Why is 0.999...9 equal to 1?

Question

They are two completely different numbers. If 0.999...9 is equal to 1 then accordingly 0.999...98 = 0.999...9 , 0.999...97 = 0.999...98 which is equal to 0.999....9 which in turn is equal to 1, thus 0.999...97 = 1. Eventually, we could get that 0.000...01 = 1 or even that all numbers equal to 1. Isn't that absurd? Mathematicians who say that 0.999...9 = 1 are not taking into account the infinitesimal difference those two numbers have (0.000....01) which in reality distinguishes each number as a unique mathematical entity.

I see. Then the same could be applied for example to 0.8(repeating), 0.7(repeating) or 1.1(repeating) equaling to 0.9 , 0.8 and 1.2 respectively. Isn't this a false distinction between decimal numbers that are followed by an infinite number of same numbers and those that are followed by an infinite number of different numbers. The infinite number of numbers could be visualised as a numerical spectre. Like the colour spectre each number is followed with a profound normality by the next number, but it retains its unique individuality. You cannot say that two numbers are equal to each other, because they are infinitely close to each other. This is such a clumsy use of infinity.

Well, I don't understand how it is possible to multiple an infinite number like 0.999... with 10 or any number since you don't really know what you are calculating. can we say 10 times infinity? No. Therefore how can we multiply an expression of infinity like 0.999...? Again the deduction 0.999... = 1 also implies that every number is equal to the other, because each number is infinitely close to the next number as well as to the previous number. We simply don't know infinity. Thank you all for your answers, anyway! : )

Answer 1

i suppose its called rounding..

Answer 2

To answer this, you need to agree with a few simple proofs and facts.

1.) ALL numbers are infinite decimal expansions.

Whenever you write "1" you are using this as a convenient shorthand for "1.0000..." in the same way that "1.0000..." is a convenient shorthand for a 1, a decimal point, and infinitely many zeros.

So, "13556" is short for "13556.0000...", "1/3" is short for "0.3333...", and "pi" is short for "3.14159265...". Shorter forms are merely useful notation because it's tiresome/impossible to write out infinitely many decimal digits whenever you want to write a number.

2.) For any two different real numbers, you can pick a third number which is between them. -This should be self explanatory-

So, if 0.9999... and 1.0000... were different numbers, then it would be possible to find a number which was between them.

But it's impossible to write out the decimal expansion of a number between 0.9999... and 1.0000... .

Therefore, they cannot be different numbers.

Therefore, they are the same number.

3.) If the difference between two numbers is zero, then they are equal. For example, 5 - 5 = 0 because 5 = 5. -Again, self evadent-

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000...
= 0

Therefore, they are equal.

To answer a few simple questions about this proof:

a} "At that first step, you're already assuming 1 = 0.9999...!"

No I'm not, I'm just doing a simple subtraction. Work it out yourself if you like.

b} "But 0.0000... should have a 1 at the end!"

No, it shouldn't. "0.0000...1" is meaningless. The "..." means the zeros go on forever. "Forever" means "without end". There IS no end for the final 1 to go on.

4.) If the assumption you start with is correct, and your lines of reasoning are correct, then the conclusion you arrive at, no matter how insane, must be correct too. -same as above, you should be able to understand and prove this yourself-

Let

x = 0.9999...

Multiply both sides by ten:

10x = 9.9999...

Subtract x from both sides:

10x - x = 9.9999... - 0.9999...
9x = 9.0000...

Divide by nine:

x = 1.0000...

5.)Most of the reason why people don't understand why point nine recurring is equal to one is because they don't fully understand what a decimal representation actually means. Take a look at the definition of 0.9999... and things become abundantly clear:

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

= 9·0.1 + 9·0.01 + 9·0.001 + 9·0.0001 + ...

= 9·10-1 + 9·10-2 + 9·10-3 + 9·10-4 + ...

n=∞
= Σ 9·10-n
n=1

n=N
:= lim Σ 9·10-n
N→∞ n=1

= lim ( 9·10-1 + 9·10-2 + ... + 9·10-N )
N→∞

= lim ( 9·0.1 + 9·0.01 + ... + 9·0.000...0001 )
N→∞ \________/
N digits

= lim ( 0.9 + 0.09 + ... + 0.000...0009 )
N→∞ \________/
N digits

= lim ( 0.999...999 )
N→∞ \_______/
N nines

= lim ( 1 - 0.000...0001 )
N→∞ \________/
N digits

= lim ( 1 - 10-N )
N→∞

= lim 1 - lim 10-N
N→∞ N→∞

= 1 - 0
= 1

If this is too hard for you, then look at the previous proofs.

6.) If your lines of reasoning are correct, but the conclusion you arrive at is definitely wrong, there must be something wrong with your assumptions.

Clearly

0.9999... ≤ 1.

Assume

0.9999... ≠ 1 (*).

Then

0.9999... < 1,

so there must be some positive number P so that

0.9999... + P = 1.

But for ANY positive P,

0.9999... + P > 1,

which is a contradiction, and definitely wrong. Therefore we are forced to conclude that the assumption (*) was incorrect, that is:

0.9999... = 1

7.) Rather simple, an elementry school student could do this with basic fraction skills.

1/3 = 0.3333...
1 = 3/3
= 3 * 1/3
= 3 * 0.3333...
= 0.9999...

Proof that 1/3 = 0.3333... is left to the reader.

8.)A sequence can only have one limit.

Observe that the limit of the sequence

0.9
0.99
0.999
0.9999
0.99999
...

is

0.9999...

That is, the sequence gets closer and closer to 0.9999..., in fact, infinitely close.

But the sequence also gets closer and closer to 1.0000..., in fact, infinitely close. So 1.0000... is a limit of this sequence too.

But a sequence can only have one limit, so 0.9999... and 1.0000... must be the same.

9.) Finally, There are NO proofs that 0.9999... and 1 are different numbers.

Anywhere.

Answer 3

0.9999....9 where the 9's end is not equal to 1. However 0.9999.... where the 9's NEVER end is equal to 1. That is the point that you have missed here. The 9's never end, so there is no way to change the last digit to an 8 or any other number; there just isn't a last digit. Thus, there is no way to find a number in between 0.9999... and 1, so they are the same number.

There are many other proofs of this fact available. If you want, I can show you a few.

edit: Steven F, mathematicians DO say that 0.9999.... is equal to 1, because they are equal. Like I said, there are many ways to prove this. Please note flunge_redouble's answer below. Note, he uses equal signs, never approximately equal or anything like that. They ARE equal.

edit again: You are very close. However 0.79999.... = 0.8, 0.89999... = 0.9, and so on. Again, the 9's go on forever. Any rational number with a finite length decimal expansian can be written with an infinite length decimal expansion.

Saying that 0.8888.... = 0.9 isn't true on many levels. There are numbers between these two, 0.89 for example. 0.888.... = 8/9, while 0.9 = 9/10, so obviously they are different.

Here, think about this one:

1/3 = 0.333333....
2/3 = 1/3 + 1/3 = 0.33333.... + 0.33333.... = 0.6666....
3/3 = 1/3 + 1/3 + 1/3 = 0.3333.... + 0.3333..... + 0.33333..... = 0.99999....., but of course 3/3 = 1. Wahlah, another proof!

edit again: Answer me this: what is 3*1/3? Of course it's 1, but you can also write it as 3*0.33333...... You are constantly multiplying fractions that have an infinite decimal expansion in real life. I don't see why this would be any different.

edit again: Saying that you are multiplying 10 times infinity isn't correct. 0.9999.... is a finite number.

You know how to find the fraction representation of repeating decimals, don't you? Try doing that with 0.99999....

0.9999.... = 0.9 + 0.09 + 0.009 + ....
= 9(0.1 + 0.01 + 0.001 + ...)
= 9(0.1^1 + 0.1^2 + 0.1^3 + ...)
= 9 * sum(i from 1 to infinity) 0.1^i
= 9 * {0.1/(1-0.1)} since |0.1| < 1 (For any r such that |r| < 1, then sum (i from 1 to infinity) r^i = r/(1-r).)
= 9 * 0.1/0.9
= 9 * 1/9
= 1.

Note that these were all equal signs. No tricks; just finding the fractional representation of a repeating decimal.

Answer 4

flunge_redouble has covered all the proofs I know. However, I would like to add a few points

1)
Saying 0.99999...= 1 implies 0.99999...998 = 0.999999...99 is false reasoning. Saying you have a series of 9's with an 8 on the end implies that the series of 9's is finite. 0.9999... is an infinite series, so the same reasoning cannot be applied so both series.

2)
Again, saying a series of 0's with a 1 on the end implies there is a finite number of 0's. 1 - 999..... would need an infinite series of 0's.

3)
0.8888888 clearly does not equal 0.9 as a number such as 0.89 between them can easily be found. However, the only "number" between 1 and 0.999... is 0.00000.....(an infinite series of 0's) with a 1 on the end, which is nonsensical.

4)
In response to your additional details
Have you ever multiplied 1/3 by a number? If so, then you've multiplied a number with an infinite decimal 0.33333........
In fact, 0.333333333....... * 3 = 0.9999999....... and 1/3 * 3 = 3/3 = 1. So 0.9999999......... = 1

Answer 5

.9999... equals 1 only if you have an infinite number of 9s. If you stop at any point, you have a number less than one

* * * * * *

In response to what you say in Additional Details:

Yeh, but I view infinity and the infinite series .9999... as two different animals. Infinity isn't a number, it's a concept. In contrast, the infinite series .9999... is definitely a number - it is the sum of the series 9/10 + 9/100 + 9/1000 + ... , etc. In fact, the sum of 1/10 + 1/100 + 1/1000 + ... is 1/9. So, 9 times that would be 9/10 + 9/100 + 9/1000 + ... = 9/9 = 1.

Answer 6

The difference between 0.9999... and 1 is infinitely small. In the normal way of looking at things, an infinitely small number is zero.

So 0.999... and 1 are just two different ways of writing down the same number.

Answer 7

0.99...9 is only approximately equal to one, as you approach its limit (calculus).. limits are what eneble you to find the area under exponential graphs and are the basis for calculus.. but they are only approximations (as it approaches a certian number.. i.e. the more decimal places.. the closer this number approaches 1. therefore the better the approximation.)

The same is not true for 0.999...98 because it does not have an infinite ammount of digits.. so the limit theory cannot be applied

Answer 8

They are not equal. The nines expansion approaches 1 as a limit, but does not equal it for any finite number of nines.

Answer 9

Tim D is the best answer so far. Mathematicians NEVER say 0.99999999...9999999... is equal to 1. They say the difference is not enough to be significant in the situation being considered. Your question implies multiplying the difference by a factor approaching infinity. At some point it becomes significant

Answer 10

What, did you just learn this.

I would rather do 9 x 1 then 9 x .99999999999999999999999999... anyday...

Now Mr. Infinitesimal Difference, nice word by the way, thanks for keeping us honest.