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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,15/01/07.

The Variables below A & B are Both Single Start Values.

A = 1

B = 0.9

Below Proves A is Infinitely > B And Gives The Variable C This Value.

C = A / B = 1.1111111111......

Below is How Infinite 0.9 is Calculated.

0.9 x C = 0.9999999999 ......

Below Proves Infinite 0.9 is <>1 By Using The Original Variables A & B.

A x C = 1.1111111111 ......

B x C = 0.9999999999 ......

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Anthony.R.Brown wrote:

0.9 x C = 0.9999999999 ......

Below Proves Infinite 0.9 is <>1 By Using The Original Variables A & B.

A x C = 1.1111111111 ......

B x C = 0.9999999999 ......

How on earth does that last part prove *anything*?

Oh, and B x C = B x (A/B) = A = 1. Since B x C = 0.99999... (from your post) we are left with only one conclusion about the equality of 0.999999.... and 1. I would tell you what it is, but you wouldn't listen to me like you haven't listened to so many others before. In fact, you probably haven't even read this far and if you have you haven't been paying attention.

Also, **THERE IS ALREADY A THREAD FOR THIS TOPIC, PLEASE STOP CLUTTERING THE FORUM**.

*Last edited by Dross (2007-01-15 02:07:38)*

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Anthony, I put to you two things:

1) in your first post of this thread, you declared that B x C = 0.99999... - and quite rightly so.

2) in the second post in this thread, I showed you how B x C = 1 (because B x C = B x (A / B) = A x (B / B) = A x 1 = A = 1).

Now, what conclusion can we make of this? How can you now say, in light of the above two facts, that 0.99999.... is not equal to 1? It might not go along with your intuition, but some things never do.

Come on, I'm not after insulting anyone here, just answer the question please.

*Last edited by Dross (2007-01-15 02:22:26)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,660

Anthony.R.Brown wrote:

Below Proves Infinite 0.9 is <>1 By Using The Original Variables A & B.

A x C = 1.1111111111 ......

B x C = 0.9999999999 ......

Not a proof ... the answers are different because A and B are different. They are different by a factor of 0.9: **1/0.9=1.111...**, so the argument is circular.

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

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To MathsIsFun

The Variables below A & B are Both Single Start Values.

A = 1 "This has to be the Start Value for 1"

B = 0.9 "This has to be the Start Value for Infinite 0.9"

C = A / B = 1.1111111111......

A x C = 1.1111111111 ......"This is the same Calculation to Make a Number Infinite! as with B"

B x C = 0.9999999999 ......"This is now Infinite 0.9"

The two Values for A and B are different! because Infinite 0.9 <> 1

If the Result for A did Equal B then Infinite 0.9 does = 1

Even If! by some Magic B does become 1 it wil still be Infinite 0.11111..... Less than A

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

im using your own results here:

A = 1

C = A / B

B×C = 0.999....

substituting youre second equation gives

B×C = B×(A/B) = A, which youre first assignment says is 1.

so using youre own reasoning, B×C = 0.999... = A = 1, 0.999... = 1

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To luca-deltodesco

I think you are missing the point of what I am saying!

A starts as = 1

B starts as = 0.9

A x C results in a number > 1

B x C results in a number < 1

The point is A is 1

and B never = 1

If we are going to Infinitely multiply a number so that is becomes 1 "if possible"

Then a number that is already 1 must always be Greater!!

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

no, i think you are missing the point of what i am saying, using youre own equations, ive shown you how they are equivelent, not different.

now, answer me this, you say B×C is less than 1? If this is true, then there must be a number that is inbetween those two, otherwise one would not be greater than the other.

for example: you can show that 0.35 < 0.36, because there exists an infinite amount of numbers between those two, such as 0.351, 0.356, 0.35126, 0.355846 etc.

so, if 0.999.... < 1, show me some numbers that are inbetween, if there exists no numbers, then 0.999... = 1

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To luca-deltode

Quote :

"so, if 0.999.... < 1, show me some numbers that are inbetween, if there exists no numbers, then 0.999... = 1"

A.R.B

That's an easy One! and tried many times! as an argument!

0.999999......(The Number in Between is Another 0.9).99999999999999... <> 1

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To " Everyone! pleae help with this PROOF! "

A starts as = 1

B starts as = 0.9

A x C results in a number > 1

B x C results in a number < 1

The point is A is 1

and B never = 1

If we are going to Infinitely multiply a number so that is becomes 1 "if possible"

Then a number that is already 1 must always be Greater!!

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,660

Anthony.R.Brown wrote:

0.999999......(The Number in Between is Another 0.9).99999999999999... <> 1

Could you write down that number clearly. The one between 0.999... and 1. On its own.

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

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To MathsIsFun

Quote:

"Could you write down that number clearly. The one between 0.999... and 1. On its own. "

A.R.B

.9

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Anthony.R.Brown wrote:

To " Everyone! pleae help with this PROOF! "

A starts as = 1

B starts as = 0.9

A x C results in a number > 1

B x C results in a number < 1

The point is A is 1

and B never = 1

If we are going to Infinitely multiply a number so that is becomes 1 "if possible"

Then a number that is already 1 must always be Greater!!

What you have shown (if anything) here is that 0.9 is not equal to 1 - we knew that already.

The number 0.9 has absolutely nothing (ABSOLUTELY NOTHING) to do with 0.99999999.... (or, as you call it, "infinite 0.9") except an *entirely coincidental* similarity in the first part of it's decimal expansion. So don't use it and hope it'll prove anything.

**Futhermore, you have not answered mine and Luca's query about how you could resolve that B x C = B x (A / B) = A, where you yourself profess that B x C = 0.999999999... and A = 1.**

Also, 0.9 is not between 0.99999... and 1.

I can't believe you actually believe what you're saying - you're either taking the mickey or you're just mechanically writing down things without actually thinking about them. There is no *way* you could believe the things you express, not least of all because the things you express contradict themselves.

Bad speling makes me [sic]

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

To Dross?

A x C = 1.11111...

B x C = 0.99999....

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

"This is Better! is? it! "

A starts as = 1

B starts as = 0.9

A x C results in a number > 1

B x C results in a number < 1

The point is A is 1

and B never = 1

If we are going to Infinitely multiply a number so that it becomes 1 "if possible"

Then a number that is already 1 must always be Greater!!

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

All you have given me here is:

1) a non-sense reply that I cannot make sense of. Yes, I agree that, as you have defined them at the start of this thread, it is indeed the case that

A x C = 1.11111...

B x C = 0.99999....

... so what? What does this show?

2)

A x C results in a number > 1

B x C results in a number < 1

This, whilst debatable*, is immaterial - so what if they do? A and B are different numbers, so it's entirely possible to find another number that gives the above result.

It's also possible to find a number D such that A x D > 1,000,000 and B x C < 1,000,000. All you've done is shown that two numbers that are not equal to each other are, in point of fact, not equal to each other. So what?

* - say that "B x C results in a number <= 1" and I'm with you.

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Anthony.R.Brown wrote:

To MathsIsFun

Quote:

"Could you write down that number clearly. The one between 0.999... and 1. On its own. "

A.R.B

.9

Oh yeah, and you still haven't resolved this issue - 0.9 is *clearly* not between 0.999... and 1. "Guess again".

*Last edited by Dross (2007-01-19 01:43:25)*

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

A x C = 1.11111...

B x C = 0.99999....

It Only takes One Calculation for A x C to = 1.1111.....Because A is! and A eqauls 1

It takes B x C to = 0.9999999..... Which is One Calculation Because B is < 1 to start with!

If B was Now to Some how now Reach! or Equal 1 by Multiplication! this would only then = 1

This would be would still make B 0.1 < A because B never started as 1 or = 1 the same as A

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

If anyone wants me to personally pick out the errors in this thread, I will.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,660

Everyone: I deleted a few name-calling posts.

Back to the "number between 0.999... and 1"

ARB: your 0.9 is **less than** 0.999..., but we are looking for a number **greater than** 0.999... but **less than** 1. What is it?

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

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Anthony.R.Brown wrote:

A x C = 1.11111...

B x C = 0.99999....

It Only takes One Calculation for A x C to = 1.1111.....Because A is! and A eqauls 1

It takes B x C to = 0.9999999..... Which is One Calculation Because B is < 1 to start with!

If B was Now to Some how now Reach! or Equal 1 by Multiplication! this would only then = 1

This would be would still make B 0.1 < A because B never started as 1 or = 1 the same as A

If I read that at all correctly, what you're trying to do is prove that 0.999... < 1, by assuming that 0.999... < 1 - this is fundamentally flawed and doesn't prove anything.

Bad speling makes me [sic]

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

LET'S MAKE THIS AS CLEAR AS POSSIBLE! USING NEW VARIABLES.

N1 = 0.9 This is a Single Number Value

N2 = Infinite 0.9 This is an Infinite Number!

N3 = 1 This is a Single Number Value

N1..............( Here between N1 and N3 there are no other Numbers! Because N1 is the Single last Number!).............N3.

N2..............( Here between N2 and N3 there are an Infinite Amount of Numbers! Because N2 is an

Endless stream of Numbers!).............N3.

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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

If a Number has to be Infinitely Multiplied! To Try and Reach a Target Value! Then the Target Value! Must be Greater than the Number being Infinitely Multiplied! Reason Being! If the Target Value was also Infinitely Multiplied! Then the Target Value would again be Greater!

Number Infinitely Multiplied! = 0.9 Target Value! = 1

[Anthony.R.Brown 20/01/07].

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

you just said that there are no numbers between 0.9 and 1... what about 0.91 or 0.95 or 0.905 or 0.984 or 0.995 or 0.9125

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Patrick****Real Member**- Registered: 2006-02-24
- Posts: 1,005

Yeah, that made me ponder aswell luca.. From what you said, ARB, there's either one decimal, or an endless amount. Please write out 1/10, 2/10 and 3/20 out, and tell me how 3/20 is not between the other two.

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