1/2+1/2)^2=1
(1/2+1/4+1/4)^4=1
(1/2+1/4+1/8+1/8)^8=1
...
(1/2+1/4+1/8+...+1/2^n+1/2^n)^2^n=1
For all n, including n->infinity.
This is a computation that has no cyclic definition and is entirely deterministically computable as a number that can be compared again other computable numbers.
Consider this, how do we know pi is larger than e? We actually have to do part of the computation and then prove that the unknown components of each are insignificant enough not to change the comparison. So once you know something like pi=3.??? and e=2.??? then you can determine pi is greater than e. This doesn't mean that pi or e were computed in their entirity, but that there's a non-zero difference that's detectable between them and the same detectable non-zero difference between .9r and 1 exists as well because we can show that the difference of 1/n or 1/2^n, even in the limit as n->infinity is still detectable via. the same techniques used in an attempt to prove that 1/n=0 (or 1/2^n=0 etc) because (1-1/n)^n for n>>0 is approximately 1/e, whereas 1^n=1, so 1-1/n, even as n->infinity never actually becomes equal to 1, it just gets close enough that if the difference doesn't become significant later we can ignore it (but that's different from an identity).
If we have two equations a>b and b>a and ask if a=c, we could do the classical 50s B-movie robot routine of "but if a=c then b>a which means b>c and a>c ..." and maybe follow up with a "Does not compute. Does not compute! Does not compute!! ... !!!" and then add some billowing smoke effects etc. while the circuits melt down or we can just realistically realize that you can't prove it deterministically and use calculus and impose a limit instead to approximate the value with a non-infinite process, 1 and skip the problem of trying to divide to relatively prime numbers in to each other (there's a reason why so much research is put into analyzing prime relationships and swapping in approximate values isn't good enough).
Oh well, most the time there's not enough of a difference to worry about, but it masks the issue that you really can't divide 1 by 9 or by 3 etc. Relatively prime values, for purposes of division, are incompatible and might as well lie in different dimensions. We can reduce 4/10 to 2/5 by removing the common multiple of 2, but the remaining 2 and 5 are effectively left undefined for division and writing this as .4 in decimal is equivalent to leaving it as 4/10. If we move to a different base representation, we could write something like 1/3 in base 3 as .1, but then we trade off having a finite way in this representation to write 2/5. Fractions provide a mixture of such prime relationships to be written together, but they don't reduce these prime relationships to a single number, though you can sum them together as products of primes but now you're left with a large fraction containing all those primes. We can call this a real number, but that means little as we could describe a computer program as a real number also (just line all the bits up in a single stream) and still fit it into a "real" value between 0 and .0000001. So the "strength" of the reals to represent information within a infinite ordered linear sequence is a tradeoff in not being able to determine precisely where those values lie within that field (it's interesting to see the uncertainty principle arise in even mathematically constructed 1-D spaces, when the field is assumed to be infinitely continuous).
But anyway, there are some interesting properties involved in this that can be correlated to physics. The prime numbers, relative to multiplication and division, create the equivalent of independent dimensions that addition allows to be connected or traversed. So if you only have the numbers 2, 5 and 17, for example, you can imagine multiplication moving along a logarithmic path from infinitesimal on one extreme to infinite on the other and these numbers lie on independent lines (Beginning from 1, no sequence of multiplications or divisions by 2 or 5 will give you 17, though if you allow motions in a "parallel direction" by using addition, you can location 17 in many places ... it could appear to be a field of reals containing complex fractal arrangements of 17, yet in the alternate space there'd be only 1 of each prime ... though there are an infinite number of primes. So imagine each prime number representing something like a quark, electron etc. (though an electron might be seen as a composite number because it contains quarks, for example if a single quark represented 2 and an electron was three dimensional, then it should contain at least 3 of these or 2*2*2, which would be interesting because of the typical 8 valences associated with chemistry ... just musing here a bit) and then the various arithmetic pathways through space that could encounter one of these and the typical light speed distance would be the minimum number of steps to reach that location, though of course, alternate pathways through space could be found as well ... just an interesting idea).
QUOTE (N O M+)
Yet again all you prove is that you don't have a clue about infinity
And, yet again, you guys offer little of anything except one liners and repetitions of common misconceptions.
QUOTE (N O M+)
.9r * .3r = .3r
Actually it's closer to .3r2(6)r7.
(1-z)*(1-z)/3=(1-2z+z^2)/3=1/3-(2/3)z+z^2
If that's a bit unclear, I'll try to describe what's happening in a pseudo decimal format like this:
(.9 * x) * (.3 * x) = .27 * x^2 (x=(1.1)r)
x^2=x*x=(1.1)r*(1.1)r=1.234567...7654321=
(There's not an easy nomenclature for this in decimal, so the equation above is better, but this should give you an idea of what it looks like)
.27+.054+.0081+...+.(0)r1*(...+81+5.4+.27)
So it may seem like the ... in a number should be ignorable, but mathematically it represents a non-zero infinitesimal value that's easy to overlook if we're just going to "roughly eyeball" the value of a number.
QUOTE (Precursor562+)
sqrt(.999...9) = .999...95
Yes, in the limit sqrt(.9r) approaches (1+.9r)/2, or it resides approximately halfway between .9r and 1. This ratio is approximately:
.9r^a~=(1*(1-a)+.9r*a)~=1-a+.9r*a
So, for example, .99999^.2~=1-.2+.99999*.2~=.8+.199998~=.999998
It's actually closer to .99999799999199995, but the approximation works. Along similar lines we can derive, e as the ratio of the error between .9r and 1.
NoCleverName
19th April 2007 - 08:01 PM
Well, E.E., at least your argument is intellectually defensible whereas StevenA, Precursor, et. al. are stuck in "calculation" mode with really nothing to add to the discussion.
I would be interested to hear if, in your view, the practical application of mathematics is flawed --- in the sense that, say, engineering results can't be trusted --- by what you view as the incorrect concept of infinity. Or, on the other hand, is it just the fact that unsettling results like .9r = 1 devolve from it with no real harm otherwise.
mmax
19th April 2007 - 08:04 PM
QUOTE (Precursor562+Apr 19 2007, 09:48 AM)
So with .9r = .9n n->infinity
.9n = 1 - 1/10^n
n->infinity
1/10^n n -> infinity behaves exactly like zero. There's a section in the link I last gave that explains why it's so.
QUOTE
Within the real numbers, 1/infinity (which I will represent in this essay with the variable E, for epsilon, a standard name for very small (but not infinitely small!) numbers) behaves just like the number zero. As an example of this, I will refute:
Variations on: .333... doesn't really equal 1/3. It is just slightly less than 1/3 in the same way that .999... is just slightly less than 1.
(Now remember, I have to put some words into objectors' mouths, since they do not present their objections precisely, which is the whole problem. I apologize if I mischaracterize an objection somehow.) The only way to make this consistent, I think, is to claim that:
1/3 = .333... + E (remember that E is 1/infinity)
This would mean that:
1 = 3/3 = 3(.333... + E) = .999... + 3E
So if you object like this, tell me about this 3E number. Is it greater than E? If it is, then .999... isn't really as close as you can get to 1, is it? "Aha, Mr. Polymath! You yourself said that you can't use normal algebra on infinity, so it's still possible that the 3E number (the residue from tripling 1/3) might be the smallest possible postive number!" Yes, indeed, I said that. But if you're going to say that E and 3E are really the same thing, then you are saying exactly what I'm saying. If E and 3E are the same, then E is behaving exactly like the number zero. Or, if you don't like using algebra on this, we can just consider it like this: if .333... lacks something when it tries to be 1/3, and if .999... lacks the same thing when it tries to be 1, then either that lacking quantity is behaving just like the number 0, or .999... is actually not as close to 1 as you can get.
http://polymathematics.typepad.com/polymat...6/06/index.html
Precursor562
19th April 2007 - 08:55 PM
QUOTE
As for sqrt(0.9r), it would be defined as limit n->infinity of sqrt(1-0.1^n) = sqrt(1) = 1. Therefore sqrt(0.9r) = 1 (assuming positive square root obviously).
Absolute nonsense. The whole validity of this argument is the assumption that 1/10^n = 0. It doesn't equal zero. It equals a number depending on the value of n. Where n is given the limit of infinity this simply means that there is no limit to how large a value of n we can have. It can equal 1, 1k, 1M, 1G etc. There is no limit.
Given any finite value for n we will receive a decimal number such as
.1 when n=1, .01 when n=2, .001 when n=3 etc.
Where the number of zeros between the decimal and the 1 depend on the value of n. Give the value of n no limit (n->infinity) and we get the same for the number of zeros. This simply means there is no limit to the number of zeros between the decimal and 1 when there is no limit to what n can be.
This is true because there is no limit to the number of digits you can have after the decimal. There are only two ways this can not be true. One is a misconception of the term infinity. Often this is associated with the counter statement, "You can't have .0r1 because there is no end for the one to be there just like you can't have infinite+1". The reason you can't have infinite + 1 is because infinity is just the concept of "no limit".
The other is if there were a limit to how many digits can be found after the decimal. If you can only have 100 then you can't have 100 0s and then a 1 because that would leave you with 101 digits passing the limit.
.0r1 simply means there is no limit to how many 0s there can be between the decimal and the 1. Just as .9r means there is no limit to how many 9s there are after the decimal. All because there is no limit to how many digits you can have after the decimal.
QUOTE (->
| QUOTE |
| As for sqrt(0.9r), it would be defined as limit n->infinity of sqrt(1-0.1^n) = sqrt(1) = 1. Therefore sqrt(0.9r) = 1 (assuming positive square root obviously). |
Absolute nonsense. The whole validity of this argument is the assumption that 1/10^n = 0. It doesn't equal zero. It equals a number depending on the value of n. Where n is given the limit of infinity this simply means that there is no limit to how large a value of n we can have. It can equal 1, 1k, 1M, 1G etc. There is no limit.
Given any finite value for n we will receive a decimal number such as
.1 when n=1, .01 when n=2, .001 when n=3 etc.
Where the number of zeros between the decimal and the 1 depend on the value of n. Give the value of n no limit (n->infinity) and we get the same for the number of zeros. This simply means there is no limit to the number of zeros between the decimal and 1 when there is no limit to what n can be.
This is true because there is no limit to the number of digits you can have after the decimal. There are only two ways this can not be true. One is a misconception of the term infinity. Often this is associated with the counter statement, "You can't have .0r1 because there is no end for the one to be there just like you can't have infinite+1". The reason you can't have infinite + 1 is because infinity is just the concept of "no limit".
The other is if there were a limit to how many digits can be found after the decimal. If you can only have 100 then you can't have 100 0s and then a 1 because that would leave you with 101 digits passing the limit.
.0r1 simply means there is no limit to how many 0s there can be between the decimal and the 1. Just as .9r means there is no limit to how many 9s there are after the decimal. All because there is no limit to how many digits you can have after the decimal.
Flawed logic.
You're logic amounts to "It's not true for 0.9r because it's not true for some other number which isn't 0.9r". Not exactly unsuprising.
Flawed logic is to misconstrue the word infinity and ignore the pattern that emerges. A pattern that will not change no matter how far you take it.
1 - 1/10^1 = .9
1 - 1/10^2 = .99
1 - 1/10^3 = .999
etc.
Do you see the pattern? Do you think that this pattern will suddenly change at some point? If so tell me at what point it changes.
QUOTE
OK, I'm beginning to see how you view it --- you are saying even if something is infinite (small, large, whatever) it is still something and it should have the properties that everything else of the class shares. For example, I'm thinking you believe mathematicians look at the "infinitely small" as something so small as to dismiss it for all practical purposes. But you would say, "so what, it's still a number, no matter how small it is".
Replace "so what, it's still a number, no matter how small it is" with "so what, it's still not zero, no matter how small it is" and you have hit the nail on the head. Though that might be what you are saying, just thought I would clarify.
QUOTE (->
| QUOTE |
| OK, I'm beginning to see how you view it --- you are saying even if something is infinite (small, large, whatever) it is still something and it should have the properties that everything else of the class shares. For example, I'm thinking you believe mathematicians look at the "infinitely small" as something so small as to dismiss it for all practical purposes. But you would say, "so what, it's still a number, no matter how small it is". |
Replace "so what, it's still a number, no matter how small it is" with "so what, it's still not zero, no matter how small it is" and you have hit the nail on the head. Though that might be what you are saying, just thought I would clarify.
However, I think the mathematical view of the infinitely small is more like "something so small it behaves like zero, even if it isn't".
I think you're right. Unfortunately behaving like or rather similar to zero doesn't make it zero.
QUOTE
In other words, mathematical infinitely small is no longer a real number so it doesn't have properties like "rational", "irrational", etc. So then your basic objection is the idea of inventing something outside the real numbers; you argue it is a sham to do so and that it creates illogical consequences to have such concepts.
If that's what it takes then that's what it takes.
Although you were addressing Eric England it sure as hell felt like you were addressing me.
QUOTE (->
| QUOTE |
In other words, mathematical infinitely small is no longer a real number so it doesn't have properties like "rational", "irrational", etc. So then your basic objection is the idea of inventing something outside the real numbers; you argue it is a sham to do so and that it creates illogical consequences to have such concepts.
|
If that's what it takes then that's what it takes.
Although you were addressing Eric England it sure as hell felt like you were addressing me.
http://polymathematics.typepad.com/polymat...6/06/index.html
Sorry but a blog doesn't make for valid proof. It could have been written by anybody.
It also only covers the same lame spoofs.
1/3 != .3 because one is a process that yields an error and the other is the result of the endless process of trying to work out that error.
x = .9r
10x = .9r
etc. has already been shown to be a spoof.
When you multiply by ten you shift over. They treat the .9r in 9.9r as the exact same as the .9r you started with. This is the same as saying .999 * 10 = 9.999.
x = .999
10x = 9.999
10x - x = 9.999 - .999
9x = 9
x = 1
Oh look I just showed that .999 = 1
NoCleverName
19th April 2007 - 09:17 PM
I'm afraid this thread has descended into pointless --- dare I say it --- repetition. On the one hand we have the mathematical establishment (and I use that word purposefully) that can "prove" in endless fashion that according to the rules they set there can be no doubt about .9r = 1. On the other, we have those who are suspicious of the establishment (the "question authority" crowd) who believe the "establishment is covering up some dirty little secret". Both sides are entrenched, and it seems that we can find no acceptable arbiter to break the deadlock.
I might observe that all of this was argued back and forth within the profession long ago; so that is that.
One final note to combatants and lurkers alike: if you get this question on a test, you know what to answer to get it "right", even if you don't believe it.
rpenner
19th April 2007 - 09:19 PM
Ouch, that's the Dedekindest Cut of all.
mmax
19th April 2007 - 09:32 PM
QUOTE (Precursor562+Apr 19 2007, 02:55 PM)
When you multiply by ten you shift over. They treat the .9r in 9.9r as the exact same as the .9r you started with. This is the same as saying .999 * 10 = 9.999.
x = .999
10x = 9.999
10x - x = 9.999 - .999
9x = 9
x = 1
Oh look I just showed that .999 = 1
You're telling me that you see a difference between the .9r in 9.9r and the original 0.9r??? If so, then please enlighten us what the difference is. This is math, if there's a difference between two values then you should be able to find out what it is. And if you say the difference is something like 1/infinity then I'll say that there is no place where the value 1/infinity behaves differently than zero.
Eric England
19th April 2007 - 09:38 PM
QUOTE (NoCleverName+Apr 19 2007, 01:01 PM)
I would be interested to hear if, in your view, the practical application of mathematics is flawed --- in the sense that, say, engineering results can't be trusted --- by what you view as the incorrect concept of infinity. Or, on the other hand, is it just the fact that unsettling results like .9r = 1 devolve from it with no real harm otherwise.
As I've stated it before in this thread, tolerance at any scale is not what I'm referring to.
Anywhere from .9 to any number of 9s after that (depending on the requirements of the application), is perfectly acceptable as 1.
Engineering (no matter the scale) need not concern itself with infinity.
Mathematics in the broader context, however, needs to "properly" concern itself with infinity.
To say it is "equal to, lesser than, and greater than" finite (limits/sets/etc.) is showing less than a proper concern.
The "infinite field of center points" I referred to, is always and in all ways GREATER THAN, the apparantly finite 1 and 0.
I say "apparently finite" quite on purpose.
Precursor562
19th April 2007 - 10:13 PM
QUOTE
You're telling me that you see a difference between the .9r in 9.9r and the original 0.9r??? If so, then please enlighten us what the difference is. This is math, if there's a difference between two values then you should be able to find out what it is. And if you say the difference is something like 1/infinity then I'll say that there is no place where the value 1/infinity behaves differently than zero.
Well lets see.
First there is .9r = .9n; n->infinity
Would you agree there? Where n represents the number of 9s?
Then
.9n = 1 - 1/10^n
n->infinity
With those then
x = (1 - 1/10^n)
10x = 10(1 - 1/10^n)10x - x = 10(1 - 1/10^n) - (1 - 1/10^n)
9x = 9(1 - 1/10^n)
x = (1 - 1/10^n)
So we have 10(1 - 1/10^n) which gives us 10 - 10/10^n.
Now you ask "what the difference between the .9r in 9.9r and the original .9r?" and I answer with the question, what is the difference between 1/10^n and 10/10^n?
Here is the difference
As n = 1, 2 , 3, 4, 5, etc. to infinity the first is
.1, .01, .001, .0001 etc. to infinity.
The second is
1, .1, .01, .0001 etc. to infinity once gain this is as n = 1, 2, 3, 4, etc. to infinity (n->infinity)
1/10^n != 10/10^n
where n = n
So you start with .999.... (.9r)
Where the 9s are repeating/recurring
Now .9r + 9 = 9.9r so with that comes 9.9r - 9 = .9r or 9.9r - .9r = 9.
Now adding 9 and multiplying by 10 will leave you with different results.
.9 + 9 = 9.9
.9 * 10 = 9
So when you have something like .9r + 9 = 9.9r this is not the same 9.9r that you will get with .9r * 10. However the spoof is treating it as the same number.
.9r + 9 = 9.9r
.9r * 10 = 9.9r
The first 9.9r has recurring 9s after the decimal. As does the second. The difference is that due to the process to obtain them the latter legs the first when it comes to the recurring 9s. It is precisely 1 digit behind when talking only about the digits after the decimal.
So if you do the following with .9r + 9 - 9 = .9r + 0 = .9r
and .9r * 10 - 9 = .9r
You get .9r != .9r
Now you're thinking 'hold on just a minute'. We can look at it from another way.
.9r = .9n; n->infinity where n represents the number of 9s after the decimal.
.9n + 9 = 9.9n
9.9n - 9 = .9n
Therefore
.9n + 9 - 9 = .9n
.9n * 10 != 9.9n
.9n * 10 = 9.9m
Therefore
.9n * 10 - 9 = .9m not .9n
Where m = n - 1
rpenner
19th April 2007 - 11:19 PM
Somebody raised the question of how do you do math with repeating decimals.
To form a repeating decimal you carry out the long division algorithm step by step until your remainder is one that you've seen before. Since the long division algorithm is deterministic, this guarantees you are in an infinite loop.
Infinite loops are equal if you get the same digit at each step. So 0.(123123)r = 0.(123)r
1/2 = 0.5(0)r
1/3 = 0.(3)r
2/3 = 0.(6)r
1/16 = 0.0625(0)r
1/15 = 0.0(6)r
41/333 = 0.(123)r
73/333 = 0.(219)r
292/333 = 0.(876)r
1/7 = 0.(142857)r
2/7 = 0.(285714)r
A repeating decimal can always be returned to a rational number by multiplication by (10^n - 1) where n = the period of the repeat.
Shifts of where the repeat starts are always legal if they follow this pattern: 0.z(abcd..wxyz)r = 0.(zabcd..wxy)r = 0.za(bcd...wxyza)r
1.(0)r always equals 0.(9)r, including cases with shifts and adds.
To add two repeating decimals, pad out the repeats until they have the same period of the LCM of the two periods and shift the repeat start until they start at the same decimal place. Then add the repeating parts and the non-repeating parts, keeping any carries separate for a another pass until you are done.
0.(142857)r + 0.(142857)r = 0.(001110) + 0.(284604)r = 0.(285714)r
0.(142857)r + 0.(857142)r = 0.(999999)r = 0.(9)r = 1.(0)r = 1
To multiply by a natural number, split into repeating and non-repeating parts, do the multiplication pairwise over the digits, creating one repeating number modulo 10, and another with the high value. Then add them together.
1/3 * 2 = 0.(3)r * 2 = 0.[(3*2)]r = 0.(6)r = 2/3
1/15 * 5 = 0.0(6)r * 5 = 0.0(0)r + 0.(3)r = 0.(0)r + 0.(3)r = 0.(3)r = 1/3
1/15 * 75 = 0.0(6)r * 75 = 0.0(6)r * 70 + 0.0(6)r * 5 = 0.(40 + 2)r + 0.0(30)r * 5 = [ 4.(4)r + 0.(2)r ] + 0.(3)r = 4 + 0.(6)r + 0.(3)r = 4.(9)r = 5.(0)r = 5
73/333 * 4 = 0.(219)r = 0.[(2*4)(1*4)(9*4)]r = 0.[(0 + 8)(0 + 4)(30 + 6)]r = 0.(030)r + 0.(846)r = 0.(876)r = 876/999 = 292/333
2 * 0.(142857)r = 0.(001110) + 0.(284604)r = 0.(285714)r
3 * 0.(285714)r = 0.(212010)r + 0.(645132)r = 0.(857142)r
---
First, I claim these rules are a self-consistent way to do arithmetic. To disprove this claim, please demonstrate that they are inconsistent.
Second, I claim that if you say 1/3 - 0.(3)r = x, and you claim x is a number larger than zero, then you cannot claim that 1 - 0.(9)r = 3x and still claim 0.(9)r is the closest number to 1. In any case, x is demonstrably less than any rational number, so if x is a number, it is not a real number. And if we aren't talking about real numbers, none of your arithmetic is valid mathematics.
NoCleverName
19th April 2007 - 11:37 PM
QUOTE (Eric England+Apr 19 2007, 09:38 PM)
Engineering (no matter the scale) need not concern itself with infinity.
Maybe I should have used a stronger argument and said, say, physics depends on treating infinity the way we do. For example, the principle of conservation of energy in an inverse square field is shown by the line integral of a closed path=0; surely the calculation of the line integral depends on the infintesimal being valid? Not to mention finding inflection points in polynomials.
I'm having trouble seeing things thru your eyes --- and if you expect people to accept your views then that has to be possible --- so I'm trying to understand the boundaries of where you think things are OK and where they are not.
My experience does not extend to the "field of center points"; I'm also having trouble seeing the significance of your allusion to the "apparantly finite 1".
Precursor562
19th April 2007 - 11:44 PM
However with repeating decimals
1.0 = 1.00 = 1.000 etc till you end up with 1.0r
The pattern continues without a hitch and you end up with 1.0 = 1.0r where zero adds nothing to the value and so 1 = 1.0r
.9 != .99 != .999 etc till you end up with .9r
The pattern here also continues without a hitch. When you multiply by 10 you create a shift in this pattern resulting in 9 != 9.9 != 9.99 etc till you end up with 9.9r
The .9r in 9.9r is not the same .9r you started with.
Remember with the process
.9 + .09 + .009 + ..... doesn't end and also doesn't have an infinite number of steps (in that the number of steps doesn't equal infinity) and all it does have is no end to the number of steps. They go on and on and on.
9 + .9 + .09 + ..... is much the same except if you were to begin these processes at the exact same time, when it comes to the repeating 9s after the decimal that will result as the sum of the process the latter will be lagging by one digit when compared to the first.
NoCleverName
19th April 2007 - 11:53 PM
QUOTE (Precursor562+Apr 19 2007, 11:44 PM)
... when it comes to the repeating 9s after the decimal that will result as the sum of the process the latter will be lagging by one digit when compared to the first.
You have absolutely no justification whatsoever for that statement. If you believe that, then you also must believe that there are "more numbers between 1 and 3 than there are between 1 and 2, even though in the reals an infinite numbers are defined between any two reals".
Do you?
AlphaNumeric
20th April 2007 - 12:12 AM
QUOTE (NoCleverName+Apr 19 2007, 10:17 PM)
I might observe that all of this was argued back and forth
within the profession long ago; so that is that.
Yep, mathematicians pondered it and turned to the tools of logic they had at their disposal. They attacked the 'problem', solved it in a consistent manner with the rest of their work and moved on. That was over 250 years ago and mathematicians continue to push on with their work.
The 'arm chair mathematicians' can continue complaining as much as they like, the real maths continues unperturbed
Steve, Precursor, if you're into doing mindless, repetitive and ultimately flawed logic, here's one for you :
You're allowed a compass, a straight edge (a ruler with no markings on it) and a pencil. Draw a circle, the radius is up to you. Then, using only the compass and straight edge, come up with a method which allows you to construct a square of exactly the same size in finitely many steps.
Another one, using the same equipment, is to draw two straight lines which intersect at some random angle. Come up with a method which trisects that angle in finitely many steps.
Post back when you're done, and not before
Nick
20th April 2007 - 12:52 AM
NOTHING EVER REACHES INFINITY. OUGHT WE BE USING IT IN THE NAME OF MATH?
THERE IS NOTHING THAT REACHES THIS CONCEPT. REAL MATH DOESN'T HAVE IT.
StevenA
20th April 2007 - 07:53 AM
QUOTE (AlphaNumeric+Apr 20 2007, 12:12 AM)
Steve, Precursor, if you're into doing mindless, repetitive and ultimately flawed logic, here's one for you :
You're allowed a compass, a straight edge (a ruler with no markings on it) and a pencil. Draw a circle, the radius is up to you. Then, using only the compass and straight edge, come up with a method which allows you to construct a square of exactly the same size in finitely many steps.
Another one, using the same equipment, is to draw two straight lines which intersect at some random angle. Come up with a method which trisects that angle in finitely many steps.
Ok, for the square, I can sweep out six segments of the circumference of the circle by using the radius. This gives me 6 equilateral triangles embedded within the circle. I can then divide two of these 60 degree angles in half by sweeping two arcs from each side of the angle, where it intersects the circumference. That gives me 4 90 degree angles from the center that I can use to embed a diamond within the circle (I don't know what you mean by "exactly the same size", but I assume you mean to encompass the circle with a square tangent to the circumference at 4 points, so the width of the square is the same as the diameter of the circle). Then I can divide those 4 sides in half again to get 45 deg angles. I know the corners of the final square will be double the distance from the center to the midpoints of the edges of the diamond, so I can sweep a circle around each midpoint of the diamond (located by passing a line from the center to the 45 deg intersection swept out earlier), then wherever the opposite end of that circle, from the midpoint of the diamond intersects the 45 deg anges, I have the vertices of the outer square.
I assume that's what you wanted.
I'll have to think a bit harder about dividing an angle in to three equal pieces.
AlphaNumeric
20th April 2007 - 07:56 AM
By 'the same size', I mean have a circle and a square whose surface are is the same. If given either a square or a circle, the algorithm/method must be able to give the other shape of equal surface area.
The trisecting the angle must be for any angle, since it's easy to things like splitting 180 degrees into 3 lots of 60, it needs to be generalised.
Precursor562
20th April 2007 - 10:10 AM
QUOTE
You have absolutely no justification whatsoever for that statement. If you believe that, then you also must believe that there are "more numbers between 1 and 3 than there are between 1 and 2, even though in the reals an infinite numbers are defined between any two reals".
Do you?
The funny thing there is that there is no limit to the number of digits you can use to make up a number 2 has 1, 4.5 has 2, 78.928 has 5 etc. There being no limit is what allows for an infinite number of possibilities. So between 1 and 2 there are an infinite number of numbers one can give that fall between 1 and 2. This merely means there is no limit, no end to how many a person can think up and write down. There are limitations though. For between 1 and 2 (and not including 1 or 2) any number you start with will have a 1.n. Where n is any number of digits and those digits can be any number (1 to 9) in any order. Now between 1 and 3 (not including 1 or 3) you have the same thing where there is no limit. However your limit is expanded to including 2.n + 2.0.
Where 1.n < 1.n + 2.n + 2.0
If you don't include the finite value of 2 then 1.n < 1.n + 2.n regardless of n representing the unlimited number of digits one can put after the decimal.
However for infinity itself. Infinity is infinity is infinity and there are an infinite number of number (unlimited number of numbers) between both 1 and 2 as well as 1 and 3. However each of those numbers have unique identities or else every number ends up equaling every other number. This includes the recurring number that too have unique identities.
NoCleverName
20th April 2007 - 10:21 AM
Precursor: Since you seem bent on avoiding answering my question, let's try a simpler one to draw out your understanding of infinity:
Do you believe there are more "natural numbers" (1, 2, 3, 4, ...) than "even numbers" (2, 4, 6, ...) ?
bm1957
20th April 2007 - 11:25 AM
Steve, you seriously don't realise that you've just been asked to find a method for finding pi as a fraction???
Wulf
20th April 2007 - 12:09 PM
QUOTE (bm1957+Apr 20 2007, 11:25 AM)
Steve, you seriously don't realise that you've just been asked to find a method for finding pi as a fraction???
I think he is still working it out through long division, this might be a while.
AlphaNumeric
20th April 2007 - 12:20 PM
QUOTE (bm1957+Apr 20 2007, 12:25 PM)
Steve, you seriously don't realise that you've just been asked to find a method for finding pi as a fraction???
Not that it's rational but that pi is algebraic (ie not transcendental) and is of at most quadratic in nature, in that the only way you'd be able to square a circle is if pi was the solution to a quadratic equation with integer coefficents. It is not and hence squaring the circle is impossible. I did specifically say so in my post but obviously Steve has a different definition of many words compared to the rest of us.
Same goes for trisecting an angle. It would mean that the cos or sin of any given angle is always the solution of some cubic equation with integer coefficents. It's not true in general. cos(60) is and hence why you can construct hexagons easily. You can't trisect the angles of a hexagon though since cos(20) is not cubic.
A detailed list and explaination of various impossible constructions :
http://en.wikipedia.org/wiki/Trisection_of_an_angleSteve, Precursor, if you want to really step it up a gear in terms of idiotic maths claims, lets see you claim you can do a couple of those constructions
bm1957
20th April 2007 - 12:34 PM
I think the fact that he failed to spot the 'joke' demonstrates that there is a lack of study time.
Seriously guys, maths can be really interesting - why not learn about it properly instead of making stuff up and guessig bits, then trying to argue about it? You might enjoy it!!!
Precursor562
20th April 2007 - 12:57 PM
QUOTE
Precursor: Since you seem bent on avoiding answering my question, let's try a simpler one to draw out your understanding of infinity:
I did answer your question, it just wasn't the answer you were expecting. As for drawing out my understanding of infinity I have repeated times have said the infinity is merely the concept of no limit, no end, not finite etc. where finite is the concept of being limited, having and end, etc.
QUOTE (->
| QUOTE |
Precursor: Since you seem bent on avoiding answering my question, let's try a simpler one to draw out your understanding of infinity:
|
I did answer your question, it just wasn't the answer you were expecting. As for drawing out my understanding of infinity I have repeated times have said the infinity is merely the concept of no limit, no end, not finite etc. where finite is the concept of being limited, having and end, etc.
Do you believe there are more "natural numbers" (1, 2, 3, 4, ...) than "even numbers" (2, 4, 6, ...) ?
It's a similar idea of regarding limitations. The limit here is being evenly divisible by two. 2/2, 4/2, 6/2 etc. Where 1, 3, 5 etc are not evenly divisible by two without going into decimals where 1 is evenly divisible by 2 but will get you .5.
So what you have is n/2 where n represents all the even numbers and n->infinity because there is no limit to how large an even number can be. In the case of n/1 it wouldn't be n it would be m where m includes all the even numbers and all the odd numbers as well. Without going into decimals you will only be able to have m/1. Where m represents all the odd and even numbers where there is no limit to how big a number can be so m->infinity.
2n = m because for every even there exists an odd and for every odd there exists an even.
bm1957
20th April 2007 - 01:09 PM
Got to agree with precursor on this. It's quite clear from set theory that there are a larger infinite number of integers than there are infinite even integers.
I don't believe that this implies that 1 - 0.9r != 0 though. As the number of 9s approaches infinty, the difference approaches 0. At the limit, the difference between 1 and 0.9r is 0. Hence 1 = 0.9r
Not really a rigourous proof, but I couldn't do any more maths than has already been repeated on here many times.
Precursor562
20th April 2007 - 03:51 PM
QUOTE
I don't believe that this implies that 1 - 0.9r != 0 though. As the number of 9s approaches infinity, the difference approaches 0. At the limit, the difference between 1 and 0.9r is 0. Hence 1 = 0.9r
The only thing there (and is why I believe 1 - .9r != 0) is the part "At the limit". The limit being infinity you can never equal infinity therefore you can never be at infinity.
Being at infinity is like reaching the end of a road that has no end. I believe this is the root problem that causes .9r to equal 1. Infinity is treated like some number that can be reached.
What 1/10^n represents is that with no limit to how large n can be, there is no limit to how small a number you will have. Doesn't that make sense? After all there are an infinite number of numbers between 0 and 1 then there is also no limit to how small a number you can have. Where such a number is still greater than zero.
NoCleverName
20th April 2007 - 04:30 PM
QUOTE (bm1957+Apr 20 2007, 01:09 PM)
Got to agree with precursor on this. It's quite clear from set theory that there are a larger infinite number of integers than there are infinite even integers.
Assuming that whatever Precursor said in his response meant that he believes there are more integers than even integers (and it's sometimes hard to tell) then both you and he would be quite mistaken. It is easy to show that there are at least as many "even" integers as there are "all" integers.
By demonstration, simply pair each of the set of natural numbers with a member of the set of "even" integers by multiplying the natural number by 2:
1 2 3 4 5 6 ...
2 4 6 8 10 12 ...
As you can see, it is clear that every natural integer can be paired with one of the members of the set if even integers, so the sets must be of equal size.
Hopefully this will illuminate your understanding of infinity.
bm1957
20th April 2007 - 04:57 PM
Precursor:
You are exactly right about it being difficult to evaluate a number which has an infinite number of 9s after it, because you can't count 'how many'. This is why rules have been developed in maths which specify how to deal with these numbers, and they result in the fact that 0.9r = 1, by all the proofs which have been posted.
NoCleverName:
Again, infinities are complicated to deal with. You are trying to use logic to reach a conclusion, just as Precursor was to justify 0.9r != 1.
Infinity cannot be defined but it can be said that one set with infinite members can be larger than another set with infinite members; two different infinties can in fact be compared to each other. I will try to remember a more convincing argument to support this...
NoCleverName
20th April 2007 - 05:21 PM
QUOTE (bm1957+Apr 20 2007, 04:57 PM)
... two different infinties can in fact be compared to each other. I will try to remember a more convincing argument to support this...
Yes, there are in fact an "infinite number of infinities"; however, this is not one of those situations as the "cardinality" of a subset of a set is the same as the entire set. The fact that there is the same number of "evens" as "all integers" is instructive; it demonstrates that you can't compare infinity to itself, you can't have "multiples" of it, you can't add or subtract from it.
Once you accept the idea that "infinity is a proper subset of itself" then the whole business about "the lagging 9" in Precursor's example is shown to be meritless. After that, it isn't all that hard to see why the sum of an infinite geometric series is a finite value and doesn't do any "approaching" to anything.
I have to say it again: Infinity is more than just a really, really big number.
bm1957
20th April 2007 - 06:36 PM
QUOTE (NoCleverName+Apr 20 2007, 05:21 PM)
Yes, there are in fact an "infinite number of infinities"; however, this is not one of those situations as the "cardinality" of a subset of a set is the same as the entire set.
After looking for the argument I thought heard for this, I realised I was mis-remembering (is that a word???). Should have done my homework first!
Yes, realising that infinity is not a number which can be manipulated by 'normal' math is the key to understanding the answer to this problem.
rpenner
20th April 2007 - 07:03 PM
QUOTE (bm1957+Apr 20 2007, 01:09 PM)
It's quite clear from set theory that there are a larger infinite number of integers than there are infinite even integers.
Actually, what set theory teachs us is if X is an infinite set, there is always a proper subset of X which is of equal cardinality.
If you are going to talk about set