# .9999999 Is Proven To Be Equal To 1!

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

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where do you get 5/5? I really do fail to see your logic in that.

and FYI look at my satrical pos post. If it repeats forever then it shouldn't matter because it technically 'ends' at the same place.

I can see that .55555 = 5/9 but...

I mean 4/9 = .444444444444444 recurring
5/9 = .555555555555555 recurring
9/9 = .999999999999999 recurring = 1

2. ### SordWell-Known Member

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ok, i see wat ur saying... but how does 9/9 equal .99?

3. ### JDonta31New Member

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x = .9 (repeating not just .9)
so 10x = 9.9 (repeating)

Subtract the value of x which is.9 (repeating)

That leaves you with 9 : (9.9 repeating - .9 repeating)

so 9x = 9

which reduces to x=1

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well

1/9 = .111111repeeting

so 9*.11111repeating = .9999999repeeting

therefore... .999999 = 9/9 = 1

5. ### devourWell-Known Member

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It's pretty simple..

.999999[repeating] is an asymptote. it will approach one forever and ever but not actually reach one.
it's like you can take half of something over and over and not ever get to 0, just really close.

when you start manipulating the numbers you change them. and you can't even run a proof on it. you do all the stuff and get 9x = 9.
however, 9 * .9999[repeating] does not = 9.

It doesn't work.

6. ### NavieroGR Fan

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When you have .9 (With the nine repeating), what can be any closer to the value of one? NOTHING. (The nine keeps going, so if you stick a five in there, then the nine will be larger, thus ufail.)

It's pretty interesting, and fairly cool.

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if that were true there'de be a number x such that

.999recurring + X = 1

care to tell me what that number is?

try this

.9 + .1 = 1
or (1- 10^-1) + 10^-1=1

.99 + .01 =1
or(1- 10^-2) + 10^-2=1

so then we can say that...
(1- 10^-x) + 10^-x=1

limit as x goes to infinity (.99 + .01) =limit as x goes to infinity (1)
limit as x goes to infinity ((1- 10^-x) + 10^-x) =limit as x goes to infinity (1)

this can be simplified to
limit as x goes to infinity (1- 10^-x) + limit as x goes to infinity (10^-x) =limit as x goes to infinity (1)

this simplifies to
.9999recurring + 0 = 1

CALCULUS FTW. not even hard calculus at that either. AS IN YOU LEARN THIS WITHIN 1 WEEK OF CLASS, AS IN YOU PROBABLY LEARNED THIS IN PRECALC AS IN I'VE HAD IT ON TESTS. AS IN IT'S A FUNDAMENTAL PART OF THE NUMBERING SYSTEM, AS IN THERE IS MORE THAN ONE WAY TO WRITE A NUMBER, AS IN 1/7(CONVERTED TO DECIMAL) + 6/7(CONVERTED TO DECIMAL) = .99999recurring = 1, AS IN 1/13+ 12/13 = .99recurring = 1

8. ### BTKWell-Known Member

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.999999999 does not = 1

it almost does

even if it goes forever it won't be 1

1=1

and if xlink can post all that confusing crap i can say

2+2=fish

9. ### devourWell-Known Member

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xlink, I would expect more mature actions from a moderator. Huge text doesn't make you cool, it doesn't get your point across, and it just makes you seem dim witted. Do you really need text that big? That's next to flaming..

When you rewrite a number you change the number. There is no way around it. .9 repeating is not 1. It just gets infinitely close to one. And no

isn't true. The reason being if the 9 is recurring you'd have to cut it off [or limit] it to actually achieve a value to add it to give you a different value. Your use of limits renders your argument completely circumstantial. Once you incorporate limits into the equation you.. lose.

you're basically doing the same thing as multiplying by one. look at it, you've got 1-x [x being 10^x] + x = 1.

Well duh. 1 - 2 + 2 = 1. So on and so forth.

Your limits.. eventually they get rounded. 10^- infinity doesn't actually exist. It's a concept. Not a fact.

10. ### blahablaheekWell-Known Member

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I still don't understand how people with hard facts (as in MATHEMATICIANS) show that .999... = 1. It just does. Just get it through your head. It's not that hard to figure out. The world's just going to end in 2012 anyways.

11. ### Bl00dFoxWell-Known Member

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LOL if this turns into a civil war

12. ### Charlie SWell-Known Member

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Woah, way to blow my mind Posts:
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isn't true. The reason being if the 9 is recurring you'd have to cut it off [or limit] it to actually achieve a value to add it to give you a different value. Your use of limits renders your argument completely circumstantial. Once you incorporate limits into the equation you.. lose.

you're basically doing the same thing as multiplying by one. look at it, you've got 1-x [x being 10^x] + x = 1.

Well duh. 1 - 2 + 2 = 1. So on and so forth.

Your limits.. eventually they get rounded. 10^- infinity doesn't actually exist. It's a concept. Not a fact.
[/b][/quote]
made the font smaller to make you happy. Sorry i was a bit of a dick there.

again stating "y + x = 1" is a legitimate algebraic argument. substituting .9recurring is within reason. you then subtract .9recurring from both sides to get x. There is no need for a limit in this example as .9recurring for one is a rational number and two it's not a limit problem. The limit example is effectively a calculus based proof made on a tangent and is derived form the fact that 10 ^ -∞ technically has no mathematical value

Also, .99recurring is more of a process than a number per se. It's effectively: add .9 + .09 +.009... you can put it into sigma notation. There's a formula too and if you put it in and do the algebra you'll get 1.

here is another way to think of it. Take a paper, cut away 9/10th s of it. Then cut away 9/10ths of the remainder. and so on. Place the cut away part in it's own pile. after an infinity number of cuts, the pile with the cuttings will ahve one whole paper and the other will have nothing. It's still a bit abstract, but it's the best visualization that I could think of.

basically there is no real number(or imaginary), X, for which it .9recurring + X = 1. In standard mathematics, everything is dealt with in terms of real numbers and in some cases, imaginary numbers. I'll be blunt i'm not in advanced enough mathetmatics to deal with other number systems outside of real numbers and imaginary numbers and in the real and imaginary sets, it DOES equal 1.

good read if you have the time.
http://en.wikipedia.org/wiki/0.999...

I guess I'm a bit anal-retentive on the matter. I'm taking 3 semesters of college calculus and I get annoyed when i get in math wars and after several explanations someone insists for one reason or another. It's akey to a 3rd grader arguing about the integral of tan(x) from -pi to pi with someone with a doctorate. I'll be blunt, on that I have a hard time seeing why it's undefined and not 0, it would make perfrect sense for it to be 0, but I accept it as undefined.

14. ### CurseSealZeroWell-Known Member

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This won't work, ill tell you why:

0.999¯ will round to 1, but therefore, it rounds to 1, but it doesnt equal 1. Now, there is a possible way if you did like 0.999 x 10 = 9.99, and divide it by itself and then you have 1. But really, it DOESNT EQUAL 1, THE GUY WHO INVENTED MATH WOULD TELL YOU! He would think your stupid, he invented 0.999¯ and if he was alive, he would say that it isnt 1.

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who's the man who invented math? ^^

I could understand you saying that it's a flaw in our number system, but as it is in standard mathematical use, it is. by definition if two numbers have a difference of 0 between them, then they are considered the same number.

1 - .99999recurring = 0

I could see someone saying

1 - .99999recurring = 0.00...1

but there is no such number. having an infinity number of zeros followed by a 1 is not definable as if it had an infinite number of zeros, then it would never reach that 1.
again call it a flaw in our number system, but as it is that's how numbers are defined. .9999recurring is just another way of writing one.

and if you don't believe me then

http://www.blizzard.com/press/040401.shtml

argue with he maker of your video games.

in the end simple arithmetic solves it too   you can do that for any number and you'll end up with the same scenario.