Is this right? .99 repeating equals 1?

Question

1/3= .33 repeating
(3)1/3= .33 repeating (3)
1=.99 repeating ?

Answer 1

Technically yes, in that 1/3 equals .3 repeating, 2/3 equals .6 repeating, and that 3/3 would thus be equal to .9 repeating with 3/3 equal to 1. A mathematician would say however that .9 repeating is infintesimally close to 1.

Answer 2

Here's another way to look at it- 1 divided by 3 equals .3 repeating. If you were to multiply .3 repeating by 3, shouldn't that equal 1, since all you're doing is reversing the equation? That tells me that .9 repeating actually does equal 1. Not in the pure sense, but the logic in the solution says it does.

Answer 3

Yes. Here's a partial FAQ:

Q: Are you positive that 0.999... equals 1 exactly, not approximately?
A: In the set of real numbers, yes. This is covered in the article. If you still have doubts, you can discuss it at Talk:0.999.../Arguments. However, please note that original research may not ever be added to a Wikipedia article, and original arguments and research in the talk pages will not change the content of the article—only reputable secondary and tertiary sources can do so.

Q: 0.9 < 1, 0.99 < 1, and so forth. Therefore it's obvious that 0.999...<1.

A: No. Something that holds for various values need not hold for the limit of those values. For example, f (x)=x 3/x is positive (>0) for all values in its implied domain (x ≠ 0). However, the limit as x goes to 0 is 0, which is not positive. This is an important consideration in proving inequalities based on limits.

Q: 0.999... is written differently from 1, so it can't be equal.

A: 1 can be written many ways: 1/1, 2/2, cos 0, ln e, i 4, 2-1, 1e0, 12, and so forth. Another way of writing it is 0.999...; contrary to the intuition of many people, decimal notation is not a bijection from decimal representations to real numbers.

http://en.wikipedia.org/wiki/Talk:0.999...

Answer 4

Yes. 0.9999... = 1.

Does not "tend towards" 1. Does not "eventually reach 1 after an infinite number of 9's are added. Is not "infinitesimally close to" 1.

Equals 1. Is the same thing as 1.

Show me a number that is smaller than 1 and I will show you a number that is smaller than 0.9999...

http://en.wikipedia.org/wiki/0.999 for more info.

Answer 5

Yes; to prove it, consider the following:

If x = .999 (9s repeating forever)
Then 10x = 9.999 (9s repeating forever)

10x - x = 9x; Therefore

9.999 - .999 = 9 (9s repeating forever - 9s repeating forever) = 0

9x = 9 Therefore x=1

QED

Answer 6

your dealign with round off error. no computer can store an infinitely long number. by it's anture it would take an infinite number of bits to store it.
1/3 times 3 = 1
.3333 times 3 does not equal 1
1 does not equal .99 repeating.

remember the computer saves it in a FINITE number of bits. usually 8, 16, 32, or 64 bits. in most hand held calculators a floating point number (one where it's represented with something point something and the POINT can move back or forth instead of a fixed point number where it's some number of bits poitn and some number of bits) is usually represented on a 8 or 16 bit method.
my hand held casio crap tastic oldie calculator displays up to 8 numbers but internally it stores a little further (don't ask, it took some fiddling to figure it out) but my ti 89 emu can display 30 some numbers but stores about 30 more.

in general work with the fractionals if you think round off will be an issue.

Answer 7

Yes
.111111111...=1/9
.222222222...=2/9
.333333333...=3/9
.
.
.
.
.
.
.888888888...=8/9
.999999999...=9/9
ans since 9/9=1,
.999999999...=1

Answer 8

.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 < 1

rounding off the number then yes but in actuality no because as long as there is no number placed in front of the decimal other than a zero it will not be 1 and since it is repeating it will repeat but never come to one.( only in dealing with infinity)

Answer 9

Yes it is.
Basically, you are pushing the deviation from 1.0 to infinity, infinitely small.
What is 1 - .999999...? 1e-infinity, which is zero.

Answer 10

NO, The value approaches one but not =