Verified answer

1/3 = 0.333...

so 3 * 1/3 = 3 * 0.333...

but 3 * 1/3 = 1 and 3 * 0.333... = 0.999...

finally 1 = 0.999...

There are many proofs that 0.999… = 1. Before demonstrating this using algebraic methods, consider that two real numbers are identical if and only if their (absolute) difference is not equal to a positive (third) real number. Given any positive value, the difference between 1 and 0.999… is less than this value (which can be formally demonstrated using a closed interval defined by the above sequence and the triangle inequality). Thus the difference is 0 and the numbers are identical. This also explains why 0.333… = 1⁄3, etc.

There are a number of demonstrations, but most of them ignore or miss some crucial issues. The first issue is what the string of symbols .9999... means. The definition is that it is the limit of the sequence

.9

.99

.999

.9999

.99999

etc.

Let x_n be the n^th term of this sequence. Then x_n =1 -1/10^n as is easily seen.

Now, when talking about the limit of a sequence, you have to go to the epsilon, delta definition. SInce we know what the limit should be, set L=1. Then, we need to show that for every e>0, there is an N so that n>=N implies that |x_n -L|<e. But if e>0, choose N so that 1/10^N <e. Then if n>=N, we have

|x_n -L|=|(1-1/10^n)-1|

=1/10^n <=1/10^N <e.

This finishes the proof.

There is another proof that is convincing to many people:

Let

x=.9999..., then

10x=9.9999....

Subtracting

10x-x=9.999...-.99999=9,

so

9x=9, so x=1.

The problem with this proof is that it assumes that .999... vaidly defines a number. It does, but that has to be proved. As an example where this logic fails, consider:

Let x=....1111111.

Then

10x=......11110., so

x-10x=...11111.-...11110.=1,

so x=-1/9.

The problem is that the string of symbols

....11111. does not define a real number, so the algebra does not apply.

There are a couple of different ways to prove this. The first one I ever learned starts with letting N = 0.999... Then:

10N = 9.999...

10N - N = 9.999... - 0.999...

9N = 9

N = 1

You could also show that the series 9/10 + 9/100 + 9/1000 + 9/10,000 + ... goes to 1. Or similarly prove that there is no positive number small enough that you can add to 0.999... to get 1.

Rock solid!?

Well, let me show you a conclusive or irrefutable proof that 0.999... is indeed equal to one.

Let x = 0.999…

Now

x

= 0.9 + 0.0999…

= 0.9 + (1/10)(0.999…)

= 0.9 + (1/10)x

Rearrange to get

x – (1/10)x = 0.9 or x = 9/10.

This reduces to

(9/10)x = 9/10 or x = (9/10)(10/9) = 1.

THUS x = 0.999… = 1.

Done!

It depends what you mean, how 0.999... is defined, and the system you are using.

For example, the sum of 9x10^-i gets closer to one as i = 1 to n becomes greater, and the limit as n goes to infinity equals 1.

Suppose

x = .999...

10x = 10*.999...=9.999...

subtract the two equations

10x-x = 9.999...-.999...

9x = 9

x = 1

this is an infinite geometric progression

sum = 0.9 + .09 + .009 + .0009 + .00009 + . . . .

r = 0.09/0.90 = 0.1 . . . . or r = 0.009 / 0.09 = 0.10

a = 0.90. . . . . this is the first term

sum = a / (1 - r) = 0.90/(1 - .10) = 0.90/0.90 = 1

so the limit of 0.99999 . . . = 1

0.999...

=9/10 + 9/100 + 9/1000 .....

a1=0.9

q=0.1

s=a1/1-q

s=0.9/1-0.1

s=0.9/0.9=1

They are not exactly equal; 0.999... is less by 0.001... than 1. It's no use proving that they are exactly the same because 0.999 is 0.999 and 1 is 1. Just because you can conveniently round off to1 0.999 and everyone accepts it that they become exactly the same.