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# 0.9 (repeated) does NOT equal 1, with proof

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posted on Dec, 12 2006 @ 10:42 PM
OK, I was wondering where you were going with this. You're just stubborn about your idea being either right or wrong and you don't care about lots of other highly qualified people having dealt with the problem and its solutions. Whatever, I guess I can deal with this odd first post of yours...

First of all, I have no idea where you get this equation. A true source would be nice, not just an equation.

This guy:

I have no idea why you are going from .9 to the next line under it:

Originally posted by Zurahn

0.9 = 9/10
(9(10^(-a-1) - 1)) / (10(10^(-1) - 1)) // 10^1 * 10^(-1) = 10^0 = 1
= (-9(10^(-a-1) - 1)) / (9) // (9)s cancel out
= -(10^(-a-1) - 1)
= 1 - 10^(-a-1)

Lim a->infinity, substitute infinity to get the value

After this point you switched from "a"s to "i"s. That's a no-no.

Originally posted by Zurahn
= 1 - 10^(-i-1)
= 1 - 10^(-i) // negative infinity subtract a number is negative infinity

Actually, it should be:
= 1 - 10^(-i-1)
= 1 - ((10^(-i))(10^(-1)))
= 1 - (1/(10^i))(1/10)
OR
= 1 - 10^(-i-1)
= 1 - 10^(-(i+1))
= 1 - 1/(10^(i+1))

Originally posted by Zurahn
= 1 - (10 / i)
= 1 - (1 / i)

Can we agree that
Lim a->infinity = 1 - (1 / i)
All I did was simplify the equation by elimination, so there's nothing changed.

I assume you're trying to say that, in limit form:
1 - (10 / i) ~ 1 - (1 / i), when Lim i->oo

Originally posted by Zurahn
Assume,
0.9r = 1
(I write 0.9 repeated as 0.9r)

The use of geometric progression is the only way to write 0.9r.

Not really, but we'll go with your idea.

Originally posted by Zurahn
Therefore
1 - (1 / i) = .9r

Well, you left out the limit notation, but OK...

Lim 1 - (1/i) = ?
i->oo

It either equals .9r, 1, or both.

Originally posted by Zurahn
Agree? Ok.
.9r = 1
.9r = 1 - (1 / i)
1 - (1 / i) = 1
1 / i = 0

.9r = 1 (assumption)
IF
.9r = Lim 1 - (1/i)
i->oo
THEN
1 = Lim 1 - (1/i)
i->oo
SO IF
Lim 1 - (1/i) = 1
i->oo
THEN
Lim 1/i = 0 (which is true)
i->oo

Originally posted by Zurahn
FALSE. Division by infinity DOES NOT EQUAL ZERO. It's an infintesimally small number, NOT ZERO.

THEREFORE 0.9r CANNOT EQUAL 1

... What's the argument against me?

Ah, this is probably your mental block. There is no such thing as an infinitesimal number. An infinitesimal is more of a placeholder or a pointer (in computer science). In math, we use infinitesimals to describe the gap between the smallest possible discrete numbers. Discrete means individual units. THEY ARE NOT NUMBERS. Remember one of my statements is that an infinitesimal, no matter how many times it is multiplied by itself, can never equal even the smallest number in existence? That is because it is not a number, but instead used to describe the space in between numbers. They do not describe how big that space is, they just represent the tiniest of gaps that separate the real numbers.

Since an infinitesimal does not have a size, other than not being 0, it is not large enough to separate two numbers. It is only used to point at the gap in between. HOWEVER, in this case, we are trying to subtract something without a real size from a number (i.e. 1). An infinitesimal distance is no distance at all. That is why I describe it as a pointer. It points to something. If you point it to 1, then your value is 1. If you point it at a gap between two numbers, it is the gap. We have to DESCRIBE it as being larger than zero, because if you call it zero, it automatically becomes as useless as dividing by zero.

I don't know if you are trying make a theory look beautiful, but all good mathematicians know that at this point you lose any significance. Any numbers smaller than 1/(2^70) is considered inconsequencial, forever. Why? Because 2^70 is roughly the number of seconds in the suspected life of the universe. Anything beyond that is believed to have no significance for any application we can conceive. So, in practicality:

Lim 1 / i = 0, because who cares if it is anything else at this point
i-> (any # > 2^70)

But in truth and in theory, the number resulting from i->oo will be a real number, and that real number will be 1. That real number will ALSO be .9r

In summary, to say .9r is an infinitesimal distance from 1 is to say that .9r IS 1, since there is no such distance as an infinitesimal distance. If you want to prove otherwise, you'll need to study the hyperinfinitesimals and prove whether or not they can prove that infinitesimals have some definable distance AND THEN prove that .9r is at least an infinitesimal away from 1. If you can't do that, then .9r IS STILL 1.

posted on Dec, 12 2006 @ 11:01 PM
If you really think you have what it takes, here's a source for you on infinitesimals (contains a reference to hyperinfinitesimals).

Henle, James M. and Eugene M. Kleinberg. Infinitesimal Calculus. Cambridge: MIT Press, 1979.

posted on May, 17 2008 @ 06:03 PM
dudes, cmon 0.9 r DOES equal 1 .

Proof: so simple:

1/3 = 0.3 r
*3 *3
1=0.9r

SO SIMPLE

posted on May, 18 2008 @ 05:04 AM
Who cares?

.9 (repeated) is infinitely close to 1, so why does it matter?

EDIT. What would 2.14343434343434343434343434343434 (repeated) equal?

[edit on 18/5/2008 by C0bzz]

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