does .999999999999999999999999999999 repeating equal 1?

Answer 1

YES, it does. It equals it exactly. You can mathematically prove it in several ways. There is no finite number that is small enough to be the difference between the two. (See the recently posted equation about the "smallest number", which doesn't exist).

You can go here for a full explanation:
http://mathforum.org/dr.math/faq/faq.0.9999.html
http://mathforum.org/library/drmath/view/55746.html
http://mathforum.org/library/drmath/view/55748.html

Once again, YES, the two are equal to each other. People who say otherwise are flat-out wrong. This is not a matter of opinion; it's proven mathematics. They can give me a thumbs-down for raining on their parade, but they're still wrong.

Answer 2

Before you are ready to accept a proof in mathematics, you need to have a clear understanding of what the terms involved mean.

What I want you to focus on is, what *is* .9999...? Or any decimal expansion, like 3.1415... or whatever.

A number doesn't change. It's not 0.9 one moment and then 0.99 the next, as if some outside person is changing it by contemplating it. It's stuck somewhere on the number line, no matter how you describe it.

And if I ask, what is the difference between 1 and 0.9999...., it's going to have to be a fixed number too, not some "infinitely small" thing. There has to be an answer; you can always subtract numbers. And it's easy to see, as many have pointed out, that the difference, if positive, would have to be smaller than any other positive number. If not positive, it must be zero.

Can it be that there is a positive number smaller than any other positive number? No, then it would be smaller than itself (or half itself). Contradiction.

Here's another perspective: do you believe that 1/2 + 1/4 + 1/8 + 1/16 + ...=1? (Classic Achilles & the Hare problem.)
That's the same as saying 0.11111.... in binary is equal to 1. Your problem is the same idea base ten.

Answer 3

Yes.

Ironduke's proof is one way to show it.
Another is to say that if .999... is less than 1, then there must exist some number that is more than .999... but less than 1.
To be larger than .999... but less than 1, it must have some digit larger than 9 in at least one place. But there is no digit larger than 9, therefore there is no number larger than .999... but smaller than 1. Since there is no number "in between" .999... and 1, .999... must equal 1.

Another way is to note that 1/3 equals 0.333.... Therefore 3/3, which equals 1, must equal 3 times 0.333...

Basic arithmetic shows that 0.333... times 3 equals 0.999...

Since 1/3 times 3 equals 1, and .333... equals 1/3, then 0.333... times 3 must equal 1. But it also equals 0.999..., and so 0.999... must equal 1.

Answer 4

Yes. Basically, it involves the concept of "the limit of 0.99.. repeating is 1.0"

Answer 5

Yes for all practical purposes.

Answer 6

Let x = .9999999999999999999..............
Then 10x = 9.999999999999999999.........
10x - x = 9.9999999999...- .9999999999999.....
9x=9
x = 1

Answer 7

No, it equals .999999999999999999999999999 repeating.. if you round it it'll be 1, but it does not equal 1.

Answer 8

It does not equal 1. It is very close but there is a finite difference between it and 1.

Answer 9

ya it does.

x = 0.99_
10x =9.9_
10 x - x = 9.9_ - 0.9_
9x = 9
x=1 = 0.9_

Answer 10

If you are working with rational numbers, no.
If real numbers or anything larger, yes.