Does 0.9 (unlimited 9s) equals to 1?

Question

Is this theory correct:

0.3333=1/3
0.3333*3=1/3*3
0.9999=3/3=1
Hence, 0.9999=1

Answer 1

no not quite.

1/3 = 0.333....... without stopping

to within an error of 1/3 /10^n
1/3 = 0.3 .....3 (n digits)
so to within an error of 1/10^n
3/3 = 0.9.....9 (n digits)

but however many digits you take 3 * 0.3....3 will never quite get there.

You can amuse yourself by asking why, if you keep adding another place, which means adding 3*10^(-n) the number does not get infinitely large!.

Answer 2

This is a really good question. But since there are infinite decimals in 0.3333...1/3 is NOT equal to 0.3333...but is it's limit value. The point is that we can not write 1/3 EXACTLY in a decimal form, while we can do that with 1, so we don't need to write 1 as 0.9999....that would be an approximation.

Answer 3

Yes.
You could also use
1/9 = 0.11111111......
9*1/9 = 0.999999.....

I guess it would be possible to make it more sophisticated than this, but this is the basis of it.

Hustolem's answer seems to take the decimal as ending somewhere, hence the discussion of error, but as you can see I've assumed you really mean infinite repeating decimals. The most sophisticated treatment of these would still amount to multiplying the infinite series
0.111111..... (which certainly sums to 1/9)
by 9. So your argument is still correct. Just put some dots at the end of the decimal number, or put a repeater sign to show that you mean infinite recurring decimal.

Answer 4

There are so many theories that prove that 0.999999999999999999999999999999999999999999999999999999999999999999999 = 1

This is a very simple one.
Let x = 0.9999...------------(1)
10x = 9.99999...

10x - x = 9.99999... - 0.9999...
All the 9s after the decimal point cancel each other out to give:
9x = 9
x = 1 --------------(2)

From (1) and (2),
0.99999..... = 1

Answer 5

do note that 1/3 is equals to 0.3333..... (recurring, i.e. endless 3's)

so it is correct that
=> 0.3333... x 3 = 1/3 x 3
=> 0.9999... = 1

however, do note that 0.9999... (unlimited 9s) is equal to one only when there is unlimited 9s.

if it is 0.9999999999999999 (limited 9), it is NOT equal to 1; it TENDS TO 1.

=)

Answer 6

Use the geometric series calculation:
S= 0.999..
Multiply by 10
10S = 9.999...
Subtract the latter from the former:
10S-S = 9.99999... - 0.99999.. = 9
9S = 9
So S = 1

.9999.. can be written as .9(1 + 1/10 + 1/100..)

You can use this strategy for any repeating number
eg. say you have 0.346346346...
To cancel out the repeating .346, multiply by 1000
1000S = 346.346346...
S = 0.346346...
Subtracting, 999S = 346
S = 346/999
If the repeating group is 2 digits long, you would have multiplied by 100, so the answer would have been 2digits/99.

Answer 7

The problem with stating that the 3's or 9's go on indefinitely without end is that we lose sight of the original problem.

Does 0.999...(without end) = 1.0000000...(without end) is the same as asking:
Is 1 -- 0.999 (without end) = 0.000 (without end)?

At first glance, we believe this to be true as we observe the following pattern:

1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001
...
1- 0.999... (without end) = 0.000...(without end) ...1

One could argue that because the 1 at the end of this pattern is pushed off to the smallest of measure that the value of 0.000...(without end)...1 should be considered the same as 0.000...(without end) which is of course = zero.

But another person could argue that in math we use infinities all the time.

If I make an equation that says that
Y = X(1 -- 0.999...(without end)), what does Y equal as X grows very very very large?

At X = 1, Y is about zero...close enough to say it is zero.
But at X = 10 to power infinity = simplified to be infinity = number of digits of nines after the decimal, Y is appears to equal 1, but in actuality, Y at this point becomes undefined.

To prove this, we look at the following.
(1 -- 0.999... (without end)) = 0.000...(without end)...1
0.000...(without end)...1 can be rewritten as being the ratio of 1 / (10 x 10 x 10... without end) = 1 / (10^N) where N approaches infinity....this simplifies to be 1 / infinity

Therefore as X ----> infinity
Y -----> infinity / infinity
We cannot simply conclude that this value is one for two important reasons:
Infinity is not a real actual number, it is an idea, it just means an undefined very large value...
Y is also undefined because it is possible to rewrite the value of Y / Y as being = (zero x infin.)/(zero x infin.) = zero / zero.

Unless we get very theoretical here, it is safe to assume any value divided by zero is undefined or in computer terminology = error.

To end this discussion, simply put, 0.999... is not equal to 1.

0.999...approaches 1 from 1's lesser side at an ever slowing rate as the number of 9's grows.

For rounding purposes, we say that 0.999... = 1.

To see where this dilemma came from it is useful to take 1 and multiply it by 10 to gain 10.

dividing 3 into 10 gives us 3.3333333...(without end) on a calculator, but in REAL LIFE dividing 3 into 10 = 3
It is simple. There is a remainder of 1. So, 3 + 3 + 3 + 1 = 10
This remainder exists because 10 is not a multiple of 3.

In fact, no simple power of 10 is a multiple of 3. Not 1, not 0.1, not 10, not 10000, not 0.0000001

Now we finally see where the problem arises, so of course 1/3 will have a remainder (or go on without end in decimal), this is because 3 does not go into 1 x T where T is any power of 10.

Why is 10 so important? Well, this is only important because we're in decimal (base 10), this means each held place symbolizes a power of ten. So for example, we could rewrite the decimal number 110.011 as:
10^2 + 10 + 10^--2 + 10^--3

This is why irrational numbers or transcendental values have numbers go on forever to the right of the decimal point....Because no power of 10 is ever a product of any integer with that irrational number...or simply because there is no way to describe the number at the ratio of two decimal integers.

I could go on and on... blah blah, I suggest looking up the topic of bases or decimal or binary or hexadecimal or all of them and you'll start to get a feel for why this phenomenon exists in decimal.

So to review, to simply state 0.999... = 1 is WRONG,
What we should state is the limit of 0.999... = 1 (Correct).

Have a great day!

~Xzaerynus

Answer 8

Yes it is most definately one, real analysis will tell you that. It's rather annoying at times because it means there are two ways of expressing the same number.

I like this conceptual 'proof', for two numbers to be distinct there must be a number between them. Can you think of a number larger than 0.9~ and smaller than 1?

Answer 9

Yes the theory is correct, although you should indicate the 3s and 9s are recurring.

You can also see that it equals one this way:

1-1=0
1-0.9999.....=0.0000000..........
so 0.999......=1

Answer 10

Yes, according to the definition of the limit, for each ε>0 there exists a positive number x such that if the number of nines is greater than x , |1-0.999999...| < ε

Take for instance, x=-log(ε)