0.9999....(recurring) = 1?

Maelin · 2007-02-22 00:48:34

Anthony.R.Brown wrote:

There are some problems that have to be solved and shown to be solved before the Argument Infinite 0.9 = 1 can be considered an Argument!
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(1) Explain how it is possible for a Number that starts with a lower value than another Number! is or becomes equal to the other Number! without using an Algebra sign.
(2) Explain how it is possible for a Number that starts with a Decimal Point! losses the Decimal Point! without using an Algebra sign.
(3) Give a Maximum Possible Value for the Amount of Infinite 0.9's. ( there has to be a Value ? ) otherwise the .9's will Continue!

1. It is not possible for numbers to "start" or "become" anything. In the context of mathematics, numbers are fixed. They never move, ever. We can look at different numbers in a logical pattern and consider what happens as WE move along them, but that is our attention moving along the numbers, not the numbers themselves moving.

2. The number does not "start" with a decimal point. A number may be expressed in a way that involves a decimal point, and it may also be expressed in a way that does not involve a decimal point. Example: 2.0 = 2

3. The number of 9s in the decimal representation of 0.999... is infinite. That means the number does not have a value that is a member of the set of natural numbers (ordinary counting numbers). There is nothing wrong with this, just like there is nothing wrong with any other recurring decimal.

Incidentally, hello everybody! I've come to join in the fun and hopefully help you folks out with your understanding of this tricky little problem.

Last edited by Maelin (2007-02-22 00:50:51)

Anthony.R.Brown · 2007-02-22 00:52:37

To Maelin

I do agree with you! Infinite 0.9 Starts as 0.9 and so Remains 0.9 forever! no matter How long it is!!

Maelin · 2007-02-22 01:03:29

Anthony.R.Brown wrote:

To Maelin
I do agree with you! Infinite 0.9 Starts as 0.9 and so Remains 0.9 forever! no matter How long it is!!

Sorry, are you suggesting that 0.999... (infinite digits long) is equal to 0.9 (one digit long)? Or that 0.999 (three digits long) is equal to 0.9 (one digit long)?

Please use standard mathematical notation. This "Infinite 0.9" terminology merely obscures your argument.

Anthony.R.Brown · 2007-02-22 01:25:21

To Maelin
Quote:
"Sorry, are you suggesting that 0.999... (infinite digits long) is equal to 0.9 (one digit long)? Or that 0.999 (three digits long) is equal to 0.9 (one digit long)?"

It is According to The Infinite Difference Remaining the Same! No Matter How Long they
Both are!

A = 1 " Single Start Value For 1 "
B = 0.9 " Single Start Value For 0.9 "
C ( A/B ) x B " Infinite 0.9 Value "
D ( A - C ) " Infinite < 1 Value "
C <> ( C + D ) " INFINITE 0.9 <> 1 "

In my Formula Above D will always have a Value! so that Only ( C + D ) or A can = 1

Maelwys · 2007-02-22 01:56:08

Anthony.R.Brown wrote:

Quote:
"Sorry, are you suggesting that 0.999... (infinite digits long) is equal to 0.9 (one digit long)? Or that 0.999 (three digits long) is equal to 0.9 (one digit long)?"
It is According to The Infinite Difference Remaining the Same! No Matter How Long they
Both are!

Can you explain the "Infinite Difference Remaining the Same" proof? I'm not quite sure I understand the mathmatical basis behind this statement. You're telling us that we can't prove 0.999... = 1 because they have a different number of digits, and then you're trying to tell us that 0.999... = 0.9? I don't get it...

Anthony.R.Brown wrote:

A = 1 " Single Start Value For 1 "
B = 0.9 " Single Start Value For 0.9 "
C ( A/B ) x B " Infinite 0.9 Value "
D ( A - C ) " Infinite < 1 Value "
C <> ( C + D ) " INFINITE 0.9 <> 1 "
In my Formula Above D will always have a Value! so that Only ( C + D ) or A can = 1

Can you explain the terms "Single Start Value" and "Infinite < 1 Value"? Also, explain your proof of why D <> 0? You can't simply disregard that case because you don't like the idea, you have to prove that it can't be true, or your statement that C <> C + D doesn't hold for the case where D = 0 (which it does, according to your definitions of C and A).

mathsyperson · 2007-02-22 03:30:30

A hearty welcome to the forum, Maelin! What a great first post. And you've got an excellent avatar as well.

Anthony, what will it take to convince you that 0.999... = 1? We've given you many proofs that this is the case, we've shown flaws in all arguments that you have put forward, so there's no rational reason why you would doggedly stick to the belief that you are right. And yet you do. What more evidence can we possibly give you?

Anthony.R.Brown · 2007-02-23 00:37:05

To mathsyperson Moderator

Thank you for your opinion!

A.R.B

Anthony.R.Brown · 2007-02-23 01:15:46

To Maelwys

Quote:
" Can you explain the "Infinite Difference Remaining the Same" proof? "

A.R.B

1 - ( 0.9 ) = ( 0.1 ) " Infinite Difference " which makes 0.9 < 1

1 - ( 0.999... ) = ( 0.001... )" Infinite Difference " which makes 0.999... < 1

The length of the above two examples! makes no difference to the actual difference remaining the same!

Quote:
" Can you explain the terms "Single Start Value" and "Infinite < 1 Value"? "

A.R.B

Infinite 0.9 has to start from somewhere! 0.9 is the Start/Stage one/Decimal Place for Infinite 0.9

The Infinite < 1 Value is the Value that always remains when 1 subtracts the Infinite 0.9 Value from the Start/Stage one/Decimal Place onwards.

Quote:
"Also, explain your proof of why D <> 0? You can't simply disregard that case because you don't like the idea, you have to prove that it can't be true, or your statement that C <> C + D doesn't hold for the case where D = 0 (which it does, according to your definitions of C and A)."

A.R.B

It is impossible foe D to ever = 0 because D is made up from the Infinite Difference
D ( A - C ) where that A started as an Infinitely greater value than C because C ( A/B ) x B
is where that C is calculated by multiplying the Value 0.9 x 1.111..... etc.
the Value C that has been multiplied by 1.111..etc.. will always be less than the Value A that starts as = 1 because 1 is a whole Number with a higher Value,and C is a value that has been divided within 1 which makes it permanently < 1

luca-deltodesco · 2007-02-23 01:37:01

ehm... how exactly is 0.1 an 'infinate difference' its very much a finite difference...., and infinate difference would imply that they are infinately apart, i.e.

Anthony.R.Brown · 2007-02-23 01:42:12

To luca-deltodesco

The term " Infinite Difference " is just to explain that any Difference is one of the same!
we are after all talking about from the Start onwards!

luca-deltodesco · 2007-02-23 02:05:13

by infinate difference, you mean, they have the same difference, then you are still talking nonsense.

immediate example, taking directly from you

1-0.9 = 0.1 (youre 'infinate difference')
1-0.99 = 0.01 (your 'infinate difference' again)

clearly, 0.1 and 0.01 aernt the same, so you are still talking nonense

Maelwys · 2007-02-23 02:19:03

Anthony.R.Brown wrote:

Infinite 0.9 has to start from somewhere! 0.9 is the Start/Stage one/Decimal Place for Infinite 0.9
The Infinite < 1 Value is the Value that always remains when 1 subtracts the Infinite 0.9 Value from the Start/Stage one/Decimal Place onwards.

Why does it have to start from somewhere? It's a number, it's not starting anywhere and it's not going anywhere... it just is. I think this concept of numbers "starting" somewhere (stage one?) is one of the confusing ones, because it makes no sense. A number is a number. 5 is 5. 15 is 15, it doesn't "start" as 1 and later "become" 15 just because you write the numbers from left to right. Similarily, 0.999... doesn't "start" as 0.9, just because you wrote that number first before you started adding extra 9s to the end. Otherwise you just proved that 1.9999.... = 1, because that's what it "starts" as, which makes no sense either.

Anthony.R.Brown wrote:

It is impossible foe D to ever = 0 because D is made up from the Infinite Difference
D ( A - C ) where that A started as an Infinitely greater value than C because C ( A/B ) x B
is where that C is calculated by multiplying the Value 0.9 x 1.111..... etc.
the Value C that has been multiplied by 1.111..etc.. will always be less than the Value A that starts as = 1 because 1 is a whole Number with a higher Value,and C is a value that has been divided within 1 which makes it permanently < 1

By using arithmatic, (A/B) x B = A, as has been shown to you many times before already. The fact that you insist that D <> 0, because C < A seems to be based entirely on inspection, that you THINK 0.999... looks smaller than 1, because the first digit of the number is smaller. And it's true that intuitively, ignoring all the math behind it (basically what you're doing) that does appear to be the case. And that's exactly why this is such a controversal proof, because it going completely against intuition. But the math backs up the proof and shows that your intuition is wrong. Just like if you'd asked somebody a hundred years ago if they thought that a 400 tonne hunk of metal would fly for any distance, they'd tell you that it's impossible because it's completely against their intuition that anything that heavy could get off the ground. Turns out their intuition is wrong, because now we see 747s and the new Airbus A380 do it with relative ease, because the math supports it. Just like the Wright brothers, you'll have to learn to get past your intuition and see that the math doesn't lie.

George,Y · 2007-02-23 02:20:52

1 is a number, however 0.999... is not.

Ricky · 2007-02-23 03:20:17

George,Y wrote:

So literally you mean Post 108 and Post 125??
All you did were embeding infinite digits(a self-contradictory concept) into your defination of r-somes and claiming you need it. Do you call that valid? Can a 1/10[sup]∞[/sup] or r[sub]∞[/sub] exist?
Hiding a problem doesn't solve it.

George, infinity is not a real number. So:

1/10[sup]∞[/sup]

r[sub]∞[/sub]

Don't even make sense.

Infinite series exist. We define decimal expansion to be equivalent to infinite series, that is, a limit representation. What is wrong with this definition? How is it contradictory?

Zhylliolom · 2007-02-23 05:49:19

George,Y wrote:

1 is a number, however 0.999... is not.

Then what is 0.999...? How about 0.333...? Is pi not a number? Define a number in such a way that 0.999... is not a number but all our notions of what a number is remain intact.

George,Y wrote:

All you did were embeding infinite digits(a self-contradictory concept) into your defination of r-somes and claiming you need it. Do you call that valid? Can a 1/10[sup]∞[/sup] or r[sub]∞[/sub] exist?

In my proof it is clear that the n appearing in 1/10[sup]n[/sup] and r[sub]n[/sub] is a natural number. I'm sure you know that ∞ is not a natural number, much less a reach number. Then there is no 1/10[sup]∞[/sup] or r[sub]∞[/sub]. n just increases indefinitely, is it so wrong to count 1, 2, 3, 4... forever? You'll always be saying a natural number.

I can rewrite my proof to give another infinite decimal representation for a real number, a representation you wouldn't argue against. This proof is equally as valid as the original; it merely gives another decimal representation for a real number, the "standard" decimal representation.

Theorem. Assume x ≥ 0. Then for every integer n ≥ 1 there is a finite decimal r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub] such that r[sub]n[/sub] ≤ x < r[sub]n[/sub] + 1/10[sup]n[/sup].

Proof. Let S be the set of all nonnegative integers less than x. Then S is nonempty, since 0 ∈ S, and S is bounded above by x. Then by the Completeness Axiom of the real numbers, S has a supremum, which we denote a[sub]0[/sub] = sup S. a[sub]0[/sub] ∈ S, so that a[sub]0[/sub] is a nonnegative integer. a[sub]0[/sub] is then the greatest integer in x, [x]. Clearly, a[sub]0[/sub] ≤ x < a[sub]0[/sub] + 1. Now let a[sub]1[/sub] = [10x - 10a[sub]0[/sub]], the greatest integer in 10x - 10a[sub]0[/sub]. Since 0 ≤ 10x - 10a[sub]0[/sub] = 10(x - a[sub]0[/sub]) < 10, we have that 0 ≤ a[sub]1[/sub] ≤ 9 and a[sub]1[/sub] ≤ 10x - 10a[sub]0[/sub] < a[sub]1[/sub] + 1. Then a[sub]1[/sub] is the largest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 ≤ x < a[sub]0[/sub] + (a[sub]1[/sub] + 1)/10. More generally, having chosen a[sub]1[/sub],..., a[sub]n-1[/sub] with 0 ≤ a[sub]i[/sub] ≤ 9, let a[sub]n[/sub] be the greatest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + a[sub]n[/sub]/10[sup]n[/sup] ≤ x < a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + (a[sub]n[/sub] + 1)/10[sup]n[/sup]. Then 0 ≤ a[sub]n[/sub] ≤ 9 and we have r[sub]n[/sub] ≤ x < r[sub]n[/sub] + 1/10[sup]n[/sup], where r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub]. This completes the proof.

It is easy to see from this theorem that we can define an infinite decimal representation of x. For x = 1, we see that a[sub]0[/sub] = 0, and a[sub]n[/sub] = 0 for all n > 1. Then 1 = 1.000....

Since these proofs are virtually identical, you can see that saying 0.999... ≠ 1 is basically the same as saying 1.000... ≠ 1.

George,Y · 2007-02-23 13:51:28

Do I need to repeat Post 107? Just by avoiding saying infinity doesn't mean you avoid using it.

Actually, "any" integer itself contains the concept of infinity, at least potentially. As Cantor himself put it, people unconciously use the concept of infinity in cases like "any" integer.

Suppose you have a basket to contain "any" amount of apples. How large is it then? It's as large as containing Infinite apples, right? It cannot contain only finite apples.

Back to the r.r1r2r3... case, how many are r's? Infinite of them. No, not finite, any finite will destroy the power of infinite digits, by which you can arbitarily pick up one finite rn larger than a particular finite 0.9999999999999999, 0.99999999999999999999999999999999999 or so. You can do so only because you have got infinite r's as backup.

Well, you may say you are not using r-infinity, you are only picking up rn or rm or any finite part of it. But the reason that you can do so is not that you can add digits arbitarily (growing) but that you have already infinite r's. And infinite r's guarantee the bizarre existence of r[sub]∞[/sub] (Do I need to use q again to replace the horrible-looking ∞ as I did in Post 107?)

Last edited by George,Y (2007-02-23 14:08:29)

Anthony.R.Brown · 2007-02-24 01:31:22

To Maelwys

Quote: " Why does it have to start from somewhere? It's a number, it's not starting anywhere and it's not going anywhere... it just is. "

A.R.B

Are all your Mathematical idea's Calculations based on the above Quote: of yours!
you have to show how you arrived at a Number to represent Infinite 0.9 before you can give any arguments about it!
Not knowing where Infinite 0.9 has come from! or how it is made! is the main reason you cannot grasp anything concerning my Formula!

From my Formula again!! the Calculation C ( A/B ) x B shows the Number ie. 0.9 is being Multiplied x 1.111... etc to achieve Infinite 0.9

If A from my Formula was Multiplied x 1.111 it will also always be greater than C
A(1) Starts as a higher Value than C(0.9) and so will always be greater! ( no Algebra signs + x etc. are used to make up the difference! )

D(A-C ) will always contain the Infinite Difference missing from C.

Anthony.R.Brown · 2007-02-24 02:01:33

To luca-deltodesco

Quote: " 1-0.9 = 0.1 (youre 'infinate difference')
1-0.99 = 0.01 (your 'infinate difference' again)
clearly, 0.1 and 0.01 aernt the same"

A.R.B

0.9 + 0.1 = 1

0.99 + 0.01 =1

Same difference!!

Last edited by Anthony.R.Brown (2007-02-27 00:17:45)

luca-deltodesco · 2007-02-24 02:14:27

0.99 + 0.001 = 0.991, not 1, if you can't even get that right...

Ricky · 2007-02-24 05:05:48

Actually, "any" integer itself contains the concept of infinity, at least potentially. As Cantor himself put it, people unconciously use the concept of infinity in cases like "any" integer.
Suppose you have a basket to contain "any" amount of apples. How large is it then? It's as large as containing Infinite apples, right? It cannot contain only finite apples.

Perhaps you misunderstand what is being said when one says any integer. It simply means that you can pick any integer you wish, and it will work for that integer.

There is no concept of size in mathematics in the way you describe. A basket is used as an analogy, but there isn't a finite amount of material in the world of mathematics as there is in the real world. By attempting to apply the real world to mathematics, you break the purity of math. Only the purity of math can be applied to the real world, the reverse is not true.

George,Y · 2007-02-24 14:32:06

I am sorry to tell you that the paragraph of apples is just to illustrate the next paragraph.
Purity of Math? You are kidding? Do I have to shrine the Pure Math?
All Maths are Applied Maths.
Otherwise they are some games played by only a few people (literally puzzles).

Last edited by George,Y (2007-02-24 15:10:56)

Ricky · 2007-02-24 15:37:20

All Maths are Applied Maths.

That is a sad view of the mathematical world.

When Euler defined his phi function and investigated its properties, was his goal to revolutionize the field of cryptography? When Galois, Lagrange, Abel, Cayley started studying abstract algebraic structures, were they preparing for their use in quantum physics 100 years later? When Boole invented and studied boolean algebra, was he just preparing for the invention of the computer?

Time and time again, we see that problems which mathematicians come up with themselves, purely theoretical, are later applied to the real world. It's truly magical in the way it works, but we can see from history that it does indeed work.

George,Y · 2007-02-24 17:57:04

Yes, mathematicians may study some problems that aren't immediately useful, but the majority of the remaining maths is useful, the rest having connections with the useful part. It's evolution.

How can maths be theoretical? Have you ever thought about it? Only when it follows logic, makes sense. How about the situation when it contradicts logic? Can illogic or irrational assumptions or definations also be a part of maths?

Last edited by George,Y (2007-02-24 18:16:30)

MathsIsFun · 2007-02-24 18:19:59

Of course mathematics has concepts that don't exist in the real world ... oh, I shouldn't be butting in like this, sorry.

George,Y · 2007-02-24 18:45:17

Well, the plane works fine when used in the triangle in a plane, the rectangle in a plane, never when used in infinitly large plane (perhaps better in any large plane). The closer the part to reality, the better the part works.

Math Is Fun Forum

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