Is .999 repeating the same as 1?

Question

Had a pretty heated debate about this today lol. I don't think it is, but the people I was arguing with say that it equals 1 because 2/3=.666 repeating and 1/3=.333 repeating and 2/3+1/3=1. So, what do you think?

Answer 1

In mathematics, the recurring decimal 0.999… , which is also written as or , denotes a real number. Notably, this number is equal to 1. In other words, "0.999…" represents the same number as the symbol "1". Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.

Please do visit the link below, it would definitely help!

Answer 2

I had doubt about the equality 0.99999999......=1
then I thought about convergence of geometric series
but something was still missing, the idea of closure under limit.
I mean does the set of decimal contain the limit point of any sequence of its elements? The answer must be yes,and it can be proven from construction point of view, if we do not want to run into contradiction that would create issues about adding or subtracting decimals.
The simple proof is
0.999999.... = Sigma [ ( 1/ 10) ^ k] with k going to positive infinity FROM 1. Here Sigma stands for " SUM OF "

Still I had doubt ,then I researched, and found on Wikipedia
a broader view of real numbers and HyperReal numbers and also the proof this type of questions can be answered in any integer base or non integer base.
Enjoy the reading @
http://en.wikipedia.org/wiki/0.999...

If it is too much reading, just read from the beginning to INFINITE SERIES AND SEQUENCES, the jump to

4.1 Impossibility of unique representation

Answer 3

i've got responded this question many situations right here, and Steven_S does it approximately to boot as I even have seen it achieved. you have 2 possibilities: you could the two settle for that .999... represents a actual variety or you do no longer. in case you do no longer settle for that it represents a actual variety then you are achieved. forget approximately approximately calculus on account which you will no longer comprehend it. in case you do settle for that it represents a variety, then that variety must be a million. in case you're saying that it is the rest, it in simple terms screws up your arithmetic. .999.. is an endless sum .9 + .09 + .009 +... you need to use the guidelines for comparing endless sums once you assign a value to .999... and those rules say that it particularly is a million.

Answer 4

As numerous people have already answered, the infinite decimal fraction 0.999… does indeed equal 1. Strictly speaking, it is an infinite series that converges to 1 as a limit.

It is correct to “sum” the series by the operation

10 * 0.999… = 9.999… = 9 + 0.999…

Subtracting 0.999… from both sides,

9 * 0.999… = 9

Dividing by 9,

0.999… = 1

However, these are just formal manipulations that beg the question since they do not prove that the expression 0.999… represents any number at all, let alone 1. It is first of all necessarty to examine whether any such geometric series as

S = a + ar + ar^2 + ar^3 + … + ar^n + …

converges at all.

It can be proved that if abs ( r ) < 1 then S converges to the sum a/(1-r); but if abs(r) is greater than or equal to 1, the series diverges unless a = 0. If a=0 the series converges to the sum 0.

In this case a= 0.9 and r = 1/10.

that is,

S = 0.9 + 0.09 + 0.009 + … etc.

Incidentally, someone attempted to prove by contradiction that 0.999… is not greater than one. This is only more question begging and proves nothing, since at that stage there is no proof the expression 0.999… even is a number.

Answer 5

No you are wrong
1/3 is not equal to .333.... because if you multiply.333... by 3 you do not get 1 as the answer.
simillarly, 2/3 is not equal to .666....
so, .999 is not the same as 1

Answer 6

A recurring or repeating decimal is a number which when expressed as a decimal has a set of "final" digits which repeat an infinite number of times. For instance 1/3 = 0.3333333... (spoken as "0.3 recurring").
More fully: a recurring decimal is a rational number whose expression in the (decimal numeral system) has some point after which the same sequence of digits repeats infinitely-many times. The repetition may begin before, at, or after the decimal point and the repeating sequence may consist of just one digit or of any finite number of digits.

for your case
In mathematics, the recurring decimal 0.999… , which is also written as 0.9' or 0.(9), denotes a real number. Notably, this number is equal to 1. In other words, "0.999…" represents the same number as the symbol "1". Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.

I hope this helps

Answer 7

No .999 is below 1.It is somewhat like the difference between 999 and 1000.

Answer 8

well 2/3+1/3 does = 1
but .666+.333 doesnt = 1 but .999

Let x = .999... (repeating)
10x = 9.99...
10x - x = (9.99...) - (.99...)
x(10-1) = 9.00...
9x = 9
x = 9/9 = 1

Answer 9

No they aren't 2/3 can be rounded off to .667 so it certainly adds to 1.

Answer 10

This is an issue of form. Look at
.5 = 1/2
Same value different form.

.99999999.... Is certainly not the some form as 1 but it is equal to 1. Not approximately 1 but 1 exactly. Here is the proof

x = .999999999...
10x = 9.999999999...

subtracting

9x = 9
x = 1