# 0.9999... != 1?

Status
Not open for further replies.

#### devire

Registered Member
this has been discussed an infinite amount of times, but i think i can show that .999.. does not equal 1.

1 - .999... = .000...1 = 1 / 1 * 10^(infinity) = 1 / infinity. now most people will say that 1 / infinity equals zero, but the constant 'e' is (1 + (1/infinity))^(infinity). if 1/infinity was really zero, then 'e' would be equal to one, but it of course isn't. so 1/infinity must have some non-zero value. and if it does, 1 != .999... since there is a non zero value between the 2 numbers.

am i missing something here?

edit: i typed in (1 - 1/1E80)^1E80 in my calculator and it gave me one, but does the function (1 - 1/x)^x really equal one as x approaches infinity?

Last edited:
Dear god, not this again. Maybe we need a sticky FAQ on the subject? The only thing more annoying than this is trying to convince people of the correct solution to the Monty Hall problem.
1 - .999... = .000...1

.000...1 is self-contradictory. The "..." means that the number in front of it goes on forever. If you say ".000..." you are saying that the zeros go on forever - you can't do that and then stick a 1 at the end. That's like saying "the zeros never end, but when they end there is a one." It's meaningless, or at the very least badly self-contradictory.

am i missing something here?
Yes. You're confusing
$$\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^{x}$$​
with
$$\lim_{y \rightarrow \infty} \left( 1 + \left ( \lim_{x \rightarrow \infty} \frac{1}{x} \right) \right)^{y}$$​
They're not the same. The former is e, the latter equals 1.
edit: i typed in (1 - 1/1E80)^1E80 in my calculator and it gave me one
That's due to rounding errors. Most calculators can only store the first 8 to 15 significant figures of a number (depending on the machine), so they'll round $$1 + 10^{-80}$$ (which requires 81 significant figures to store) down to 1. Then when you raise this to the power $$10^{80}$$, of course you get 1.

Last edited:
Dear god, not this again. Maybe we need a sticky FAQ on the subject? The only thing more annoying than this is trying to convince people of the correct solution to the Monty Hall problem.

.000...1 is self-contradictory. The "..." means that the number in front of it goes on forever. If you say ".000..." you are saying that the zeros go on forever - you can't do that and then stick a 1 at the end. That's like saying "the zeros never end, but when they end there is a one." It's meaningless, or at the very least badly self-contradictory.

ok, actually i realize that, but i thought since (1 + 1/infinity)^infinity equals 'e', that .000...1 in some cases is not really zero.

You have to undertsand what a limit is. Do not every try to memorize the result of an expression involving infinity, like (1/infinity)^infinity.

I have a headache now. :bugeye:
Thanks guys...:bugeye:

ok, actually i realize that, but i thought since (1 + 1/infinity)^infinity equals 'e', that .000...1 in some cases is not really zero.

Hi devire,
$$(1 + {1 \over \infty})^\infty$$ is undefined, just as $$0 \over 0$$, $$\infty \over \infty$$, and $$1 ^ \infty$$ are undefined.

This expression, on the other hand:
$$\lim_{n \rightarrow \infty } ( 1 + {1 \over n}) ^n$$
is defined, and equal to e... but you can't just write it as $$(1 + {1 \over \infty})^\infty$$. That's sloppy and ambiguous.

Hi devire,
$$(1 + {1 \over \infty})^\infty$$ is undefined, just as $$0 \over 0$$, $$\infty \over \infty$$, and $$1 ^ \infty$$ are undefined.

This expression, on the other hand:
$$\lim_{n \rightarrow \infty } ( 1 + {1 \over n}) ^n$$
is defined, and equal to e... but you can't just write it as $$(1 + {1 \over \infty})^\infty$$. That's sloppy and ambiguous.

i thought someone might say this. i do know there is a difference, but in the case of this function though, i would think they are the same since the graph has already passed one, and by it's nature won't just start to go back to one after it reaches a certain point.

i guess since it can never reach infinity in the first place means it doesn't matter if it will ever go back to one, and thus the expression is undefined? but if the expression is undefined, like 1/infinity on it's own is undefined, wouldn't that make 0.999... on it's own undefined as well, and thus not really equal to 1?

edit: how would 1 ^ (infinity) be undefined? i thought it was just 1. thanks.

1 / infinity aproaches zero in infinity.

edit: how would 1 ^ (infinity) be undefined? i thought it was just 1. thanks.
$$\frac{1}{\infty}$$ is shorthand for $$\lim_{x \rightarrow \infty} \frac{1}{x}$$, which equals zero.

$$\left( 1 + \frac{1}{\infty} \right)^{\infty}$$ is more ambiguous. In general it would mean something along the lines of: "the limit of $$\left( 1 + \frac{1}{x} \right)^{y}$$ as $$x$$ and $$y$$ tend to infinity". The problem here is that you get different results depending on how you calculate the limit. If $$x$$ and $$y$$ approach infinity together (with $$x = y$$), you get $$e$$. If $$x$$ approaches infinity first with $$y$$ fixed, then $$y$$ approaches infinity, you get 1.

$$\frac{1}{\infty}$$ is shorthand for $$\lim_{x \rightarrow \infty} \frac{1}{x}$$, which equals zero.

$$\left( 1 + \frac{1}{\infty} \right)^{\infty}$$ is more ambiguous. In general it would mean something along the lines of: "the limit of $$\left( 1 + \frac{1}{x} \right)^{y}$$ as $$x$$ and $$y$$ tend to infinity". The problem here is that you get different results depending on how you calculate the limit. If $$x$$ and $$y$$ approach infinity together (with $$x = y$$), you get $$e$$. If $$x$$ approaches infinity first with $$y$$ fixed, then $$y$$ approaches infinity, you get 1.

oops. did u think i meant 1 divided by infinity? i actually meant 1 to the infinte power. i thought that would be 1.

.9999... is an impossible number.

Mass is made up of energy, which is finite.

Yeah.
After the post above this thread needs deleting.

i thought someone might say this. i do know there is a difference, but in the case of this function though, i would think they are the same since the graph has already passed one, and by it's nature won't just start to go back to one after it reaches a certain point.
Try these:
$$\lim_{n \rightarrow \infty } ( 1 + {1 \over n^2}) ^{(2^n)}$$
$$\lim_{n \rightarrow \infty } ( 1 + {1 \over 2^n}) ^{(n^2)}$$

i guess since it can never reach infinity in the first place means it doesn't matter if it will ever go back to one, and thus the expression is undefined? but if the expression is undefined, like 1/infinity on it's own is undefined, wouldn't that make 0.999... on it's own undefined as well, and thus not really equal to 1?
I don't know that 1/infinity is undefined... no matter what expression is used to reach infinity, 1/infinity will always equal zero.
More formally:
For all $$f(x)$$ where: $$lim_{x \rightarrow \infty} f(x) = \infty$$
$$lim_{x \rightarrow \infty} 1/f(x) = 0$$

edit: how would 1 ^ (infinity) be undefined? i thought it was just 1. thanks.
This is a bit beyond my knowledge of maths theory, but I think of it like this:
When you write $$\infty$$ is a shorthand way of expressing a limit. There's no way of knowing exactly how that limit is approached.
It might be not as easy to see that 1 can also be a shorthand way of expressing a limit... So similarly, there's no way of knowing in the given expression whether "1" means precisely 1, or if it means a limit, such as $$lim_{x \rightarrow 0} (1+x)$$

Similarly, $$0 \times \infty$$ is undefined, and 0^0 is undefined.

I'm going a bit beyond my knowledge here. I could be mistaken.

0^0 = 1, I was taught. (@ Pete)

.9999... is an impossible number.
0.999... is 1. Since 1 is a possible number, .999... is also a possible number...because they're the same number.

.9999... is an impossible number.

Mass is made up of energy, which is finite.

This is a mathematics problem. not a physics problem. Even then, we would rather talk about "1" then 0.999...

Status
Not open for further replies.