Is .999...repeating exactly the same as 1?

Question

The arguments being these:
1) Let n=.999...
100n=99.999....
100n - n = 99
99n = 99
n = 1 = .999....?

or how about this?
2) 1/9 = .1111....
2/9 = .2222....
3/9 = .3333....
4/9 = .4444....
..
..
8/9 = .8888....
9/9 = .9999.... = 1?

Yes or No?

Answer 1

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 2

I've answered this question many times here, and Steven_S does it about as well as I have seen it done.

You have two choices: You can either accept that .999... represents a real number or you don't.

If you don't accept that it represents a real number then you are done. Forget about calculus because you won't understand it.

If you do accept that it represents a number, then that number has to be 1. If you say that it is anything else, it just screws up your arithmetic.

.999.. is an infinite sum .9 + .09 + .009 +...

You have to use the rules for evaluating infinite sums when you assign a value to .999... and those rules say that it's 1.

Answer 3

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Answer 4

Most say yes, however, there is much debate about this.

One of the arguments is that in your first equation you multiply .999... by a number but what is .999.. continues infinitely. Infinity is not a number and can't be multiplied.

That is just to give you a perspective of some peoples views.

You can read about why it is equal to 1 here:
http://mathforum.org/dr.math/faq/faq.0.9999.html

Check the bottom for more links on why it is equal to 1.

Answer 5

Both points are valid proof for proving .9 repeating to be the equivalent to equal 1.

Should you decide to study mathematics further, you will take a class called Calculus. In that class, you can actually prove, using limits, that .9 repeating is in fact equal to 1.

Answer 6

Yes, 0.9999999... = 1
Your proof is correct, but if it helps here's another one involving infinite geometric series:

0.99999999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...
0.99999999... = 0.9 + (0.9)(10^-1) + (0.9)(10^-2) + ....

t1 = 0.9 = 9/10
r = 1/10

Sum of an infinite geometric series = t1 / (1-r)

S = (9/10) / (1 - 1/10)
S = (9/10) / (10/10 - 1/10)
S = (9/10) / (9/10)
S = 1

Also, contrary to at what at least one previous answerer has said, there is no dispute or controversy over this.

Answer 7

It reflects the frequent incompatibility between reality and the measurement thereof.

How exactly can you have an infinite number of 9s? You'd need an infinite amount of paper to write them on.

Believing one has a grasp on infinity is always delusional.

Answer 8

argument 1 doesn't make sense, 100n-n= 99n not 99.
100n/n-1= 99

also 99n= 98.999...

Answer 9

YES it is, both proofs are nice.
To the nay-sayers: think of 1.000....1 with the final 1 infinitely far - does that equal 1 ?
See also:
http://mathforum.org/dr.math/faq/faq.0.9999.html

Answer 10

NO
.9999 repeating to infinity is not nor shall it ever be equal to one. Can it be rounded up to 1? Yes
Is it equal to one? No and never will be, but it is darn close
Give you an A for effort