Controversial math question [Archive]

verty

Mar25-04, 11:47 AM

Think of the sequence with the n'th term given by '1 - 1/n'. When n = 10, the number is 0.9. When n = 100, the number is 0.99. As n grows large, the value of the term gets closer to 1. In the limit when n reaches infinity, the term 1/n becomes 0, so the limit of '1-1/n' as n tends toward infinity is 1. That's why 0.99999... is = 1.

Zurtex

Mar25-04, 03:21 PM

Let a be any real number. We can say that:

a * 0 = 0

Dividing both sides by 0,

a = (0 / 0)

Hurkyl

Mar25-04, 06:41 PM

Let a be any real number. We can say that:

a * 0 = 0

Dividing both sides by 0,

a = (0 / 0)

What grounds do you have for asserting that dividing a valid equation by zero yields a valid equation?

Janitor

Mar25-04, 09:11 PM

I think Zurtex was just having his fun anticipating the invalid proof that MC was going to provide.

ShawnD

Mar26-04, 02:34 AM

x = 0.99999999.......
10x = 9.999999........
10x - x = 9.9999....... - 0.999999......
9x = 9
x = 1

There's always the way of expressing repeating decimals.
0.777777.... is 7/9
0.484848...... is 48/99
0.674674674... = 674/999

Therefore, 0.9999..... = 9/9 = 1

That might be a stretch for believability though so try this one

0.333333..... = 3/9 (go ahead and check with your calculator)
3(0.33333.....) = 3(3/9)
0.9999999...... = 9/9
0.9999999...... = 1

[;)]

chroot

Mar26-04, 05:13 PM

http://www.mathforum.org/dr.math/faq/faq.0.9999.html

Bob3141592

Mar26-04, 11:53 PM

Can anyone tell me why, or why not,.\bar{9} is equal to one?

I think of it this way. What's \frac{1}{3} * 3 ? If you just simplify the fraction, it's simply 1. But if you write \frac{1}{3} as .3333... then the answer is .9999....

Right?

Zurtex

Mar27-04, 07:23 AM

What grounds do you have for asserting that dividing a valid equation by zero yields a valid equation? No grounds, in fact even if we ignore that it's really easy to show the proof is self contradicting, I've assumed 0/0 = 1 to prove that 0/0 = any real number :rolleyes:

Zurtex

Mar27-04, 07:38 AM

Owww sorry for the double post but I just thought of a good one.

Let a be any real number.

a*0=0

Dividing both sides by 0.

a \frac{0}{0} = \frac{0}{0}

Dividing both sides by \frac{0}{0}

a = \frac{\frac{0}{0}}{\frac{0}{0}}

a = 1

All real numbers are equal to one :eek: .

matt grime

Mar27-04, 07:52 AM

Zurtex

Mar27-04, 08:03 AM

Pah, childs play. Surely you can see the easy generalization of this idea to demonstrate that every real function is identically equal to 1. In fact I think I can demonstrate that everything is equal to 1, and I can also show 0=1 so we're really in trouble now. Well I'm fairly new to maths and it amuses me :tongue:.

Speaking of which would anyone know some good links to learn various aspects number theory? I am very interested in it but the maths course I am on covers nothing of it.

matt grime

Mar27-04, 08:08 AM

There's a small cheap book by Baker on Number theory that's worth a look, or Le Veque's Fundatmentals of Nember Theory was available in a Dover reprint a few years ago too. I presume you mean algebraic (the nice elegant stuff) and not analytic (arguably more powerful but nasty with it).

As for the above, as there is nothing that allows us to divide by zero, yet we do it, there is nothing to stop us declaring 0*x=0 for any conceviable x, be it a number, a function or absolutely anything, and we're doing no more damage than we were before. After all, surely no lots of anything are nothing? So x=1 and I didn't even say what x is.

Integral

Mar27-04, 04:40 PM

Here is (http://home.comcast.net/~rossgr1/Math/one.PDF) a pdf I put to gather with a couple of proofs. The first simply uses the expression for the sum of a geometric series.

The 2nd however is a bit more fundamental. It uses a method which I feel gives a very good insight as to why they must be equal.

If HallsofIvy or one of the other real mathematicians on the board should look at this I would appreciate feedback on how to clean it up and finish it off better.

Integral

Mar27-04, 04:50 PM

matt grime

Mar27-04, 06:31 PM

Algebraic manipulation of infinitely long decimals is well defined.

The pdf seems far too long given the information it attempts to convey. Perhaps you ought to simply deal with the finite partial sums to explain why th infinite one exits, even if that requires you to explain the very basic analysis you are eliding, which is only a definition after all.

Integral

Mar27-04, 07:00 PM

Algebraic manipulation of infinitely long decimals is well defined.

Sure it is, but it is not appropriate for a proof. As I said it is a fine demonstration.

JonF

Mar27-04, 07:00 PM

There's always the way of expressing repeating decimals.

Therefore, 0.9999..... = 9/9 = 1

Oh really? Proof please. To say that .9999=1 is to assume what you are trying to prove, before you prove it.

I’d also like to see a proof that all repeating decimals can be expressed as a fraction. A few examples hardly proves anything. And even if that is true, how do you know that .99999 corresponding fraction is 9/9.

0.333333..... = 3/9 (go ahead and check with your calculator)
3(0.33333.....) = 3(3/9)
0.9999999...... = 9/9
0.9999999...... = 1

Oddly enough my TI8-89 claims that .33333 is an approximation of 3/9...

ShawnD

Mar27-04, 07:18 PM

I'd also like to see a proof that all repeating decimals can be expressed as a fraction. A few examples hardly proves anything.

Show me 1 example where a repeating decimal number cannot be expressed as a number divided by a series of 9's.

JonF

Mar27-04, 07:57 PM

But that is not what your previous post suggest. You made it out to be that all repeating series can be express as: A/B where “A” and “B” are integers. That’s the foundation of your whole “proof”, now please prove it.

ShawnD

Mar27-04, 10:18 PM

I did prove it. (3/9) x 3 = 1. What is so complicated?

Mar27-04, 10:50 PM

"I’d also like to see a proof that all repeating decimals can be expressed as a fraction"

Geometric series: If |r|<1 then the sum of r^k from k=0 to 00 is 1/(1-r): Sum(r^k) = 1/(1-r)

Example:
0.9999... = 9/10 + 9/100 + 9/1000 +... =
(9/10)[1 + 1/10 + 1/100 +..] =
(9/10)[ 1 / (1-1/10)] = (9/10)(10/9) = 1

Integral

Mar28-04, 02:33 AM

Here (http://home.comcast.net/~rossgr1/Math/decimal.pdf) Is a quick demonstration that any repeating 2 digit pattern is a decimal representation of those 2 digits divided by 99. I think if one worked at it a bit the method could be generalized to show that ANY repeating decimal can be represented as the repeating digits divided by as many 9s as there are repeating digits.

It is also pretty straight forward to show that any rational number can be represented as either a finite length decimal expansion or a infinitely repeating decimal expansion.

Consider the process of long division. There are exactly 10 possible results to the multiplication of the divisor by a single digit of the quotient. The number of possible results to the subsequent subtraction is limited by the size of the divisor since the result is always smaller then the divisor. As the process is carried out there are only 2 possible outcomes, the process terminates or at some point a result of the subtraction will be repeated. This must happen because there are a finite number of possible results and the process can be repeated indefinitely. If the process does not end it must repeat on or before the number of digits equal to the size of the divisor.

A great number to look at this with is 1/7 =.142857...

I believe that this concept could be written up as a formal proof of the fact that rational numbers can be represented as either a terminating decimal or an infinitely repeating pattern.

Muzza

Mar28-04, 03:36 AM

Oddly enough my TI8-89 claims that .33333 is an approximation of 3/9...

Did you type in an infinite number of threes?

cookiemonster

Mar28-04, 03:57 AM

I think bringing calculators into this is a bad, bad, bad idea.

cookiemonster

Integral

Mar28-04, 04:19 AM

I think bringing calculators into this is a bad, bad, bad idea.

cookiemonster
My thoughts also.

matt grime

Mar28-04, 04:49 AM

Integral

Mar28-04, 05:22 AM

I thought I'd posted a proof for Jon, but it ain't here so:

let z be an eventually recurrent decimal, that is z = x+y where x is terminating (and trivially rational) and y is a recurring decimal, say with period n. Then y*10^n - y is a terminating decimal, and hence rational, thus y is rational, and it follows z is rational too.

It is not clear to me how this proves y is rational, could you please expand on this? It looks to me like y*10n -y is an integer. (since y is repeating period n) Example
y=.242424...
n=2

y*10n= 24.242424...

24.242424... - .242424...=24

Thanks

matt grime

Mar28-04, 05:29 AM

If it is an integer it is rational. It doesn't need to be an integer. Try it with .000012121212.... with the repeating12 pattern. you get .001200000.... which is rational. It just produces from a recurrent decimal y, a terminating decimal, r, ie rational, satisfying y(10^n-1)= r hence y is a rational divided by an integer, thus a rational.

Organic

Mar28-04, 08:04 AM

What is the result of 9.9999.../0.9999...?

Zurtex

Mar28-04, 08:28 AM

What is the result of 9.9999.../0.9999... ? Well as 0.9999... = 1 and 9.9999... = 10 * 0.9999... = 10 * 1 = 10

Then you are asking what the result of 10 / 1 is.

Edit:

I suppose another way of looking at the problem is saying that it is the same as:

\lim _{x \rightarrow \infty} \frac{10 - 10^{-x}}{1-10^{-x}}

Where x is a natural number.

Organic

Mar28-04, 08:51 AM

9.9999.../0.9999... = 10 , which is the ratio between 9.9999... and 0.9999...
9.9999 .../1.0000... = 9.9999... , which is the ratio between 9.9999... and 1.0000...

Now let us check this arithmetic:

x = 0.99999999.......
10x = 9.999999........
10x - x = 9.9999....... - 0.999999......
9x = 9
x = 1

A question: How we can be sure that the result of 0.99999.... - 0.99999...
is exactly 0 where there is no right side to begin the subtraction operation?

matt grime

Mar28-04, 08:55 AM

Because addition and subtraction can be put into an algorithm for infinitely long decimals if you need to do so. However as you're asking what the result of x-x is it doesn't depend on x having a nice form, it is zero by definition. Why do people confuse numbers and their decimal representations?

matt grime

Mar28-04, 08:57 AM

Or if you don't like that answer then how about doing it in terms of Cauchy sequences, which is after all the most useful construction of the Real numbers as a mathematical object.

Organic

Mar28-04, 09:02 AM

Please prove that there is no connection between a number and its stuctural represention upon infinitely many scales.

matt grime

Mar28-04, 10:02 AM

I didn't say there wasn't a connection (though what you mean by that is unclear) but that the addition of two real numbers is independent of the choice of decimal expansion, should it have two. The simple proof of this fact follows from the definition of the real numbes as the completion of the rationals. Go and get a basic analysis book. Just because you do not know it, Organic, does not mean it is not true or known by other better informed people. Proof: let x_n and y_n be two equivalent Cauchy sequences. This means x_n-y_n converges to zero. Let w_n and v_n ba any other pair of equivanlent cauchy sequences.

then the element of R that [x_n-w_n] coresponds to is the same as the class (real number) [y_n-v_n]

proof: we are to show x_n-w_n-y_n+v_n converges to zero, but that is trivially true since |x_n-w_n_y_n+v_n| < |x_n-y_n| +|w_n-v_n| and both those terms can, be made arbitrarily small by hypothesis,a nd we have proved subtraction of two real numbers is indpendent of the Cauchy sequences we pick to represent them. OK?

Organic

Mar28-04, 10:33 AM

What you show is a rough jump that forces infinitely long sequence to become finitely long, and than you use subtraction after you created an artificial right side (which cannot exist in infinitely long sequence) in a non-logical way for your own purpose.

matt grime

Mar28-04, 10:53 AM

No, this is the rigorous proof that addition is well defined on the Real numbers. Perhaps you ought to go and learn some mathematics?

Zurtex

Mar28-04, 11:47 AM

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

Janitor

Mar28-04, 11:57 AM

I would anticipate that Organic's objection to your suggestion is that "there is no right side to begin the base-conversion operation."

Organic

Mar28-04, 12:23 PM

Hi Zurtex,

This is exactly my point of view on this case, a number is not just a quantity but has also an internal structute that cannot be ignored, for example look at this paper:

http://www.geocities.com/complementarytheory/Complex.pdf

Perhaps you ought to go and learn some mathematics?

I cannot agree with mathematics which is based on forcing methods.

Integral

Mar28-04, 01:00 PM

matt grime

Mar28-04, 01:11 PM

Organic, the question was posed in the field of real numbers with their mathematical definition. Your opinions as to whether that is not the correct object are irrelevant to the question, and its answer. Or would you like to point out where the proof that 0.99999 =1 is wrong when working in the usual definition of the real numbers as equivalences classes of cauchy sequences? I do not need to use all the things that are true about some object to prove things about it. For instance, every digit in 0.99999... is a prefect square, I didn't use that fact. I didn't use the fact that considered as curves embedded in the plane the digits involved all have fundamental groups that aren't trivial. I didn't use the fact that 9 is 6+3 where 3 is the smallest odd prime and 6 is the smallest order of a non-abelian group. If you aren't prepared to learn what mathematics involves then how can you possibyl answer questions about it? I mean, there is a theory where 0.9999... is not equalt to 1. Perhaps you want to learn about Abraham Robinson's non-standard analysis? Whereof you do not know do not speak?

Integral, yes, but, there's no need for the r in there to be an integer, which is what you wanted originally I seem to recall. And ignores the x. Example .011111 = 1/90, arguably a recurrent decimal and not consisting eintrely of 9s in the denominator.

matt grime

Mar28-04, 01:13 PM

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

In base 3 one has other problems such as .22222222... =1

Forget decimals, or any other system of representation like that. Just operate with the definitions of the real numbers. That's how mathematics works, practically.

Integral

Mar28-04, 01:15 PM

Matt,
Perhaps now you are beginning understand why my proof does not perform operations on non finite digits. People such as Organic, who should be restricted to posting in Theory Development, simply will not accept any proof you can provide that non finite operations are allowable. Beyond that I was taught in my analysis courses that such operations should not be included in fundamental proofs.

Organic is special in that he has his own number system which he cannot separate from the Reals that the rest of us use.

Integral

Mar28-04, 01:41 PM

Integral, yes, but, there's no need for the r in there to be an integer, which is what you wanted originally I seem to recall. And ignores the x. Example .011111 = 1/90, arguably a recurrent decimal and not consisting entirely of 9s in the denominator.

So simply factor out the non repeating part.

.0111... = .111... x 10^{-1} = \frac 1 9 x 10^{-1}

So we have a multiple of 10 and a rational with 9's in the denominator. This is validation of the methods mentioned up thread using 9's in the denominator. We have shown that every repeating decimal can be represented as a factor of 10 and the repeating portion over 9s.

Integral

Mar28-04, 01:51 PM

In base 3 one has other problems such as .22222222... =1

Forget decimals, or any other system of representation like that. Just operate with the definitions of the real numbers. That's how mathematics works, practically.

Matt,
The Reals are not base dependent, a base 2 or 3 or 16 system is the same as a base 10 representation as far as the real system is concerned. Yes, different bases have different rationals as repeating decimals. For example

.1 (base 10) = .0001100110011...(base 2) (I think I got the right number of leading zeros). This is of significance because it means your computer must round off .1 .

EDIT: Opps, Matt I just reread your post, I got different meaning the 2nd time. I am trying to say the same thing you are. The Reals are Base independent.

matt grime

Mar28-04, 01:56 PM

Where did I say the reals are base dependent? If everyone remembered what the real numbers acutally are then none of these recurring (pun intended) nightmares would happen. It's amazing how often this question comes up, isn't it?

ShawnD

Mar28-04, 02:03 PM

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

That's sort of how the fraction representation works. 1/3 x 3 = 1.

Organic

Mar28-04, 03:05 PM

[quote]
Organic is special in that he has his own number system which he cannot separate from the Reals that the rest of us use.

Let us take the circle's equation: (x-h)^2 + (y-k) = r^2
http://www.xavierhs.org/departments/Mathematics/PreCal/Conics/conics.htm

solid is a "one piece" state

r=radius

h=x center

k=y center

But the interesting variables are x and y, where x is the entire x-axis and y is the entire y-axis.

x-axis or y-axis are "actual form of infinity" as we can see in this model:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

To construct the circle we have to break the solid states of both x-axis and y-axis and define a sequence of unique pairs of R members, which are used as x,y coordinates of the circle.

The point here is that we have no R members before we break the solid state of x-axis and y-axis, and only after we break them we get R members.

The same state is an information form of, for example, 0.9999999...

It cannot be in both states of finite and infinite sequence of non-zero values upon infinitely many scales.

Therefore there is a XOR condition between 1.0 and 0.9999... exactly as there is a XOR condition between a solid state and a broken state.

Integral

Mar28-04, 03:12 PM

Organic,
Please restrict your posts to the topic at hand.

Organic

Mar28-04, 03:21 PM

Hi Integral,

My previous post fits exactly to subject of this thread.

Please read all of it and see for yourself.

Integral

Mar28-04, 04:38 PM

Organic,
We are all talking about the Real numbers, you are talking about the organic numbers, you are off topic. Please take your garbage back to theory developement where it belongs.

Zurtex

Mar29-04, 04:42 AM

That's sort of how the fraction representation works. 1/3 x 3 = 1. I know but organic seems to be struggling with the concept of converting decimals into fractions when they recur.

In base 3 one has other problems such as .22222222... =1

Forget decimals, or any other system of representation like that. Just operate with the definitions of the real numbers. That's how mathematics works, practically. Well convert 0.111111.... to decimal to get 0.5, multiply by 2 to get 1 :tongue:

Zurtex

Mar29-04, 06:13 AM

O.K, merging what I put earlier with my work on base numbers.

Lets say we are working in base b, and lets us say that x is a number in base b and that x is a natural number. n is a digit in the decimal number x such that n = b -1.

So if:
x = 0.n = 1 - 10^{-1} = \frac{b-1}{b}

Or if:

x = 0.nnnnn = 1 - 10^{-5} = \frac{b^5-1}{b^5}

If n occurs p number of times:

x = 0.nnnnnnnn... = 1 - 10^{-p} = \frac{b^p-1}{b^p}

If n occurs an infinite number of times:

x = 0.nnnnnnn... = \lim _{p \rightarrow \infty} 1 - 10^{-p} = 1
Or x = \lim _{p \rightarrow \infty} \frac{b^p-1}{b^p} = \lim _{p \rightarrow \infty} (b^{-p})(b^{p}) - 1(b^{-p}) = \lim _{p \rightarrow \infty} 1 - b^{-p} = 1

Now your equation x - x: x - x = x(1-1) = x*0 = 0 As long as x has a numerical value regardless of how it is expressed.

matt grime

Mar29-04, 06:42 AM

Don't think you're going to get any sympathy there, Organic, especially given the God-awful choice of font, colour, and spelling. You are posting off topic, something we're probably all guilty of at some point admittedly, but you are also posting utter tripe that has no place in a mathematics thread. I'm amazed that, as a moderator, Integral didn't just delete your incoherent rubbish. You aren't some innocent posting a silly question and being dismissed out of hand. You are a recidivistic poster whose answers are tantamount to vandalization and are entirely unmathematical and plain wrong, yet you keep making them.

HallsofIvy

Mar29-04, 06:46 AM

Well, if it were me instead of integral, I certainly wouldn't have used the word "garbage"!

(The word I would have used would have gotten all "knocked off the air"!)

Organic

Mar29-04, 08:41 AM

Matt and HallsofIvy,

I invite you to show your skills and prove that my ideas have nothing to do with Math language development.

here it is again:

Let us take the circle's equation: (x-h)^2 + (y-k) = r^2
http://www.xavierhs.org/departments/Mathematics/PreCal/Conics/conics.htm

solid is a "one piece" state

r=radius

h=x center

k=y center

But the interesting variables are x and y, where x is the entire x-axis and y is the entire y-axis.

x-axis or y-axis are "actual form of infinity" as we can see in this model:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

To construct the circle we have to break the solid states of both x-axis and y-axis and define a sequence of unique pairs of R members, which are used as x,y coordinates of the circle.

The point here is that we have no R members before we break the solid state of x-axis and y-axis, and only after we break them we get R members.

The same state is an information form of, for example, 0.9999999...

It cannot be in both states of finite and infinite sequence of non-zero values upon infinitely many scales.

Therefore there is a XOR condition between 1.0 and 0.9999... exactly as there is a XOR condition between a solid ("one piece") state and a broken state.

matt grime

Mar29-04, 08:53 AM

We only need to show that what you wrote doesn't have anything to do with the question as asked.

Item 1. Your bloody post which has nothing to do with the question "why does 0.9999..=1?"

Item 2. Your bloody post which has nothing to do with the question "why does 0.9999..=1?"

Strictly speaking they are the same, but I thought it important enough to mention twice. This is a question about the real numbers. Do you know what they are? Evidently not judging by your bilge of a repost.

Edit: deleted silly thing about bandwidth. Still think you're a moron though.

matt grime

Mar29-04, 09:16 AM

Can some moderator please lock this thread? I think it has been adequately answered as to why there is nothing reomtely controversial about 0.9999.. being the same as 1, and all mathematical issues arising therefrom seem to have been sorted.

Organic

Mar29-04, 09:26 AM

matt grime

Mar29-04, 09:39 AM

It is very hard to explain to someone why their post has nothing to do with the question in hand if they refuse to accept the bleeding obvious as the last post on page 4 attempts to demonstrate. No one, not even someone with a first in mathematics can see the slightest bearing your ill-conceived opinion has on the original question, idiot boy.

Organic

Mar29-04, 10:17 AM

Matt,

Well, you have a PHD title in Math so please use it, I am waiting to you.

HallsofIvy

Mar29-04, 10:25 AM

Waiting for him to do what? The only thing a person with a dozen "PHD titles" in math could do was explain exactly what he already has.

matt grime

Mar29-04, 10:28 AM

I don't have PhD in mathematics. I am doing one; come back in October.

Zurtex

Mar29-04, 10:33 AM

Yeah this is getting a little tiring Organic, it has been proved over and over again why 0.9999... = 1 and you have not managed to disprove it or provide a counter example in terms of real numbers.

Organic

Mar29-04, 10:35 AM

HallsofIvy,

I invite you to show your skills and prove that my ideas have nothing to do with Math language development.

here it is again:

Let us take the circle's equation: (x-h)^2 + (y-k) = r^2
http://www.xavierhs.org/departments/Mathematics/PreCal/Conics/conics.htm

solid is a "one piece" state

r=radius

h=x center

k=y center

But the interesting variables are x and y, where x is the entire x-axis and y is the entire y-axis.

x-axis or y-axis are "actual form of infinity" as we can see in this model:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

To construct the circle we have to break the solid states of both x-axis and y-axis and define a sequence of unique pairs of R members, which are used as x,y coordinates of the circle.

The point here is that we have no R members before we break the solid state of x-axis and y-axis, and only after we break them we get R members.

The same state is an information form of, for example, 0.9999999...

It cannot be in both states of finite and infinite sequence of non-zero values upon infinitely many scales.

Therefore there is a XOR condition between 1.0 and 0.9999... exactly as there is a XOR condition between a solid ("one piece") state and a broken state.

Organic

Mar29-04, 10:37 AM

Hi Zurtex,

Please prove that I am not talking about R members.

matt grime

Mar29-04, 10:42 AM

The point is that this has nothing to do with the question that was asked. The decimals in your post could be replaced with anyting and it would still make as little sense, therefore you've not actually done anything remotely interesting or applicable. Not to mention the mistakes in the post anyway such as asserting that a variable IS the x-axis. No it isn't. It's a variable, which may take values in the set of real numbers, which isn't the same thing. 'Breaking the solid states' is a meaningless sentence, it doesn't even have a valid metaphorical interpretation, the real numbers aren't eggs. The idea of being an actual infinity is vague and fluffy and garbage that has nothing to do with the question in hand (I could rewrite it so that there were only a countably infinite number of points involved, and we know you get confused then.

Oh, and you're evidently not using any definition of the real numbers that makes sense to the rest of us. You do understand that they are, as mathematical objects in which one does analysis the completion of the rationals? In non-standard analysis they aren't, and don't even begin to think about the p-adics.

Zurtex

Mar29-04, 10:47 AM

I'm only really starting at mathematics, I certainly am not studying PhD in fact I am only at A level (doing maths equivalent to 1st year degree). But from my understanding of mathematics and this includes my extra research as I find the whole area fascinating you have proved nothing and you make no sense in more than one sentence.

matt grime

Mar29-04, 10:58 AM

Hi Zurtex,

Please prove that I am not talking about R members.

He (or she) doesn't have to do anything of the sort. It is up to you to demonstrate that you are using the Real Numbers correctly, which you evidently aren't. SImple way to show you are would be to explain clearly if you think 0.99.. and 1 are the same real number. A simple, yes they are, or no they're not. It appears that you're saying they're different, but given your misuse of XOR it's hard to tell.

Organic

Mar29-04, 11:01 AM

Ok Matt,

...Not to mention the mistakes in the post anyway such as asserting that a variable IS the x-axis. No it isn't.

All you proved is you cannot go beyond the convetional point of view of Math langauge.

Dear Zurtex,

Please reply more to the point, prove that my point of view is an illegal one from Math languge point of view.

Zurtex

Mar29-04, 11:05 AM

Ok Matt,

All you proved is you cannot go beyond the convetional point of view of Math langauge.

Dear Zurtex,

Please reply more to the point, prove that my point of view is an illegal one from Math languge point of view.

Well seemingly this means you are talking in a slightly different maths language to the rest of us. Care to explain what this language is and how it works? In maths you must rigorously prove something, not write something down and wait for somebody to disprove it and if they point out a seeming mistake just say they have no idea what they are talking about.

Please by all means explain to us simpletons who do not understand your maths why exactly it is correct and how it proves that 0.999.. and 1 can be different numbers under mysterious and strange circumstances.

matt grime

Mar29-04, 11:06 AM

The point is that your langauge is not the langauge in which one does mathematics if one wants to be understood, or say things that are not stupid and wrong. Remember if you go to a foreign country it might be advisable to learn to speak their language, and a very good idea not to tell them they are speaking it incorrectly and yours is the correct way of doing it. It smacks of arrogance and crass stupidity.

If any moderator wants to delete all these pointless posts please do so, I won't bat an eyelid.

Organic

Mar29-04, 11:07 AM

Matt,

Again you dont prove anything about my point of view.

misuse of XOR it's hard to tell.

Please show where is the mistake of using XOR to show that one case prevents the other?

matt grime

Mar29-04, 01:35 PM

Your mistake here is to use badly constructed sentences such as that which uses a mathematical symbol in its centre when it isn't what you ought to write . Here's a simple one: are 0.999... and 1 different real numbers (where we are using decimal expansions)?

Organic

Mar29-04, 02:48 PM

Matt,

The difference between my view and your view is this:

Matt's view: A one eye view where a number is a "quantity-only" information form.

Organics view: Two eyes view where a number is at least structural/quantitative information form.

Organic can see Matt's one eye view.

Matt cannot see Organic's two eyes view.

In one eye view 0.999.... = 1

matt grime

Mar29-04, 03:00 PM

No, Organic, the point is that BY DEFINITION 1 and 0.999... (when considered as base 10 expansions) are the same real number (didn't you read the bit about non-standard analysis? not that I'd expect you to understand), as has been proven in this thread MANY times. That you cannot seem to understand this is because you are an ill-informed mathematical-illiterate who isnt' prepared to learn what is necessary to talk about mathematics. Nor are you prepared to answer simple questions. Now someone please close this thread!

Organic

Mar29-04, 04:17 PM

Matt,

I am talking about 2Iview.

Non-standard analysis of A.Robinson and standard analysis of Epsilon-Delta Definition are both 1Iview.

Integral

Mar29-04, 04:35 PM

Organic,
As long as you have been posting to these forums you should know and understand that the place for your ideas is in the Theory Development forum. I will request once again that you please refrain from posting your non standard concepts in the Math forums. This is the place where a student can come to learn the STANDARD ACCEPTED Mathematics. Your contributions only serve to confuse those who are uncertain.

I have requested that the Math mentors lock this thread. I have no special privileges here or it would have been locked already. Preferably would be to shovel out Organics posts and let Matt and I finish the perfectly good conversation we were having.

JonF

Mar29-04, 04:36 PM

ok if

.\bar{9}= 1

then

\lim_{n\rightarrow \infty} .\bar{9}^n =1

correct?

Tom Mattson

Mar29-04, 04:40 PM

ok if

.\bar{9}= 1

then

\lim_{n\rightarrow \infty} .\bar{9}^n =1

correct?

No, because .999... does not equal limn->∞0.9n

JonF

Mar29-04, 04:42 PM

doesn't
\lim_{n\rightarrow \infty} 1^n =1

Integral

Mar29-04, 04:46 PM

ok if

.\bar{9}= 1

then

\lim_{n\rightarrow \infty} .\bar{9}^n =1

correct?
You need an n on the RHS of the second expression for the limit to have any meaning.

it is correct that

\lim_{N \rightarrow \infty}9 \Sigma_{n=0}^N .1^{-n}=1

Edit: Opps, I some how missed your superscript n on my first go round.

Tom Mattson

Mar29-04, 04:48 PM

I had granted Integral and Matt's request that the thread be locked, but as I was doing it people were still posting. Make up your mind, guys! :biggrin:

doesn't
\lim_{n\rightarrow \infty} 1^n =1

Whoops, I hadn't noticed the bar above the 9 in the limit.

JonF

Mar29-04, 04:51 PM

I’m just now about ½ way through my first semester of calculus. I have no clue what an RHS is. What I’m trying to say is:

Isn’t it true that .9999…. to the power of infinity = 1

I’m not trying to troll or bait here, it’s just that 1 = .9999… goes against all the math intuition I’ve have so far.

JonF

Mar29-04, 04:53 PM

Sorry just got home from school and have been thinking about this problem all day... Is it ok if they finish explaining it to me really quick?

Tom Mattson

Mar29-04, 04:53 PM

I’m just now about ½ way through my first semester of calculus. I have no clue what an RHS is.

RHS=Right Hand Side

I'll let Integral continue.

matt grime

Mar29-04, 05:07 PM

Tom, and others, I started an answer to some of JonF's question in a new thread if you want to clean it up a little. Though I was doing it from memory in case this should get locked, so it might not be an accurate recollection of the question

square_root_beer

Mar29-04, 05:32 PM

A mathematical issue about this arose when I tried to find a constructive bijection between IR^1 and IR^2 (or IR^k for that matter) (how do I get LaTeX here?). My idea is best illustrated with the following example:

f: 573491.14387469712 |--> (741.48491, 539.137672)
f inverse: (741.48491, 539.137672) |--> 573491.14387469712

The only problem is that certain (pairs of) real numbers map to the same 2-space point.

.0391949195999296959... (digits from pi interspersed with 9's)

and .1301040105090206050...

both map to (.314159265..., .1)

So we don't have a bijection at all.

BUT, we do get a surjective function from IR^1 to IR^2 which is INTERESTING, in that we can think of a different surjection from IR^2 to IR^1, and the combination of these is all we need to prove that A BIJECTION EXISTS.

This double representation of the same real actually CAN be a problem!

Any thoughts would be appreciated (such as a constructive bijection).