Is .9999 repeating an actual number?

Question

My math teacher says it isnt a number it just equals 1

Answer 1

your math teacher is WRONG

Answer 2

It is 1.

.111 repeating = 1/9

.222 repeating = 2/9

.333 repeating = 3/9

etc

.999 repeating = 9/9 = 1

Answer 3

it is actually the converging series 9/10 + 99/100 + 999/1000 + 9999/10000 ...........

it is actually the number 1 written in a different form

the easiest way to prove this is ....

what is 1/3 in decimal?  .3333333333333333333333333.....

what is .33333333333..... * 3?   .9999999999...........

what is (1/3) * 3?  1

therefore

.9999..... = 1

it is the same exact number as 1  

Answer 4

nearly equal to one but actual number

Answer 5

yes, .9999999999999999... = 1 exactly

not sure what your math teacher meant by saying it isn't a number. 1 is a number

ps. a few things

1. many here talk about how it "rounds" to 1. This has nothing to do with rounding.

2. we do not just "take it" as 1. It is 1. No matter if you're in the "math world" or the "real world"

3. There are a variety of real proofs that demonstrate this fact

Answer 6

It is a real number.

It is so close to 1, most the time we take is as 1.

But, not exactly 1.

Answer 7

It rounds up to 1 but is a real number indeed

Answer 8

let X = .9999 repeating

then 10 X = 9.9999 repeating

                    10X  = 9.9999.......    

                  -    X  =    .9999,,,,,,,,,,

                 ---------------------------------

                      9X = 9

                      

                       X=1

Answer 9

My math teacher says the same thing he showed me

.9999=x                         times by 10

9.9999=10x                    subtract the top by the bottom

9.999 - .9999 = 10x - x  simplify

9 = 9x  therefore

x=1

Answer 10

You are supposed to round up to 1 just because it's close enough.

How many .999999999999999's must you go through before you just call it "1" ? lol.

In reality .999 continuing is a real number, not equal to 1.  But close enough to 1 that in standard math classes they expect you to just round.

Answer 11

in math 0.9(bar) is always taken as 1

Answer 12

Between any TWO numbers, there is another number.

Assume that 1 and .999999 repeating are TWO numbers (with .999999.... less than 1)

Let X be the number between them.

If X is a terminating decimal, it must consist of only 9s (or it would be less than .99999999999.....  X is less than .999999999.... as they have the same start, and .99999999... has more places after.

If X is non-terminating, at least one of its places must be NOT 9 (or it would be .99999999.....).  If it isn't 9, it's < 9, so X is less than .9999999999....

So if X is between 1 and .999999999.... it must be less than .99999999... repeating.

This is a contradiction.

So there is no number between 1 and .99999999999....

1 and .999999999..... are not TWO numbers.

Therefore, 1 and .999999999..... "are" ONE number.

Pick one of the descriptions as the name for general usage :)

Answer 13

i actually have seen a pretty intriguing argument on the history or discovery channel once about how three thirds can be considered less than one b/c .333333333333333333333... + .3333333333333333333... + .333333333333333... = .999999999999...  and if you don't round it, then you don't get one.

it's pretty neat idea, and while .99999999999... is in the real world considered one, it could theoretically be a different number and represent a value that is slightly smaller than 1