0.99999.....=1

David, it is good to see you have a sense of humor lurking within your empirical intellect. May the climate of your new domain be advantageous for your health.

Now guys; I have been called many things, such as, ‘Hey hillbilly, haul your fat ass off my lawn – and your truck too.’
Butt I have never been called a matherma wotsits.


Quote: “But infinitely many 9's after the decimal WILL get you to 1.”

I demure!

Do we agree that when you subtract a number from itself, the result is zero?

For instance, 4 – 4 = 0.

Good, I can see we iz going to getalong just fine.

So what is the result when you subtract 0.999... from 1?

For shorter subtractions, you get:

1.0 1.00 1.000 1.0000 1.00000
0.9 0.99 0.999 0.9999 0.99999
__ ___ ____ ______ _______
0.1 0.01 0.001 0.0001 0.00001


Then what about 1.000... – 0.999...?
You'll get an infinite string of zeroes. But you ask, "what about that '1' at the end?"

Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.
Then 1 – 0.999... = 0.000... = 0, and 1 = 0.999....

But if two numbers are different, then you can fit another number between them, such as their average. Am I write or am I right!

But what number could you possibly fit between 0.999... and 1.000...?

The fundamental question is whether 1/3 = .3333....

[LOL at what precedes this except. ]



Yup. That's where the "0.999 < 1" crowd gets into trouble. If they are different then their average should lie between them. If they are real numbers, so is it. As a real number it should have some kind of decimal representation. But what would that be, to fall between 0.999... and 1.000... ?

Pardon the digression...



I should add something to the remarks I made earlier about hyperreal numbers.

If you make .999... the kind of hyperreal that would put it between 1 and any real number smaller than 1, then you should be consistent and make .333... the kind of hyperreal that would put it between 1/3 and any real number less than 1/3.

In other words, you should define .333... as (.3, ,33, .333, ...). It would then be less than 1/3 by an infinitesimal amount.

In hyperreals, you can actually add and subtract infinitesimals just like normal numbers. (1/30, 1/300, 1/3000, ...) is the hyperreal form of an infinitesimal, as is every sequence of reals that approaches zero in the limit.

There is one more thing about hyperreals that ... anyone who cares about them ... should know -- without which you can't really understand how they work. You know that one hyperreal is less than another if all except a finite number of the terms in the sequence of reals that represents the first hyperreal are smaller than the corresponding terms in the sequence of reals that represents the second hyperreal.

So for example (0.9, 0.99, 0.999, ...) < (1, 1, 1, ...),
because 0.9 < 1, 0.99 < 1, 0.999 < 1, etc.

But for any real number x < 1,
(x, x, x, ...) < (0.9, 0.99, 0.999, ...),
because eventually the terms in the second sequence overtake the terms in the first sequence for good as they rise above x.
So if x were, say, 0.999997, the first five terms in the first sequence would be greater than the corresponding terms in the second. But except for those five, all the terms in the second sequence would be larger than the corresponding terms in the first.

Okay, enough about hyperreals. Back to your regularly scheduled debate...

"except" should be "excerpt"

Also, I meant the "0.999... < 1" crowd.

Not all math is arithmetic; all arithmetic is math.





David

The essence of your argument
is that ALMOST good enuf, but not quite IS good enuf
and shud not be challenged by Orthodox Tea Party types like ME.

Therefore, if an impoverished worker who curries your favor
leave work an hour early, then that is EXACTLY what he is supposed to DO
and it is perfect execution of his contract because O.99 = 1.

BULLoney! How many bridges r u selling ??





David

Why do you insist on using finite series in your arguments instead of the infinite series which is the topic of discussion? Nobody claimed that any finite number of nines to the right of the decimal equals one.

Practical applications of mathematical evil purported in O.9999 = 1:

Defendant is arrested for using pliers to remove the first finger of victim
(without his consent) on a given Tuesday, but only up to its first knuckle.
It is stipulated by all counsel that as of Monday, the victim was left un-disturbed
and that he then had 1O fingers intact. The victim 's testimony is taken.
He inculpates the defendant. On cross-exam, he admits that he lost
only 1 of his fingers to defendant's violence, and even that finger
remains intact, except beyond its first knuckle.

Defense counsel qualifies an expert witness in evil liberal math,
who attests that O.9999 = 1.
Defense counsel argues in exculpation that .9 = 1 and indeed,
the victim still had MORE than
.9 of all of his fingers remaining intact. Since .9 + more of victim 's fingers
remain intact: ALL of his fingers were left alone, un-troubled and intact.
Hence, there was no crime.

An evil, liberal judge will dismiss the case, taking judicial notice
of the expert 's mathematical testimony and judicially declaring
that no harm was done and that the victim remains fully intact because .9 = 1.

Defendant is liberated; he leaves jail
and joining the ranks of recidivism, he finds new victims n he does it again,
repeating the cycle in criminal litigation.





David

Yet another finite example.

Except nobody here is attesting that. There are an infinity of numbers that fit between .9999 and 1.

The fundamental question is whether the decimal format
will enable u to express one-third, or not.

Well, there's your problem.

I can express 1/3 in decimal notation just fine. Stick that in your vinculum and smoke it.

http://en.wikipedia.org/wiki/0.999%E2%80%A6

That 's a lie; a liberal lie.
Its not MY problem. If that format is dysfunctional,
then I will not persist in beating a square peg into a round hole,
as u r obsessively-compulsively doing.

I can do this: 1/3.





That 's another liberal lie; (DD contradicts himself)
"don 't look behind that curtain", DD screams. "Pay no attention to that charlatan back there."





David

I see you've been reading 1984 again....

Oldthinker unbellyfeels newmath.

If that number is subtracted from 1,
then according to u, the result will be O, right, DD???


http://en.wikipedia.org/wiki/0.999%E2%80%A6


Its not just "counterintuitive"; its false. 1 does not equal .9





David

Question David---If I'm standing ten feet from a wall and with every step I halve my distance to the wall, how many steps will be required to get to the wall?

Rap

Question raprap -- If I cut a length of string up into pieces of roughly equal length shouldn't those pieces add up to the original length of the string?

.999... = lim n->∞ (1- (1/10^n))
.999... = lim n->∞ 1 - lim n->∞ (1/10^n)
.999... = 1 - lim n->∞ (1/10^n)

Now lets look at the term lim n->∞ (1/10^n). The limit L = 0 as n->∞. But if we cut up a string of length 1 into roughly equal lengths based on the expression (1/10^n) for finite values of n we always can sum the pieces up and get the total string length back. So if n= 1, I make 9 cuts in the string and each string length is 1/10 of the total and there are 10 pieces at that length. If n= 2 then there are 100 pieces with a length 1/100. This can be done for any value of n. Then we can sum the individual pieces to get the original length of the string. When you apply that logic as n->∞ a problem occurs. You start with a string of length 1, but the length of each cut by the limit laws must be zero [lim n->∞ (1/10^n) = 0], so when you sum up an infinite number of zero length pieces you get nothing for your total string length. So there is a contradiction in equating the quantity lim n->∞ (1/10^n) with zero I think.