Does .9-repeating equal 1?

[Voltaire]

Yes it is equal to 1. Not for all practical purpose, or close enough, it is exactly equal to one.

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

[SlowwHand]

http://tinyurl.com/fn44j

[KrazyHorse]

Originally posted by civman2000
This is such a silly question. It assumes that things like "1" and ".999..." actually have some independent existence from what we say they are (OK, I'll accept for the sake of argument that "1" might, but ".999..."??). The reason that .999...=1 is that that's what we define it to mean. There's no reason we couldn't define it to mean something else though.

No, .9999... is defined to be the sum(i=1 to inf) 9*10^-i

Let an = sum(i=1 to n) 9*10^-i

1.0 - an = 10^-n

0.999.... = lim(n->inf) an

Therefore 1.0 - 0.9999... = lim(n->inf) (1.0 - an) =
lim(n->inf) 10^-n = 0

If a - b = 0 then a = b

[civman2000]

Originally posted by Winston
No it isn't, it's as straight as can be. Please tell me that you disagree that 1/3 is .3 repeating?

I don't disagree. What I'm saying is that that statement needs proof just as much as the statement that 1 is .9 repeating. Why should you believe that 1/3 is .3 repeating if you don't believe 1 is .9 repeating?

[Solly]

No. That extra 0.0001 makes all the difference you know.

[Jaguar]

They equal each other, mathematically. But they are different concepts. The sum of a series can be .999..., but you'd never say that there are .999... people in a room.

[civman2000]

Originally posted by KrazyHorse


No, .9999... is defined to be the sum(i=1 to inf) 9*10^-i

Let an = sum(i=1 to n) 9*10^-i

1.0 - an = 10^-n

0.999.... = lim(n->inf) an

Therefore 1.0 - 0.9999... = lim(n->inf) (1.0 - an) =
lim(n->inf) 10^-n = 0

If a - b = 0 then a = b

Well, yes. I didn't mean that we literally defined .999... to be 1. I meant that it followed from the way we defined it that it was 1.

[KrazyHorse]

Originally posted by Jaguar
They equal each other, mathematically. But they are different concepts. The sum of a series can be .999..., but you'd never say that there are .999... people in a room.

Just because you'd never say it doesn't mean it isn't true. They are two different ways of representing the same number.

[KrazyHorse]

Originally posted by civman2000

Well, yes. I didn't mean that we literally defined .999... to be 1. I meant that it followed from the way we defined it that it was 1.

ah. Just seemed you were making some sort of odd claim there.

Anyhow, I don't think it's reasonable to ask that things be true regardless of how we define them....

[Kuciwalker]

Originally posted by SlowwHand
I said Yes, but in strictest theory the answer is No.

Wrong.

[Kuciwalker]

Originally posted by Ninot
I vote no, because it's techincally not.

Wrong.

[KrazyHorse]

Look, people, the concept is simple:

If a does not equal b then there exists a number c not equal to 0 such that a - b = c

Find me the number c such that 1 - 0.9999.... = c

[SlowwHand]

I stick by what I said. It's quite close to what jaguar says.

[Ramo]

This is such a silly question. It assumes that things like "1" and ".999..." actually have some independent existence from what we say they are (OK, I'll accept for the sake of argument that "1" might, but ".999..."??). The reason that .999...=1 is that that's what we define it to mean. There's no reason we couldn't define it to mean something else though.

There are some nuances to this question that go beyond mere semantics. It's not totally trivial that this series (the infinite sum over n of 9*.1^n) should converge to 1.

[Kuciwalker]

In the strictest sense the answer is yes. Only with the fuzzy I'm-a-liberal-arts-major logic is it ever false.