.9999999 Is Proven To Be Equal To 1!

Discussion in 'General Discussion' started by Aurori, Sep 12, 2007.

1. wefaWell-Known Member

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As soon as you tell me what .999 recurring is exactly without the number growing, i'll tell you what the x equals to in ".999~+x=1"
The number .999~ is not defined, it can only be expressed as a limit.

http://en.wikipedia.org/wiki/Rational_number
.99recurring is the ratio of what two integers? as far as i know 9/9 is 1

1xlink, please disprove this. I can't find any response to this.

There is a zero; an infinite amount of zeros is automatically given for any real number. 1.0000 recurring = 1
So 1.5968*10 = 15.96 = 15.960 = 15.96000 and so on
You don't write 1*1* (1/1)*x for x just as you do not add ten thousand zeros behind a number.

1: I'm sorry but this post is absurd. It's incorrect to assume that previous posts do not explain themselves, have you read any of xlinx's posts?
2: Apparently you did not even read the first post. The point of this argument is whether or not that .999 is equal to zero. Your statement is not baseless at best because what you have just said is equivalent to one's word against the other. ".999~ is less than zero" ".999 is equal to zero" How do we compare these statements?
3.Actually i can't think of anything for this because it is actually your only valid point. I would request that xlink refute this one seeing as how he quoted the next post immediately after.

2. xlinkGR's Tech Enthusiast

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it can be defined as a series quite easily in fact, it's more accurate to some degree to view it as a series or a process than an integer.
which works out to
.9 + .09 +.009 ... for infinity.

now if you took any low level algebra classes, you should hopefully recall that an infinite geometric sum can be solved. It will either be a number or it will be infinity. unless you meant with a finite number of decimals. well .9 with an infinite number of decimals has an infinite number of decimals by definition. The reason why I states such before was to demonstrate that the only number which works in that case is 0. Two numbers are equal if you can add 0 to one and get the other.

again .999recurring equaling one is an integral part of the fundamental theorem of calculus, or at th very least an extension of it. basically it works out such that some integral(LRAM) <=the given sequence<=another integral(RRAM). When it's worked out both integrals are 1 and by the squeeze theorem it's found that .999 does in fact equal 1. too lazy to work it out. Take college calculus yourself should be pretty easy, heck I even learned that in high school(had to repete calc as my college wouldn't take my AP score, good news though is that calc is an easy A)

note to self: buy wacom tablet...

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second part.
if 1 = 9/9
and .9999recurring = 1
then 9/9 = .9999recurring

and low and behold
1/9= .111recurring
9* .11111recurring = .99999recurring
therefore
9/9 = .9999recurring

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to hilsee -
it's only heavily debated on the intarweb by people aged 2-6. I'm apparently in this group as well since I took the time to post multiple proofs.
what is debated is whether or not the number system needs to be redefined to better represent infinitesimals. As it, decimals are in a sense another way to write a fraction. in the case of .999recurring this is just a byproduct of changing a numbers base. Every decimal can be rewritten as a fraction. In fact the higher you go in math, the more common fractions become and the rarer decimals become. Decimals are inaccurate. This includes the decimal form of pi and e

3. devourWell-Known Member

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The problem is.. you can't actually.. check your answer.

you're saying 9/9 = 1
and .999recurring = 1

multiply both by 9
low and behold you can an incorrect answer

.899recurring1 = 9

How about we just say our number system is flawed and you can really manipulate it some way to get a different answer?

Sounds good.

4. legenddurnWell-Known Member

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Mind blowing lol.

5. xlinkGR's Tech Enthusiast

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I'll agree that the number system is flawed in the sense that you cannot write an infinitely small number, or a number subtracted by it

I hope you mean
8.9999recurring and not .8999 recurring.

also 8.999recurring is the same as 8 + .9999recurring
and since .999recurring is 1, then 8 + 1 = 9

I assume you failed to read the following

wikipedia
Blizard's official response to the forum post.
the page in your algebra II text book explaining it.
the 50000 proofs I've posted.

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7. reanimation-06Member

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um, no this isn't proving .99 to the equal to 1...
with the steps you took, almost any number can be proved to be 1

8. Xero0Well-Known Member

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1/3 = 0.33333 repeated
2/3 = 0.666666 repeated
3/3 = 1 when it should be .9999999 repeated

9. xlinkGR's Tech Enthusiast

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once in 2nd year algebra
once in precalc
another time the next time I took calc
once in my first semester of college calc
and again in my 2nd semester of college calc.

no

.777777recurring is not equal to 1. It's equal to 7/9
.88888recurring is not equal to 1, it's 8/9
.333333recurring != 1, it's 3/9

.77777recurring
+ .333333recurring = 1.1111recurring = 10/9

.7777recurring
+.2222recurring = 9/9 = 1 = .9999recurring

again while it's tempting to think that .999recurring is a number which is just slightly off from 1, that's incorrect, as in the decimal system, there is no way of expressing a number which still exists but is indefinately small and as such, there is no way of representing a number with an indefinitely small number subtracted from it.

to put this simply, when you're dealing with infinity, normal logic might not always apply and often indirect approaches must be used when doing analysis.

EDIT: just used it to save my ass on my calc test today.

"use a reiman sum for the integral from 0 to 1 for 3x" HAHAHAA.
use one of the laws of integration to extract the 3 from the 3x, to get (3)(int 0 to 1 | x dx) the bull---- away with
(3) limit as n approaches infinity Sigma(n on top, i =1 on bottom) [9 *10^(-n)]/2

that my friends is some serious bull----ting since I didn't study at all for that section, neglected to learn how to sigma summations all to well back when I was 14 and well... .999 recurring equaling 1 and the BS I put up in this thread basically just saved my A test average.

10. Jimit87Well-Known Member

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I will varify this with a professor.

11. AG MountaineerWell-Known Member

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Alright I've now realized your x cannot equal 1 because your x variable already equals .99999 repeating, so the equations you've used cannot be relevant with a changing variable.

12. Soulja BoyWell-Known Member

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Learn about this a week ago, which I came to this topic earlier could have surprised my teachers !

Thanks for sharing the information here with everyone!

13. CasanovaWell-Known Member

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wow, so much complicated mumbo jumbo in this thread....

i like to think of it like this:

1/3 = .333 (repeating)

2/3 = .666 (repeating)

3/3 = .999 (repeating)....wait
3/3 = 1 right?

14. xlinkGR's Tech Enthusiast

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if .999recurring = 1

then that wouldn't matter now would it?

if 100 penies = 4 quarters
and 4 quarters = 1 dollar

then it's only natural for 100 pennies to be equal to 1 dollar.
in essence, the way a value is represented does not alter the value.

so
if x = .9999 recurring
and x = 1
then .999recurring would = 1

========================

another example.

supposed that x=1 - .9999recurring

then
x + .999recurring = 1

then one could say that
x + 9 + .999recurring = 1 + 9

one could also say that
10(x + .999recurring) = 10(1)

therefore you get
x + 9.999recurring = 10
and
10x + 9.999recurring = 10

you could set them to each other

10x + 9.999recurring = 10 = x +9.999recurring

so
10x + 9.999recurring = x +9.999recurring

then subtract .999 recurring
10x = x

for the above to be true, x would need to be zero

go back to the original equation
x + .999recurring = 1
substitute 0 in for x
0 + .999recurring = 1

.999recurring = 1

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now, in calculus an integral is defined to effectuate the area of a region.
The integral of f(x)= 2x from 0 to 1 = 1
if you draw that graph you will find it's a right triangle and it has a base of 1 and a height of 2 and an area of 1.

before integrals were used to find area, reman sums were used. basically the triangle or whatever graph it was was split into many equal segments on the x axis and a rectangle was formed and the area of these rectangles was added and the result was an approximation. The more rectangles used(hence the smaller the interval) the more accurate the result was.

if f(x)= 2x is split into four rectangles (RRAM method) you'de get
x<sub>0</sub>=0
x<sub>1</sub>=.25
x<sub>2</sub>=.5
x<sub>3</sub>=.75
and
f(x)<sub>0</sub>=0
f(x)<sub>1</sub>=.5
f(x)<sub>2</sub>=1
f(x)<sub>3</sub>=1.5

for rectangles and add them...

for two rectanges(forumula is width of rectange * height)
1/2(0) + 1/2(1) = 1/2

for four
1/4(0) + 1/4(.5) + 1/4(1) + 1/4(1.5) = 3/4

for eight
.... = 7/8

for 16
.... = 15/16

for 32
.... = 31/32

In time, it was discovered that if you took the antiderivitive of a function and used the two numbers by which the region was bounded, it was the same as taking an INFINITE(and therefore 100% accurate) number of rectangles

integral from 0 to of (2x) = F(2) - F(0)
=1

for infinity
.... = 99999999999999recurring / 99999999999999recurring = .99999recurring = 1

wish i could represent it better i need either a scanner or a wacom tablet.

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