0.9 repeater = 1

[whitefork]

In other words. 0.999999999999999...... NEVER equals 1

This is true.
We can make it as close to 1 as you require.
If you give me a fraction, as small as you please, I can show how, in a finite number of steps, to produce a repeating decimal that is closer to 1 than that fraction. That's one of the things a limit is. You're not permitted to give me 0 as the fraction because there's no way to produce zero with a non-zero numerator.

 

[Terry]

Since 1 == 0.999 repeating, there are two representations for (at least) one real number.

How does this affect Cantor's diagonal argument for the unlistability (if that's a word) of the real numbers? Couldn't you come up with a number different in every digit that never the less represented the exact same real as something already in the list?

 

[whitefork]

Cantor

Margin too small - must provide link

http://www.wikipedia.org/wiki/Cantors_Diagonal_argument

 

[ceptimus]

Also....DON'T CHANGE THE SUBJECT

regards,
BillyJoe

[Sheepish]Sorry - I was responding (sort of) to this:

Can you use a more complicated version of the same proofs to do it with other numbers? Could you eventually prove, say, that 1=2?

[/Sheepish]

Did you notice that your 'quoted' version of my post doesn't match my original? Looks like vbulletin can't handle the 'squared' symbol properly. Oops, changing the subject again.

ceptimus.

 

[Nucular]

In other words. 0.999999999999999...... NEVER equals 1

Surely by this logic, 0.999... never equals anything. Neither does pi, or any other infinite decimal.

Surely 0.999... = 1 is as true as saying pi = circumference/diameter?

 

[patnray]

Brown beat me to my favorite proof of the concept, which I learned in college many years ago.

There seems to be some confusion over the nature of a repeating decimal. 1 = 0.9 is clearly false. 1 = 0.99 is also false. Both of those decimals terminate, and Brown's "multiply by 10" trick fails because there are only 0s after the last 9, not more 9s. But 1 = 0.999... (i.e. repeating 9s) IS true because the sequence never terminates (there is always one more 9).

In the same class I learned a trick for turning any repeating decimal into a fraction, but I've forgotten it. I'll see if I can look it up.

 

[Brown]

In the same class I learned a trick for turning any repeating decimal into a fraction, but I've forgotten it. I'll see if I can look it up.

You're right, the same trick can be used for other repeaters.

Suppose x= 0.142857142857142857142857....
Then 1000000x = 142857.142857142857142857142857....
So 999999x = 142857
x = 142857/999999 = 1/7

 

[whitefork]

Brown's trick works for that dunnit?

eg x= .123123123......
1000x = 123.123123123..... (where the multipler = 10^period)

999x = 123

x = 123/999 = .123123123...

in fact this should work for any number base.


edited after seeing prior response> Guess it does.

 

[patnray]

Well, there's the handy fact that if you have 0.abc...n with the whole thing repeating, is has the same value as (abc...n)/(999...9). So 0.9 repeating = 9/9 = 1.

rwald is correct. This can be proven using the same technique as Brown's method for proving that 1 = .999... Multiply by 10^n, where n is the number of digits in the repeating sequence, subtract, and divide by (10^n) - 1.

Which means every repeating decimal is a rational number...

 

[rwald]

BillyJoe, the thing is, when we say 0.999..., we mean that the 9's go on forever to infinity. It's implicit. So, it is valid to say that 0.999... is infinitely close to 1. That is to say, the difference between the two is infinitely small. I think that we can say that the difference between the two is infinitely small, than the two are in fact the same.

 

[jimlintott]

Here is one that I've seen. It's not as good as the ones here but it does make people think.

1/9 = .111...
2/9 = .222...
3/9 = .333...
4/9 = .444...
5/9 = .555..
6/9 = .666...
7/9 = .777...
8/9 = .888...
9/9 = 1 or is it .999...

Tonight on FOX "When irrational numbers become integers".

 

[bjornart]

In the end x.abc<repeating> is just an annoying way of representing a rational number.

jimlintott you've got it wrong, we're not dealing with irrational numbers here, they would be the ones _without_ repeating digits.

Then again, as I've been lead to believe, FOX would get that wrong.

 

[LibraryFox]

For .999... to equal 1, then
.1^x must equal zero when x is infinately arge....
Also, just because traditional notation doesn't make a number easy to represent, doesn't mean it doesn't also fit in the continuum. So by the .999...=1 logic, then
.999...8=.999...9 and .999...7=.999...8 and so on...
-LF

 

[rwald]

So, your notation 0.999...8 means "an infinite number of 9's followed by an 8"? The thing is, that makes no sense mathematically. When I was much younger, I used to think about the number 0.000...1, which was "an infinite number of 0's followed by a 1." Than I realized that this number was equal to 0. In the same sense, 0.999...8 has no meaning distinct from 0.999... If it's only different by an infinitely small number (0.1^(infinity), as you said), than for any purposes it's the same number.

 

[pgwenthold]

For .999... to equal 1, then
.1^x must equal zero when x is infinately arge....
Also, just because traditional notation doesn't make a number easy to represent, doesn't mean it doesn't also fit in the continuum. So by the .999...=1 logic, then
.999...8=.999...9 and .999...7=.999...8 and so on...

Here's the problem: if 0.999...8 is indicating that 8 is the last number in the squence, then in fact it is not the same as a number that repeats infinitely.

Not the same thing. The actual comparison is that

.999...89999... = .999...9

which is true (assuming the the middle ... replace the same thing in both nmbers).

 

[LibraryFox]

You are correct a repeater that ends in something different is irrational, but your argument is then that .1^infinity is in fact zero... which it isn't. by limit theory, it gets close and makes a useful aproximation but they aren't equal.
-LF

 

[rwald]

OK, I'm going to do this properly, with limits.

Taking the limit of (1/10)^n as n -> infinity:

(1/10)^n = (1^n / 10^n)

Plug in infinity.

1 raised to any value is still 1. So, (1^(infinity)) = 1
10 raised to infinity is infinity. So, (10^(infinity)) = infinity

So, the limit of (1/10)^n as n -> infinity = 1/infinity. And 1/infinity = 0.

And as I said above, limits are implicit when you use the ... notation. By definition, 0.999... is the infinite sum of 9*(10^(-n)) as n goes from 1 to infinity.

Any more questions?

 

[Jarom]

Isn't the correct statement that the limit of 0.999... is 1?

~~ Paul

No, this is not a correct statement. Limits are defined only for sequences of numbers, and 0.999... is one number, not a sequence of numbers.

Here are two correct statements which look similar to yours, but are slightly different:

"The limit of (0.9, 0.99, 0.999, 0.9999, ...) is 1."

"0.999... is 1."

Notice that these second statement is just a rewrite of the first, since "0.999...", by definition, has the same meaning as "the limit of (0.9, 0.99, 0.999, ...)".

- Jarom

 

[pgwenthold]

1 raised to any value is still 1. So, (1^(infinity)) = 1

Isn't 1^(infinity) actually undefined

See it this way

log [1^(infinity)] = (infinity)(log 1) = infinity*0

I seem to recall that infinity*0 is undefined (0/0)

 

[whitefork]

Yeah me I have a question. Speaking of powers of infinity, this one caused some amusement a couple months back.

evaluate (1 + 1/n)^n as n approaches infinity.

Intuitively, 1/n approaches 0 as n approaches infinity, so 1 + 1/n becomes closer to 1, and 1 to any power is 1.

So, the limit has to be 1, right?