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#926 2008-04-24 10:50:31

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

Excuse me for a moment, on what basis does the ZCF and Peano's 5th postulate stand? Sheer logic? Or some unintuitive assumptions plus logic? Ironically, the assumptions themselves might be illogical. Hence, the whole rational may just start on an arbitary basis.

George, if you were truly interested, you would look up a term like "ZFC" which I have mentioned countless times.  And if were were to spend 5 seconds reading the wikipedia page on ZFC, you would know that ZFC is that "basis" which you asked about.

But you aren't interested, you only seem to want to argue.

Why it needs to go so far as to distort a number as set of numbers in ZCF to get your "only" way to gain some results? Haven't you seen the absursity in it to introduce unnecessary clumsy assumptions similar to a white elephant?

This is how mathematics was constructed (yes - all mathematics) in the 1920's and it has been so successful that alternate axiom systems have pretty much been left in the dust.  There are probably better axiom sets, but no one knows of them, at least not yet.  We go with what we have, and a PhD thesis can be made showing a better way.  But no one has done that.  All you want to do is sit and moan that you don't like it.

And what is so wrong to go intuitive? After all, intuition observes the truth as best as possible.

That is so incredibly wrong, and I'm fairly certain Einstein just rolled over in his grave.  If you really think intuition is the best way to go, the only thing I can tell you is that you haven't always studied mathematics.  No matter how much mathematics I study, intuition will always tell me that:

is not a real number.

The link in 922 by thedude also tells few physicians believe in infinity and infinitesimal idealism. And they came up with results while disbelieving your only method. I guess at least they disagree with you, Ricky.

You're quote mining, George.  It is yet another thing which shows your not truly interested in learning or debate, but rather just arguing and getting people to agree.  The link says:

It is therefore assumed by physicists that no measurable quantity could have an infinite value[citation needed] , for instance by taking an infinite value in an extended real number system (see also: hyperreal number), or by requiring the counting of an infinite number of events. It is for example presumed impossible for any body to have infinite mass or infinite energy. There exists the concept of infinite entities (such as an infinite plane wave) but there are no means to generate such things.

It is that no physical quantity can be infinite, and this completely agrees with mathematics and the real numbers.  There are an infinite amount of real numbers, but every real number is of finite value.

And physicists use real values, and not only that, but irrational and transcendental values all the time.  The fundamental of integration, which only work because of the completeness of the reals, is the basic concept of which most of physics uses.  For example, Maxwell's Equations are all dealing with integrals.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#927 2008-04-30 12:49:13

moxiuming
Member
Registered: 2008-04-20
Posts: 7

Re: 0.9999....(recurring) = 1?

first we should know that 0.999...... is not a number.
Note


[math0.99\1-\frac{1}{10^2}math]
so

Let

when we say that 0.999......=1,we just mean that

Here is a intitutive proof:




we can see that as n goes bigger and bigger,0.999...... goes more and more close to 1,but it's not accurate in mathematics.I will give a precise proof.
obviously 0.999......<1,if 0.999...=A where A between 0 and 1,then we can easily find some N such that

a contridition derived.

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#928 2008-04-30 14:02:14

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

obviously 0.999......<1

No, it's not obvious.  In fact, it is not even true.

Now let me suggest an alternate proof:

1/3 = 0.333...

Multiply both sides by 3, what do you get?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#929 2008-05-01 03:29:55

Dragonshade
Member
Registered: 2008-01-16
Posts: 147

Re: 0.9999....(recurring) = 1?

moxiuming, please fully understand the concept of infinity......

0.999999.... is represented as, you can even consider it to be a expression which holds the value 1
 

Last edited by Dragonshade (2008-05-01 03:33:47)

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#930 2008-05-01 05:13:59

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: 0.9999....(recurring) = 1?

moxiuming wrote:

if 0.999...=A where A between 0 and 1,then we can easily find some N such that


a contridition derived.

It's true that if you could find some number x such that 0.999... < x < 1, then 0.999... would undeniably have to be less than 1.
However, in your above proof, you've assumed that 0.999... is less than 1 in order to find your number. Therefore it's a circular argument and so doesn't prove anything.


Why did the vector cross the road?
It wanted to be normal.

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#931 2008-08-12 09:07:23

MooseofDoom
Member
Registered: 2008-08-10
Posts: 13

Re: 0.9999....(recurring) = 1?

hmm this is a vey interesting subject, indead!

As Ricky said if you have 1/3=0.333...  & you multiply both sides by 3 then you would get 1=0.999...  But then doesn't that make it true that 1 is not equal to 0.999...  ?  Wow amazing how great the span of math is, it is similar to space, how we cannot grasp it being Infinite


π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...

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#932 2008-08-12 13:52:15

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: 0.9999....(recurring) = 1?

if you have 1/3=0.333...  & you multiply both sides by 3 then you would get 1=0.999...  But then doesn't that make it true that 1 is not equal to 0.999...  ?

um.. no? Multiplying both dies of  1/3=0.333 gives you 1 = 0.999..." which says they are equal. It does not say they are unequal. So how would that suggest they are unequal?


A logarithm is just a misspelled algorithm.

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#933 2008-08-13 07:52:46

MooseofDoom
Member
Registered: 2008-08-10
Posts: 13

Re: 0.9999....(recurring) = 1?

Hmm yeah... I guess I was thinking about the problems where you have to determine if an equation is true such as 2x-1=2, x=1 which would be false... I am just wondering how 0.999... can be equal to 1 if they are not the same, because in theory 0.999... would get infinitely closer to 1 but would never actually reach 1. Such as in Physics when someone touches someone else, in theory we are really never touching them because the atoms electron clouds would prevent the nucleui from thouching each other, but we do get an electrical signal to the brain signifying touch... Another example is when an oblect moves halfway to another object then halfway again, then halfway again to infintity but the object can in theory always move another half closer to the other object... 0.25(0.5)=0.125; 0.125(0.5)=0.0625 then it would move on to infinity... I guess in theory 0.999... is not equal to 1, but in reality we set it equal to 1 to make computing easier, just like in theory we never actually touch each other but in reality we feel as if we do because of what our brain is telling us...dizzy

Last edited by MooseofDoom (2008-08-13 07:56:40)


π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...

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#934 2008-08-13 08:56:14

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 0.9999....(recurring) = 1?

You're thinking of 0.999... as if it started as 0.9, then it changed to 0.99, then 0.999, and so on, continually growing more nines and getting closer and closer to 1.

In fact, 0.999... already has all of its nines. It's a constant number, not changing or getting close to anything.


Why did the vector cross the road?
It wanted to be normal.

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#935 2008-08-13 14:39:27

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

Such as in Physics when someone touches someone else, in theory we are really never touching them because the atoms electron clouds would prevent the nucleui from thouching each other, but we do get an electrical signal to the brain signifying touch...

This is incorrect.  You are trying to use it as an example of things that get "infinitely" close, but it is not.  The two nuclei always have a positive, finite, distance apart.

Another example is when an oblect moves halfway to another object then halfway again, then halfway again to infintity but the object can in theory always move another half closer to the other object... 0.25(0.5)=0.125; 0.125(0.5)=0.0625 then it would move on to infinity...

As mathsyperson said, this is not what happens with 0.999...   Nothing is "moving", and nothing is "changing".  It just... is.

I guess in theory 0.999... is not equal to 1, but in reality we set it equal to 1 to make computing easier, just like in theory we never actually touch each other but in reality we feel as if we do because of what our brain is telling us...

No.  0.999... = 1 in theory.  Two real numbers are not equal when they are a finite distance apart.  This is actually, in part, how they are defined.  0.999... is not a finite distance from 1, and thus, it equals 1.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#936 2008-08-14 15:34:07

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: 0.9999....(recurring) = 1?

1 = 0.9 + 0.1
1 = 0.99 + 0.01
1 = 0.999 + 0.001
1 = 0.9999 + 0.0001
...
1 = 0.99999... + 0.0000...
Where'd the 1 go?  Lost forever?

Last edited by John E. Franklin (2008-08-14 16:00:57)


igloo myrtilles fourmis

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#937 2008-08-15 01:23:26

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 0.9999....(recurring) = 1?

Pretty much. The 1 appears at the end of the zeroes, but in the infinite case the zeroes don't have an end and so the 1 can't be there.


Why did the vector cross the road?
It wanted to be normal.

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#938 2008-08-15 09:20:54

careless25
Real Member
Registered: 2008-07-24
Posts: 555

Re: 0.9999....(recurring) = 1?

infinity is a paradox, try to explain it and u get more paradoxes!:D

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#939 2008-08-15 16:22:15

MooseofDoom
Member
Registered: 2008-08-10
Posts: 13

Re: 0.9999....(recurring) = 1?

Wow this is a crazy concept to grasp...


π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...

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#940 2008-08-15 18:46:02

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

Re: 0.9999....(recurring) = 1?

Just don't push the button if it ain't quite one, cause then your dumbweed brain has contributed enough to mankind.


I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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#941 2008-08-15 20:40:10

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: 0.9999....(recurring) = 1?

mathsyperson wrote:

Pretty much. The 1 appears at the end of the zeroes, but in the infinite case the zeroes don't have an end and so the 1 can't be there.

To add to that, if you were to ever put a 1 at any point in the chain of zeroes, you would thereby be ending the chain of zeroes, and so the number of zeroes would be finite; so any argument along the lines of 'an infinite number of zeroes, then a one' makes no sence.


The Beginning Of All Things To End.
The End Of All Things To Come.

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#942 2008-08-15 20:57:14

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

Re: 0.9999....(recurring) = 1?

Add to that That Is Correct!


I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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#943 2008-08-16 11:36:34

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: 0.9999....(recurring) = 1?

Don't forget that you can have
a river with banks on both
sides and the river could run
forever.  And hence you could
have 7water7, where 7 are the
banks, and the water could
fill up forever.  But maybe forever
is not long enough to reach
infinity.


igloo myrtilles fourmis

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#944 2008-08-16 21:14:20

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: 0.9999....(recurring) = 1?

that is a terrible argument that doesn't describe the same thing, to say 7water7 with 7 being the banks, you would have to instead say that the river is infinitely wide not infinitely long. which ofcourse is not possible.

Last edited by luca-deltodesco (2008-08-16 21:14:38)


The Beginning Of All Things To End.
The End Of All Things To Come.

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#945 2008-10-26 16:33:11

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: 0.9999....(recurring) = 1?

Hi guys. I just submitted one article to philosophy of science. Basically I argue arbitary fraction cannot exist.


X'(y-Xβ)=0

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#946 2008-10-26 21:02:17

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

What is the title of the journal?  And do let us know if it gets published.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#947 2008-12-12 13:19:44

mrphysics
Member
Registered: 2008-12-12
Posts: 1

Re: 0.9999....(recurring) = 1?

Hi. New member here. I have been discussing this problem with some students of mine and have reached a bit of a wall. I have a way of thinking about this and I hope someone can point me in right direction.

We started with the following "proof".

let x = 0.99999999.......

10x = 9.9999999........
10x - x = 9.999999....... - 0.999999......
9x = 9
x = 1

My problem is with the following line:

9.999999....... - 0.999999...... = 9

The kids are happy to accept this and gave the following response (among others).


"Consider 9.999... - 0.999....

If this is subtracted digit by digit it become clear that the answer is 9:

First we have 9-0 so we have 9 'units'.

Then 9-9 so 0 tenths; then 9-9 so 0 hundredeths....

As it is an infinte decimal this will continue forever so the answer will be:

9.00.... = 9"

Now my response to this was to write down each term for the subtraction as follows:

9.9999.... = 9 + 0.9 + 0.09 + 0.009 + ....
0.99999... = 0.9 + 0.09 + 0.009 + 0.0009 + ....

So far so good. If we then combine these then we get

9.9999... - 0.99999... = 9 + 0.9 - 0.9 + 0.09 - 0.09 + ....

It therefore appears that each term apart from the 9 will cancel giving the answer 9 (exactly). However, I have a problem with this and it is as follows:

If we group the terms together in a different way then I think there is a different solution.

9.999.... - 0.9999.... = 9 x [(1-0.1) + (0.1 -0.01) + (0.01 - 0.001) +....]


9.999.... - 0.9999.... = 9 x [(0.9) + (0.09) + (0.009) + (0.0009)....]

                                          = 9 x 9 (0.1 + 0.01 + 0.001 + 0.001 + ...)

                                          = 81 x (0.1 + 0.01 + 0.001 + 0.001 + ...)
                                          = 8.1 + 0.81 + 0.081 +0.0081 + ...
                                          = 8.999...

Therefore the alternative line in the original proof should be

9x = 9.999... - 0.9999.... = 8.9999...

which gives the trivial answer that x = 0.9999.... which is where we started. It has not been shown that 0.9999.... = 1.

My reason for grouping the terms in this way can be understood if we again consider the series for each value.

9.9999.... = 9 x sum (n=1 to n= ∞ )[10^-(n-1)]
0.99999.... = 9 x sum (n= 1 to n= ∞ )[10^-(n)]

Apologies for the notation but I hope that this is clear.

Now if these expressions are grouped together we get

9.999... - 0.9999... = 9 x sum (from n = 1 to n = ∞ ) [(10^ -(n-1)) - (10^ -n) ]

which indeed gives the expression that does not allow 0.9999... to equal 1.

(   9.999.... - 0.9999.... = 9 x [(1-0.1) + (0.1 -0.01) + (0.01 - 0.001) +....]    )


My logic is that even though there are an infinite number of different terms in each original series the statement (from n = 1 to n = ∞ ) also suggests there should be the same number of terms in each expression. For the expression to be written in a form that allows all the non-9 terms to cancel the series for 9.9999... must contain one extra term than the series for 0.9999... If this is the case then the two series cannot be combined to give the expression

9.999... - 0.9999... = 9 x sum (from n = 1 to n = ∞ ) [(10^ -(n-1)) - (10^ -n) ]                                   

and thus,

9.999... - 0.999... = 8.999

and not 1.

So finally here is my problem. If the symbol ∞ is defined in the system to mean infinite then this surely must have a definite meaning. My thinking is that the symbol ∞ cannot mean infinity and (infinity + 1) in the same axiomatic system. I am happy to accept that infinity + 1 is still infinite but is it possible that ∞ and ∞ + 1 are actually different numbers. I am aware that Cantor defined different classes of transfinite numbers but know very little about his methods and am wondering if anyone can elaborate. If these numbers are different then all the terms other than 9 will never cancel and it is surely true that trying to prove 0.9999.... = 1 (by this proof at least) is a fallacy.

Apologies for the overly long message and thanks in advance.

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#948 2008-12-12 13:57:45

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,552

Re: 0.9999....(recurring) = 1?

Sorry I don't have time to answer you in full. But I can mention these points:

∞ = ∞ + 1

∞ is not a number in the usual sense (it is not a "real number")

See: What is Infinity?


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#949 2008-12-14 20:08:09

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.9999....(recurring) = 1?

If these numbers are different then all the terms other than 9 will never cancel and it is surely true that trying to prove 0.9999.... = 1 (by this proof at least) is a fallacy.

If you keep taking math classes, you will eventually deal with series and convergence in a rigorous setting.  If you have taken calculus classes, you may have already seen some theorems about series, though probably not proved them.  In any case, there is a theorem:

If a series of terms b_n is such that

[align=center]

[/align]

Then the any rearrangement of the series will converge to the same number.

Now "rearrangement" is rigorously defined, but I am attempting to avoid getting lost in the definitions, and it pretty much means exactly what you did in your post.  In other words, both ways you summed the terms converge to the same number.  In essence, you proved that 8.9999 = 9.

This may be a bit tricky logic to work with, but hopefully this explanation makes sense.  By the above theorem (assuming that you accept it, and you certainly don't have to, I can prove it if you would like) we have that your first way of summing to get 9 is valid.  Therefore, 0.999... = 1 by the original proof.  Now, by the same theorem, the way you summed your the terms the 2nd time, they have to agree with the first way.  Thus, we conclude that 8.999... = 9.

In any case, if you still don't accept it, then here is another version using the same technique, but a whole lot cleaner:

1/3 = 0.333...
1/3 * 3 = 1
0.333... * 3 = 0.999...

As for the stuff on infinity, it will take quite a bit of study to come to a good understanding.  We consider infinity to be the same size as infinity + 1 based on the following theorem:

If A and B are two sets containing a finite number of elements, then |A| = |B| (the number of elements in A and B) if and only if there exists a function from A onto B that is 1-1 and onto.

The existence of this function with special properties (if you're not familiar, either ask or wikipedia) can only occur when two sets of finite size have the same number of elements.  And having this map will mean that the sets have the same number of elements.  We say that these two statements are "equivalent", and what we mean is that they represent the same idea.

The great thing about this "definition" of size is that it is easily extended to the infinite set case:

If A and B are two sets (of possibly infinite elements), we say that |A| = |B| means there exists a function from A onto B that is 1-1 and onto.

Now to show you that "infinity" = "infinity + 1", let's take the natural numbers and find a 1-1 map that is also onto and maps into the natural numbers as well as the number 0.

It shouldn't be too hard to prove that this map is both 1-1 and onto.  Just ask if you want to see it.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#950 2008-12-26 17:17:47

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: 0.9999....(recurring) = 1?

mrphysics, i used the similar method to locate the last 9 in 0.999...
you see, the new 0.9999... has exactly one more 9 in it so we wanna find where it is by compare and contrast
0.999...
0.9999...

and through a proof I posted in previous thread the one 9 previous 0.999... lacks has no suceeding 9 after it or it would be ...09
in other words, it has a last 9

Start from here,
and think it this way,
try to sum up 9's from the last 9 backwards, you will get 0, or infinitesimal, which doesn't matter. And no matter how many or much you have summed up, you will never be able to come up with an even small number larger than 0.
Is it a proof
0.9+0.09+0.009+...+last9=0?
This is my paradox

I know Ricky you would argue the cardinality (only a quible name for amount when it comes to infinity) of the digits of 0.999... is N0. But I know how this Hebrew alpha 0 is defined by Cantor, and I have a similar disproof for so called mapping integers to evens.


X'(y-Xβ)=0

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