Here's a good maths question - is 0.9 recurring equal to 1?

Question

0.99999999999999999999... = 1?
personally I'd say yes...

Answer 1

Yes, it is equal to one. The point is that you have to know what the strong of symbols
.999999....
actually means. It means the limit of the sequence
.9
.99
.999
.9999
.99999
...

The *limit* of this sequence is 1, so that means that
1=.99999.....

In the same way
pi=3.14159265358979.....
*means* the limit of the sequence
3
3.1
3.14
3.141
3.1415
...

and .333333....
means the limit of the sequence
.3
.33
.333
.3333
.33333
...
which is 1/3.

The upshot is that some numbers have more than one decimal expansion. In fact, any number with a finite decimal expansion (like .234) will also have an infinite decimal expansion (like .233999....).

Answer 2

Dan,

This question is itself infinitely recurring! But some of the answers you are getting are depressing and dire. I don't think that people should be answering technical questions unless they actually KNOW the answer.

The answer is YES, YES, YES. They are exactly the same number. One of the above gave an algebraic proof, which is good (based on multiplying by 10 not affecting the recurring bit). Someone else gave a really nice proof - the idea that between any two non-identical numbers there is at least one other number... and (in spite of what someone else wrote, there IS no number between 0.9rec and 1). Two other points to help you to understand:

1. If they are different, calculate the difference. ie what is 1 - 0.9rec ?

2. (this often helps!)
Consider the fraction 1/3. Its decimal form is 0.3rec
Now consider 2/3. Its decimal form is 0.6rec
Now consider 3/3. By adding the above (or by common sense), its decimal form is 0.9rec

But, of course, 3/3 is 1. So 0.9rec is 1.

IGNORE ANYONE WHO TELLS YOU OTHERWISE and ignore all b/s about "close to" or "not quite".

Hope this helps

Perspy

Answer 3

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 4

No. Someone who said 'yes' contradicted his own answer. He said something like 'What can you get between .999 and 1?' Well there are of course an infinite amount of numbers that come between these two numbers, one of which is .9999. You can, of course apply this rule again and again.

Another way of looking at this is to ask the question, "What is the difference between 0.9 recurring and 1?" The answer is 0.00000000000001 approx. Will this ever eventually equal 0? The answer is 'no' because however many zero's you put in 0.000000000001, you can always add another zero!

Answer 5

Oh not again.

Yes, it is 1. Not "infinitely close to", not "eventually reaches after an infinite number of digits", not "infinitesimally smaller than" 1.

Equal to 1. Indentical. The same thing as 1.

If you have doubts, think about this: Between two real (or rational for that matter) numbers there is always one in between. So if 0.999... is smaller than 1, how would you write a number that is between the two?

You will find that this number cannot be anything other than 0.999... as well. therefore, there is no number that can be inserted between 0.999... and 1. Ergo, they are the same number.

More here: http://www.wikipedia.com/wiki/0.999...

P.S.: Martyn, you don't nearly have enough 9's in there...
To Kit Fang (below), you said "no, because to be equal to 1 it would have to be one" ... Guess what! It is.

Brainyandy, you pretend I said 'What can you get between .999 and 1?'. No, no, no. I was talking about 0.999..., not 0.999. That makes all the difference in the world.

Answer 6

I think it depends on how accurate you wish to be!
The precise answer has got to be no, because no matter how many '9's you put in a line, you will never reach 1- you will always be able to put in another 9, and STILL not reach 1.
The practical answer is "yes". If you use numbers to quantify something (say speed or length or time) no human would be able to distinguish between a sufficient number of 9's in your sequence and the number 1- the difference between the two would be too small to recognise!

Answer 7

In mathematics, equal is defined as identical. If you are one foot away from a wall and halve the distance an unlimited number of times you will never (mathematically) reach the wall. At some point you will be there for all practical purposes because real walls have irregularities larger than the remaining distance. But points, lines and surfaces represented by thick pencil points and lines can actually be ideal in an imperfect world and your question was about math which is logically ideal.

Answer 8

Aha, this old chestnut again. Yes it is, in the same way that 4.9 recurring is equal to 5.
x=.9 recurring
10x=9.9 recurring
9x=9
9/9=1
therefore, x=1

I think it's got something to do with the fact that any number less than 9, when divided by 9, is that number recurring, i.e 2/9 = .2 recurring, but 9/9 = 1. I stopped thinking about it because it was doing my head in!

Answer 9

since the limit of infinity is in fact infinity, it can be said that 0.9* = 1 but it can be also said it isn't.

so teh answer is Yes and No.
until infinity can actually be judged and represented then as has been said b4 since 0.9* can not be divisable in respect to 1 it has to equal 1

only teh otehr hand true decimalists argue that 0.9* is never 1 because it never reaches it.

its the same as the limit of the university.

there are no limits or there are: since noone knows they assume there is.

Answer 10

No, maths isn't an absolute science and this is why we can't define infinity