Who can prove that 0.9999999999...(0.9 recurring) is equal to 1?

Question

i can im wondering how many other people can?
prove that you can prove it!

well done jsoos!

dont we all love maths!!!

keith j - if i included my own proof, how would i know whether people knew how before or had copied my proof!
but since many people have already answered here is my proof:

lets say:
x=0.9999999999999...
multiply by 10
10x=9.99999999999999999...
if we minus the first from the second we get:
9x=9
if we divide this by 9
x=1

weird but it seems to work!

Answer 1

1/9 = .111....
2/9 = .222...
9/9 = .999...
1=.999...

OR

x=.999...
10x=9.999...
10x-x=9.999... - .999...
9x=9
x=1

Answer 2

The answer to your question is generally yes, 0.9 repeating = 1. Only in very specialized (or specialised, for your UK friend) situations involving infinitessimal arithmetic would there be a different answer. Infinitessimals can be useful, but they would not be taken as the normal mode of operation. We define 0.9 repeating to be the limit as N goes to infinity of 0.9999...9 (string of N 9's), which is a finite length decimal equal to (without controversy) 1 - 1 / 10^N. By the basic limit theorem, if lim N-> infinity f(N) exists and is finite and lim N->infinity g(N) exists and is finite, then lim N->infinity ( f(N) - g(N) ) is simply the difference. This is pedantry, sorry, but the objectors demand it. In our case, f(N) = 1, the constant, and the limit is obviously 1. Our g(N) = 1/ 10^N. lim N->infiinity 1/10^N = 0, because, for any epsilon greater than zero, (chosen by an adversary), I can produce an A, depending on epsilon, such that for any N > A, the inequality 1/10^N < epsilon holds. It is left to the reader to find one of many schemes for choosing A that works. I will now cease writing N->infinity with my lim's, because they are all the same but you should imagine that it is written. This is the definition of the limit, and the result is a static quantity. We may be sloppy and say, "the limit approaches such a value, " but we mean, "the expression next to the limit symbol approaches such a value as the variable underneath does what the arrow says it does." So, 1/10^N approaches 0 as N approaches infinity, AND lim 1/10^N is equal to 0, full stop. Note that it is possible to tinker with the definition of limit, and get a different number system involving infinitessimals. Then one has to resolve many consequences for calculus, analysis, and topology. As mentioned initially, for specialized pursuits, this may be the way to go, but as an objection to the question at hand, it is sophistry. Applying the basic limit theorem, .9 repeating = lim 1 - /10^N = lim 1 - lim 1/10^N = 1 - 0 = 1. QED

Answer 3

There is also a proof in terms of an infinite series.

0.999(repeating) is the same as

9/10 + 9/100 + 9/1000 + 9/10000 + .....

The sum of an infinite series is a/(1 - r), where a is the first term of the series, and r is the ratio of the corresponding geometric sequence. In this case, the sequence is
(9/10, 9/100, 9/1000, .... )
The first term is a = 9/10.
The ratio is any term divided by its previous term (it is 1/10). That makes the sum

S = (9/10)/(1 - 1/10)
S = (9/10)/(9/10)
S = 1

Therefore,
0.9999(repeating) = 1

Answer 4

Here's something different. Proof by contradiction:

Assume that 0.9999.... < 1. Then there is a number x between the two, that is

0.99999... < x < 1.

What is the decimal representation of x? The unit digit must be 0 (otherwise it ain't smaller than 1).

The first decimal must be 9 (otherwise it ain't bigger than 0.999...)

The second one must also be a 9.

Ditto the third one. Etc.

So all the decimals must be 9. In other words,

x = 0.99999...

which contradicts our assumption. Ergo, we reject the hypothesis that 0.9999... < 1.

Answer 5

IT IS SO NOT 1
FIRST OF ALL YOUR YOUR n=0.999... is NOT 1. WHAT ON EARTH
10 times 0.999... is NOT 9.999
It's 9.999...0, due to the Rule of Infinity!!!! I COMPLETELY agree with the GENIUS that thinks it is less than 1.
9.9999...0 the zero is NOT annexed
minus 0.9999...9 as the ROI makes all the repeating digits go one longer- is 8.99999....1, NOT 9! Who came up with that idea? It's sour and BUSTED! Go to themathpro.wordpress.com! ~=[,,_,,]:3

Answer 6

1> x=0.99999
10x= 9.99999

10x-x= 9.999-0.999=
9x = 9
x=9/9
9/9=1

2> People say that 0.9999 is 0.0001 short of 1. However if the numbers are recurring and therefore go on forever there is no "one" at the end because there is no end and its just 0.0000... recurring all the way. therefore 1-0.999.... = 0, which means 1=0.999...

people get confused with the idea that 0.999 equals one- it DOESNT, 0.999... (recurring) equals one as there is no "one" at the end of the zeros, as there IS no end.

if that makes sense?

Answer 7

There are a few ways of looking at it:

1) By definition, if x does not equal y, there exists a z that is between x and y.

Therefore, if .999999 (continued on forever) does not = 1, there must be a number that is between them -- of course, since .9999 continues to infinity, there does not exist a number that is between the two, thus .99999...=1.

2) (1/3) = .3333...
+ (2/3) = .6666...
= (3/3) = .9999...
= 1 = .9999...

Answer 8

It isn't. 1 minus 0.9999999999999 recurring will always be 0.000000000001, with an infinite amount of zeros before the 1 and after the decimal point. Therefore there is a difference between the 2 numbers.

Answer 9

Jsos is incorrect. It cannot be proven since regardless of how many 9s follow the decimal point, it will never reach one.

1/9 = 0.111111111r
2/9 = 0.222222222r
3/9 = 0.333333333r
4/9 = 0.444444444r
5/9 = 0.555555555r
6/9 = 0.666666666r
7/9 = 0.777777777r
8/9 = 0.888888888r
9/9 = 1

At most it will only ever be a rounding up exercise.

And you didn't include your own proof...

Answer 10

0.999999.../ 3 = 0.333333...

0.333333... = 1/3 no one will argue with that.

3 * 1/3 = 1

Not a true 'proof', but a convincing explanation.
.

Answer 11

x = .99999...
100x = 99.99999...
so 100 x - x = 99
99x = 99, so x = 1