What is 0.9999... as a fraction?

Question

i tried to work it out the way she taught us in class:
x=0.9999
10x=9.99999

10x=9.9999
-
x=0.99999
________________________
9x=9
x=9/9 <
i worked it out as 9 times 0.11111
to its 9*1/9 but i dont know if this is right

HOW CAN IT EQUAL 1? i dont understand- maths has got to be flawed. HOW can 0.999.. be 1? then it would 1? im confused. ouch my head hurts now.

HOW CAN IT EQUAL 1? i dont understand- maths has got to be flawed. HOW can 0.999.. be 1? then it would 1? im confused. ouch my head hurts now.

omg- tbls- the way you describe it- i understand- but i DONT understand how they're the same. omg im gonna have a meltdown HOW HOW HOW?

Answer 1

What I think would be helpful to point out is that 1 = 0.99999.... is not the only example. Any number with a finite decimal expansion also has an infinite decimal expansion. What I mean is

1.25 = 1.249999999...
0.58 = 0.5799999999999....
10 = 9.999999...

And so on. You can do this for any of these types of numbers, and you basically prove it in the same way as you did.

Answer 2

10x=9.9999
-
x=0.99999
________________________
9x=9
x=9/9 <
This is 100% correct. You've done it right. Don't doubt yourself! That's the easiest, smartest, and standard way to solve this problem and you worked it out on your own! : )

To provide further explanation, think of it like this:

Let x = .9999...

There are three possibilities:
x > 1
x = 1
x < 1

It's clearly not the case that x > 1.

If x < 1, you can find some number between x and 1 (the average of x and 1, for example). There is no number bigger than x but still less than 1. Such a number would have to "beat" x in one of the digits to be bigger than x, and you can't "beat" a 9. This means it can't be the case that x < 1.

The only other possibility must be true: x = 1.

Answer 3

Both of your ways of doing it would be fine, PROVIDED that you included those important symbols '...' following all of your repeated digits. That is, you need to end your repeated series of numbers as follows: '999...' and/or '111...' . That ending symbol of '...' literally means "these digits go on forever."

That is the only minor snag with how you wrote down your arguments. It was clear that THAT was what you were THINKING, it was just that you weren't WRITING IT DOWN explicitly. (But DO note: whatever anyone else says, flatteringly, WITHOUT that '...' tail to your repeated digits, your arguments would only be APPROXIMATIONS, while WITH it, they would have been RIGOROUSLY TRUE.)

Had you retained the '...' symbols, you would have shown by two different, solid arguments that the quantity 0.9999... was both

9/9 (via the "10 x - x" route),

AND

9*(1/9) (via the "9 x one-ninth" route).

Also, of course:

Both 9/9 and 9*(1/9) simplify to 1/1, that is, 1 !

Now the only problem you have is BELIEVING, or "GETTING" (that is, INTERNALLY "getting") your answer.

You have to learn to look at 0.9999... and say to yourself:

"That's not merely CLOSE to 1, it IS 1. And WHY? :

Because I PROVED IT!"

So, in summary, the problem lies not in your ability to DO what was required, in fact, but rather only in your LACK OF CONFIDENCE in yourself and your own arithmetic abilities. That lack of confidence made you rebel at the idea that you understood it and had obtained the right answer in two different ways!

[And I now see, from your later comments, that you still cannot accept the CORRECT IMPLICATIONS from your own arguments. You DO HAVE quite a mental barrier, of your own making, to overcome, there! I sympathise.]

I hope that you will take this comment, not negatively, but rather as encouragement to believe in yourself and your abilities.

Live long and prosper.

P.S. Just for completeness, there's another way of seeing that the answer must be ' 1.' It's very easy to say in words, but takes a little longer to write out mathematically.

In words: 0.9, 0.99, 0.999, 0.9999 etc. is a sequence of numbers getting increasingly closer to 1 (in fact reducing the difference from 1 by a factor of 10 in each step) but never able to exceed 1. So, in the end, it must in fact be 1.

Mathematically, one can write the argument as follows. (It's simply a way of putting that key idea, just presented, into a rigorous mathematical form. Frankly, the IDEA is more important than its mathematically rigorous expression of it.)

Consider the sequence a_n = 0.9, 0.99, 0.999, 0.9999, ... where n = 1, 2, 3, ... . (Here, '...' is a writer's or printer's way of saying "and so on." Notice that the COMMA and SPACE after that last 0.9999 number in the first sequence, or after the ' 3 ' in the second sequence [before the '...' symbol itself] means that the '...' "operator" doesn't JUST apply to the 0.9999 entry alone, or to the 3 alone, but rather to each of the entire preceding sequences that are being developed there. This distinction is worth noting. Who would have thought that a comma and an empty space could make such an eloquent and important distinction?)

O.K. now consider the sequence b_n = 1 - a_n for
n = 1, 2, 3, ... .

Then b_n = 0.1, 0.01, 0.001, 0.0001, ... .

[Comment: In a sense we have "tamed" what could be going on down the line in 0.9999... , by showing explicitly that every successive term in the new sequence b_n has one more '0' inserted in it between the decimal point and the final, finitely positioned '1'.]

It's now clear that as n --> ∞, b_n --> 0. [Since b_n
= 1/10^(n - 1).] Thus

a_n = 1 - b_n --> 1 - 0 = 1. QED

Answer 4

I assume that you are talking about 0.9999... as a repeating decimal. This is an interesting exercise for you because it is meant to show you that 0.9999... is exactly equal to 1. Many students have trouble with it because they don't understand how 0.9999... can be equal to 1; usually they think it is really close but still a little bit less.

What you did was do one of the mathematically proofs that 0.9999... = 1. It is one of the more interesting facts in mathematics, and you used math to prove it.

Answer 5

How many 9's are there? If it goes on forever, then the answer really is 1/1 = 2/2 = 3/3 ... 9/9 = 1.

If you have only four 9's, then
x = 0.9999
10x = 9.999
100x = 99.99
1000x = 999.9
10000x = 9999
x = 9999/10000

Answer 6

0.999... does NOT equal 1. Infinity is not a number, it is a concept. Mathematical operators ( + , - , x, / ) only applies for numbers and infinite is NOT a number.

The 10x = 9.9999... concept is flawed because it can't actually be represented no matter how long you choose to write it. By choosing to write it as 10x = 9.999... you are merely giving it a limit. x will only be 1 when u choose to end your number string which you can't because it is infinite

Answer 7

The answer is 1. There is no fraction if the 9's are endless.

Answer 8

This is correct. An infinite sequence of 0.999999... IS equal to 1.

Your proof is good, but here's a more fundamental proof. The real numbers are defined as a continuum. That means if you have two numbers X and Z that are actually different, there is always some number Y that's between them (X < Y < Z). Since there is clearly no number that's between 0.999999... and 1, they are therefore the same number.

Answer 9

Just as .1111111111 =1/9
.22222222 = 2/9
.33333333= 3/9
..
..
.888888888= 8/9
.999999999 = 9/9 =1
This is true if you think of the limit as the the number of digits becomes infinite.

Answer 10

0.9999 is (by definition) 9999 / 10000, as a fraction.

Any decimal can be expressed as a fraction. For example 3.33 is the same as 3 and 33 / 100. See how it works?