.99999 repeating = 1?

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I was reading a post on ar15.com, and thought it was a great discussion, ended up being 11 pages of math. so here is the question?

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9X

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

So is this an example of limits?

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I was reading a post on ar15.com' date=' and thought it was a great discussion, ended up being 11 pages of math. so here is the question?

X=.99999999999 repeating

10X = 9.9999999999 repeating

10X - X = 9X

9X = 9

X = 1

Therefore

.9999999999 repeating = 1

So is this an example of limits?[/quote']

I think this is an example of rounding isn't it?

9x = 9? 2x = 2?

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10x = 9.9(bar)

1x = 0.9(bar)

10x - 1x = 9.9(bar) - 0.9(bar) = (9 + 0.9(bar)) - 0.9(bar) = 9 + (0.9(bar) - 0.9(bar)) = 9

so

1/9 = 1.99999999

2/9 = 2.99999999

what would

9/9 = 1?

2 differ logics, 2 differ answers.

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10X - X = 9X

9X = 9

At a quick glance, if I'm not mistaken, that step should be X(10-1)=9X or 9=9. Nothing too profound.

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Oh come on guys.... that equation is used to convert recurring decimals into fractions....

and it is right that 9.9 ('bar') = 1

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But in the first post, the author raised the very good question about limits. Yes this is because of limits, and limits are how the real numbers are defined (or at least, that is one of the equivalent definitions - being able to take limits in a well behaved sense is the defining characteristic of the reals), and in the decimal representation of real numbers, .9 recurring and one are different representations of the same real number, jsut like 1/2 and 2/4 are different rational representations of the same rational number.

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The problem comes from the way the repeating is used and the assumptions made.

I know a few people detest the use of infinity but sorry, its the only way to explain it.

Imagine the number 0.99999999.... The rule of multiplying by 10 is to shift the decimal. 9.99999999..... But lets pretend infinity is a finite number, after multiplying by 10, the number at the right end of 9.9999999... is unknown.

e.g. 0.999999 .... 999| (where | denotes the end of the number)

10 x 0.999999 .... 999| = 09.99999 .... 99x|

x is actually unknown or undefined as it comes from outside the original number, we just take it as 9 because of the repeating factor.

if we then introduce this into the 10X-X = 9X formula we'd get:

09.99999 .... 99x|

- 0.999999 ....999|

=09.00000 .... (x-9)|

therefore 9X = 9 .... (x-9)| => X = 1 .... (x-9)/9|

Because the difference is at the "end" of the repeating number, it is insignificant in size (infact by assuming x is 9, it equates to zero) but none the less shows that .99999 does not equal 1.

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Please, for the love of mathematics, can I ask you never to use that explanation again?

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So if ive got £10,999,999,999,ive really got £11,000,000,000.Somehow it seems illogical 0.999=1,

We know from primary school that another integer is required.

Why all the maths to 0.999=1

When its 0.999+0.001=1

I believe that using the equations in this topic as is,will not yield a correct formula for calculating distance in regards to our universe.Math is fine as it is without leaving out integers.

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0.99999999... = x

Therefore

9.99999999... = 10x

Subtract them and preserve equality!

(10x) - (x) = (9.99999999...) - (0.999999999...)

Thus,

9x = 9

x = 1

and 1 = 0.99999999...

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no x=1 because x can be made to equal anything you want.

All your doing is giving a value for x,then reversing it to represent an integer.

99p=x .............................................. £1=x

£9.99p=10x ............................................£10=10x

(10x) -(x)=(£9.99)-(99p) ..........................(10x)-(x)=£10-£1

thus ................................................... thus

9x=9 ....................................................9x=9

x=1 ....................................................... x=1

and 1 =99p ..............................................and 1=£1

?? We arrive at the same integer for the value of x , but 99p isnt £1 is it ??

I realise that the calculation your all going on about is mathematically correct,but i only see conflict,( 0.99+0.01=1.......0.99=1)this doesnt make sense when we want to apply this to finding extreme distance of objects in the universe,the final answer will be flawed.

Or am i just not getting this!!

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99p times 10 certainly isn´t 999p

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Or am i just not getting this!!

You're not getting this. You're doing arithemetic with finitely long strings of 9's and drawing a conclusion about an infinite string that doesn't necessarily hold (and indeed doesn't) since they are *different*.

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Thanks matt,

yes athiest i know 99px10 is £9.90p:-) It was a quick attempt at giving an example of the value of x being = to whatever integer we want it to be.

I meant to say 99.99999......p

Or am i still wrong,in which case i take my leave of this thread.Dignity still intact,and bid you fairwell.

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there are 107 posts here:

which were "Ending the 0.999~ = 1 debates"

It's one of those things where mathematically 0.9 recurring does equal 1, at the same time how can one number equal another different number?

I'd say 0.9 recurring = 1 is correct (see maths proof in about 5 other posts in this thread and loads more in the other thread i gave a link to!)

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"How can one number equal another different number"? It can't if they are *different*. But, again, like so many people, you're confusing a *real* number with its representation.

Why can people accept that 1/2 and 2/4 are the same number but cannot do that for 0.99... and 1?

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for crying out loud people....

Oh come on guys.... that equation is used to convert recurring decimals into fractions....

say for example, you want to convert 0.16161616.....161616 into a fraction

let 0.16 be x

let 16.16 be 100x___(this is valid because the 16 decimals areinfinite, no matter how long it is....

so, 100x = 16.16161616.....1616

minus, x = 0.16161616......1616

99x = 16

x=16/99

there you go. proving that 0.9999...999 = 1 goes the same way any other recurirng decimals undergo to beconverted to fractions. ...(What i've written earlier before i replaced this is just some miscalculation...please ignore it...) anyway......recurring 9's are a special case, because what i've seen so far there's no fraction to equal that one....so that makes it a special case...

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Matt if you can come up with a better explanation, be my guest.

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It is just a technicality problem that 0.999...99 = 1' date=' because 1/9 = 0.999999...999. So there we go, deviding the contraversial "1" into 9 makes it equal to the contraversial 0.9999...999.[/quote']

1/9=0.111111...111 ..... did you just mistype it or am I not understanding what you're saying?

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My favorite game company has taken time out of its development schedule to solve this mathematical conundrum once and for all:

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

Is there anything Blizzard can't do?

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oops... sorry, i should've not written that one... srry about that

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Matt if you can come up with a better explanation, be my guest.

I have written many explanations of the proof of this simple fact, some even in this thread I beleive, certainly in another one linked to from here.

Understand the terms you're using.

0.99.. is the limit of the partial sums of the series

0.9 + 0.09+0.009+...

that is it is the limit of the cauchy sequence of rationals x_n = 0.9...9 with n nines.

Let y_n be the constant cauchy sequence y_n=1.

Then |x_n-y_n| = 10^{-n} hence in the reals, which is the completion of the rationals wrt euclidean distance, the limits are equivalent.

OK? If you think the proof is "too hard" sorry, but that's what the proof is. All the other arguments provided are illustrative at best, but at least they don't introduce spurious nonsensical things such as your "proof", such as "supposing infinity is a number", and when we multiply by ten we get something in that place holder "infinity". Plus you also state that 1 and 0.99... are different (presumably as real numbers, though you never mention anything about the set in which the argument takes place).

There is no need to introduce infnity at all. (Even analysis can be done entirely without using the word infinity. Limits are things defined in terms of finite objects that get arbitrarily "large")

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can it be done in this way

0.999999999999999.............. = 0.1111111111111111..... x 9

= 1/9 x 9

~1

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can it be done in this way

0.999999999999999.............. = 0.1111111111111111..... x 9

= 1/9 x 9

~1

you can't do that because .1111111111111111infinity isnt 1/9 it approaches it, but it never is actually 1/9

The proof with the x=.99999

10x=9.999

I dont believe that you can multiply something with infinite numbers by ten, because it would almost be like adding a number, a number would just appear....for some reason it just seems illogical.