# 0.99~=1

Discussion in 'Mathematics and Physics' started by rumiam, Apr 17, 2007.

1. ### ljcoxWell-Known Member

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Thank you.

2. ### rumiamNew Member

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As you can see..I just did add them

3. ### bloody-orcNew Member

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Originally Posted by ljcox
Your error is in adding the decimal values for 1/3 and 2/3

These have an infinite number of decimal places and so I don't see how you can add them.

Why can't you add two totally normal numbers as:
Code (text):

[U]1[/U]  and  [U]2[/U]  ?
3       3

There are no infinite numbers i can see in them... just 1,2 and 3...

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5. ### GlyphNew Member

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Proof that 0.99~ is not 1

Rewrite 0.99~ as (1-x)

(1) Therefore (1-x) =0.99~

(2) Therefore x = 0.00~1

(3) Assumption (to be proven later) x != 0

(4) Therefore, although x may be infinitely small is still not 0

(5) so if 0.99~ = 1 (as asserted by OP) then substituting into (1) the statement (1-x) =0.99~ becomes (1-x) = 1.

(6) Rearranging gives 1-1 = x

(7) therefore x = 0

(8) VIOLATION of original assumption (3) where X !=0

Proof by "Reductio ad absurdum" that 0.99~ does not equal 1

---------------------------------

Proof that assumption (3) in the above proof is itself correct, where x is not 0

(1) Axiom: 1^infinity = 1

(2) (1+0)^infinity = 1 since (1+0) = 1, therefore from (1) it must be 1

(3) base of natural logaritms "e" is give by (1+1/n)^n as n approaches infinity as given by http://en.wikipedia.org/wiki/E_(mathematical_constant)

(4) let x = 1/n

(5) as n approaches infinity, x will approach 0, therefore x = 0.000~1

(6) substitute (4) into (3) to give (1+x)^n = e since x =1/n and from (3) (1+1/n)^n = e

(7) if x = 0 then (1+x)^n = (1+0)^n = 1^n

(8) as n approaches infinity 1^n = 1 from statement (1)

(9) VIOLATION: if x = 0 then from (3) the base of natural logarithms must be e = 1 from (7) and (8) but this violates (3)

Proof that x =0.000~1 is not zero even though it is very small

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Therefore from the above two proofs it is shown that 0.99~ is NOT 1

if it were true then from the above proofs the base of natural logarithms should be 1.... you can check your calculator and spreadsheets to show that it is not 1.

If this proof doesn't satisfy you i honestly don't think any proof will.

Last edited: Apr 26, 2007

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I don't care how much of a math geek you are. If you can't come to the realization that .99~ is not 1 without prompting.... There's just no hope for you as a thinking human being

7. ### arodNew Member

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But you are wrong
http://en.wikipedia.org/wiki/.999

8. ### arodNew Member

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0.000~1 = 0
Reason:
1/(10^n) represents .000~1 as n approaches infinity

limit as n -> infinity(1/(10^n)) = 0

Therefore, .00~1 = 0.

These are fundamental laws to mathematics. This is not a debate.

9. ### GlyphNew Member

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Ok i think i'm seeing where the confusion lies.

there are certain unusual aspects in various number systems that appear as inconsistentcies when those systems are compared against each other. An obvious example is that 1/3 while being simple and discrete appears as the rather absurd 0.3333~ when put into the decimal based system. Likewise putting the same number into the base-2 number system gives even stranger constructs.

The thing is, they are all equivalent if properly defined.

The problem arises within the definition of a number system or number set. But if we're going to get into a discussion of the finer points of number system theory we're all going to need to explicitly define what types of number systems we're using and speak in the language of mathematicians. Concepts such as real analysis and the like need to be explored. Limits and the like need different approaches than other equations to be properly expressed.

I think the problem with this thread is one of communication, we're not on the same wavelength so to speak in discussing our numbers.

Realizing this i think i'm going to withdraw from further debate.

As a final note.

It is true: 0.99~ can equal 1. It is also true that 0.99~ does NOT equal 1

the statements are not contradictory, they just apply under different circumstances that are mathematically self-consistent.

10. ### ljcoxWell-Known Member

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11. ### ljcoxWell-Known Member

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Of course you can add 1/3 + 2/3. My problem is in adding or subrtacting 2 numbers that have an infinite number of decimal places.

The proof that convinced me was the sum of an infinite geometric series, ie. ar/(1-r)

For 0.9999~, a = 9 and r = 0.1. Thus the sum is = 1.

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It makes a bit more sense logically when I read
It just doesn't make sense to me intuativly, one of the reasons I don't like math I guess =>

13. ### RoffWell-Known Member

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One of you guys in the ".999... not equal to 1" gang should rewrite the Wikipedia entry. Thousands, yea, millions, of people are being misled by it.
You might want to read it first.

Last edited: Apr 30, 2007
14. ### 3v0Coop Build CoordinatorForum Supporter

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Perhaps they should just use BCD.

15. ### phalanxMember

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That's not a proof. That's math a 3rd grader would laugh at. When your first remainder is the same size or larger than your divisor it means you used the wrong value in the quotient. 9 goes into 9 one time, not zero.

16. ### tkbitsMember

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Which means you didn't notice the disturbing part.

The division pattern follows one of the patterns of valid repeating decimals. The remainder is the same at each step. In the more obvious invalid divisions, the remainder will grow at every step.

Last edited: May 4, 2007
17. ### ljcoxWell-Known Member

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I don't understand why it is a problem if the remainer is the same at each step since that is what happens with any division where the result is a recurring figure, eg. divide 1 by 3.

18. ### tkbitsMember

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It's not a problem for me.

The objection is that I didn't reduce the first significant digit fully, as is normally done. Does that make the division invalid? Or, because of the repeating remainders, is it a valid division?

19. ### Electronics4youMember

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1-0.9999999... will NEVER EVER be zero
That's wrong due to the laws of math
The right way to do it is to take the limit of the function
That's zero but only when the x-value is undefined

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20. ### RoffWell-Known Member

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I wrote this in a previous post:
Perhaps you would volunteer for this task?

21. ### 3iMaJNew Member

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For those of you too lazy to read the wikipedia article, here is your proof. Now lets stop arguing about something that is clearly TRUE.

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