0.99999.....=1

Yes

0.999... = 1.

Didn't know the answer would be that simple, but okay. My appreciations.

This post shines forth the essence of liberalism: mendacity & misrepresentation.
The motto of liberalism is: "that 's close enuf."

O.999999999 does NOT = 1.

1 = 1.





David

It looks like u let him fool u.
U fell for his liberalism. That 's sad.





David

I don't how far along in math you are, Logicus, but here are a couple proofs that .999... = 1.

First Proof:

Let x = .999...
Then 10x = 9.999...
9x = 10x - x = 9.999... - 0.999... = 9
So x = 1.

Second Proof:

Consider what "0.999..." actually is.
It's the sum of the series .9 + .09 + .009 + .0009 + ...
Let S represent the sum of the this series.
Then S = .9 (1 + .1 + .01 + .001 + ...)
= .9 (1 + (1/10)^1 + (1/10)^2 + (1/10)^3 + ...)
= .9 [1/(1 - (1/10))] (using the standard rule for summing geometric series)
= .9 (1/0.9)
= 1

Here are a few sites providing useful corroboration:

http://math.berkeley.edu/~gbergman/misc/numbers/9999.html

http://mathforum.org/library/drmath/view/53339.html

http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

No matter how the liberal slices it: its still BULLoney.
He urges us to accept his hoax upon a foundation of deception: no sale.





David

Off hand, I can think of two reasons why some people might have trouble accepting GOD's own truth that 0.999... = 1.

(1)

When people think of the number 0.999... and what it represents on the number line, they imagine a point starting at 0.9, moving to 0.99, then to 0.999, then to 0.9999, always getting closer to 1, but never reaching it, instead staying slightly to the left of 1, forever.

However, they are wrong to see the number "0.999..." this way. Like all numbers it occupies a fixed position on the number line. Do we see 0.333... as a moving point on the number line that comes closer and closer to 1/3? No, we understand that 0.333... holds a fixed position on the number line -- specifically it equals 1/3. Likewise 0.999... occupies a fixed position on the number line; it equals 3/3 (that is, 1).

(2)

One reason some people have trouble accepting that the number 1 has TWO valid representations as a repeating decimal (namely 1.000... and 0.999...) probably has to do with the source of where repeating decimals come from. Repeating decimals arise naturally when you divide one integer by another integer.

Now the long division algorithm is a deterministic process and should not give us TWO different repeating-decimal representations for a given dividend and divisor. Long-divide 1 by 3, and you will always get 0.333... . Long-divide 2 by 3, and you will always get 0.666... . But if you long-divide 3 by 3, you most certainly will not get "0.999..."; instead, you'll get 1.

But here's the thing. There are NO two integers a and b such that dividing a by b will give you 0.999... . The number simply does not occur as a result of long division. Ever. So we have to interpret the repeating decimal 0.999... in some other way.

I have already written about one way of interpreting it. You can consider it to be the sum of the infinite series 0.9 + 0.09 + 0.009 + ... . That series sums to 1, as I've shown. You can, if you like, consider it to be the limit of the infinite sequence 0.9, 0.99, 0.999, 0.9999, ... . In that case, we again find that it equals 1.

So if you are willing to accept that "0.999..." even represents a real number at all, then it represents the number 1.

You need proof. I, quote, "fell for it", because he offered an explanation to his answer, while you did not.

If they're not equal, then offer up a number that lies between them.

.OOOOOOOOOOOOOOO1 lies between them.

What do the 1O Commandments say
about taking the Name of the Lord thy God in vain ?


What people THINK or IMAGINE has no effect upon the laws of math.



Is this being offered as a PROOF,
or as idle speculation and naked guesswork???


the source of where repeating decimals come from. Repeating decimals arise naturally when you divide one integer by another integer.

Now the long division algorithm is a deterministic process and should not give us TWO different repeating-decimal representations for a given dividend and divisor. Long-divide 1 by 3, and you will always get 0.333... . Long-divide 2 by 3, and you will always get 0.666... . But if you long-divide 3 by 3, you most certainly will not get "0.999..."; instead, you'll get 1.

But here's the thing. There are NO two integers a and b such that dividing a by b will give you 0.999... . The number simply does not occur as a result of long division. Ever. So we have to interpret the repeating decimal 0.999... in some other way.

I have already written about one way of interpreting it. You can consider it to be the sum of the infinite series 0.9 + 0.09 + 0.009 + ... . That series sums to 1, as I've shown. You can, if you like, consider it to be the limit of the infinite sequence 0.9, 0.99, 0.999, 0.9999, ... . In that case, we again find that it equals 1.


That is a non-sequitur.
This conclusion does not result
from the premises offered by this charlatan.
.9 does not = 1 no matter how many 9s he writes there. 1 = 1.
Liberalism deserves no credence; it shud be peremptorily dismissed.





David

It is enuf that I debunked his fakery.





David

You exposed your ignorance of mathematics. That's about it.

Oh the ignorance. 0.99... goes on indefinitely, there for, 0.000..... cannot end in a 1, because the 0's would have to end.

Proof #3 that 0.999... = 1 (by contradiction):

(Note: in what follows, logs are base-10.)

Start by assuming the negation of what we want to prove.

Assume 0.999... 1. Then either 0.999... > 1 or 0.999... < 1. We can (obviously?) rule out the possibility that 0.999... > 1,
so it follows that 0.999... < 1. So 1 - 0.999... > 0. Let x = 1 - 0.999... . Then x is some positive real number between 0 and 1,
so log(x) is a negative real number. The integers are unbounded above or below, so let -n be the largest integer that is less than log(x).
(For example, if log(x) = -45.657, let -n = -46.) Since -n < log(x), and since the log function's inverse, f(x) := 10^x,
is an increasing function with range (0, ∞) , we have 0 < 10^(-n) < x. Thus -x < -10^(-n) < 0. Adding 1 to all terms gives:
0.999... = 1-x < 1 - 10^(-n) < 1.

Thus our assumption that 0.999... 1 has led to the conclusion that there is a number, 1 - 10^(-n), between 0.999... and 1.
What is this number? Well 10^(-n) is 0.00...01 (where there are n zeroes total, including the zero before the decimal point).
So 1 - 10^(-n) = 0.999..9 (where there are n nines after the decimal point). We have concluded that this number,
a zero and decimal point followed by n nines, is greater than "0.999...", a zero and decimal point followed by infinitely many nines.

That is absurd. Therefore, our assumption was false, which means 0.999... = 1. QED.

(I felt I should include this proof, since it's in the spirit of what you and Mark were arguing: if 0.999... < 1, there would have to be a number between them.)

Given the mathematical definition of "between" in the context of real numbers
(namely, that x is "between" a and b, if a < x < b)
what you have just said sounds pretty silly.

You've just made the statement that 0.999... < .OOOOOOOOOOOOOOO1 < 1.

That would be wrong even if we were talking about 0.999 instead of 0.999... .

9 does not = 1.

.9 does not = 1.
.99 does not = 1.

No matter how many 9s u put on the right of the decimal,
it will always be less than 1.

Your mathematical conclusion is a hoax. It is liberal mathematics.

INCIDENTALLY: I thought u had me on Ignore!

What r u doing addressing me
or even reading my posts, if u have me on Ignore???? That 's cheating.





David

True. As long as that number of 9's is finite. But infinitely many 9's after the decimal WILL get you to 1.