Originally posted by: DrPizza
For the love of God...
.999 repeating *IS* a rational number...
a rational number means: "can be written as a ratio"
a repeating decimal is a rational number.
Are some of you arguing that .333repeating does not equal 1/3 ?!?
.999... is a repeating decimal. ALL repeating decimals are rational numbers.
If your intuition tells you that .999 repeating doesn't = 1, well, your intuition is wrong. Get over it.
And, for the next person who posts that .999... <> 1 for the reason of "IMO" or "IMHO," I hope you get struck by lightening.
I wish I could say this for the final time... .999 repeating EQUALS EXACTLY 1.
RossGr (and others who know what they're talking about) are correct.
For those of you without the math background, that would be like me arguing with someone fluent in Japanese about the definition of some word in Japanese. Only an idiot continues to argue that he's right when countless people have shown valid proof that he's wrong. Some of you are as bad as those people who still believe the earth is flat and moon landings are a big conspiracy.
/end rant
We do.Sorry no point for stating the obvious. That is pretty much the definition of .999... that I have been pushing. It is a simply fact that every postion of a real number is finitly indexable
This is where you go wrong, it is not necessary to prove .1^N = 0, all I need to do is show that the relationship holds for all N.
Originally posted by: sao123
If we agree this far...
9 * E( .1 ^ N) [from 1 to N] {N > 0} + (.1^N) = 1
And Quote
We do.Sorry no point for stating the obvious. That is pretty much the definition of .999... that I have been pushing. It is a simply fact that every postion of a real number is finitly indexable
Then
This is where you go wrong, it is not necessary to prove .1^N = 0, all I need to do is show that the relationship holds for all N.
If you do not prove this point then your set of rational numbers does not obey the identity law of addition "That being that zero is the only value which may be added to a term and still equal the same value." (This law must apply to all rational set addition: it must physically be there...if the property does not hold, then your rational number set is invalid)
Secondly...as for your long division with a cheat.
Since the set of rational numbers has a true 1 to 1 mapping. (IE. any rational number operation performed on any chosen pair of rational numbers in the entire set of rational numbers, must have exactly 1 unique solution in the set of rational numbers.
Since 1/1 is already defined as 1, by the law of multiplicative identity, you cannot map it to .999999999... without destroying the 1 to 1 mapping, which once again voids your set of rational numbers.
(IE. any rational number operation performed on any chosen pair of rational numbers in the entire set of rational numbers, must have exactly 1 unique solution in the set of rational numbers.
Anyway, the thing is that we're not doing "one" operation on a "chosen pair" of rationals. A repeating decimal like 0.999.... (or any other) is implicitly an infinite series.
Now for the people who want to insist that 0.999.... differs from 1 by the number 0.000.......1 then there is only thing to say. There is no such number as 0.000.....1, that notation is never used because it is superfluous - any such number (that is, an infinite number of zero's followed by some nonzeros) is more compactly represented as zero, which is what it is.
Prove this statement.Correct, but numbers do exist in the form such that as N->oo they become indeterminant.
The idea is that you can make this equal different values by
approaching 1^infinity in different ways. First, we know that 1 to any
finite power is 1. But on the other hand, we know that any number
other than one, raised to an infinite power, would be infinite if the
base is greater than 1, and zero if the base were less than 1. It may
seem natural to you to define
1^infinity = lim[x->infinity] 1^x
which would be 1, since 1^x is always 1; but in order to have a solid
definition, it turns out that we have to allow both numbers to vary,
and define it as
1^infinity = lim[x->1, y->infinity] x^y
But this is defined only if we get the same limiting value regardless
of how we approach 1 and infinity. If we hold x constant at 1, we get
1 as before; but if we hold y constant at infinity (which isn't really
legal, but it gives us a good extreme case to picture), while x
approaches 1 from above or from below, we get infinity or zero. If we
let x and y simultaneously vary, we can get any answer we like,
depending on which moves faster toward its destination.
Therefore It could equal any value whatsoever.
if you have "1^infinity" what you really have is some kind of limit:
the base isn't really 1, but is getting closer and closer to 1 perhaps
while the exponent is getting bigger and bigger, like maybe (x+1)^(1/x)
as x->0+.
The question is, which is happening faster, the base getting close to
1 or the exponent getting big? To find out, let's call:
L = lim x->0 of (x+1)^(1/x)
Then:
ln L = lim x->0 of (1/x) ln (x+1) = lim x->0 of ln(x+1) / x
So what's that? As x->0 it's of 0/0 form, so take the derivative of the
top and bottom. Then we get lim x->0 of 1/(x+1) / 1, which = 1.
So ln L = 1, and L = e. Cool!
Is it really true? Try plugging in a big value of x. Or recognize this
limit as a variation of the definition of e. Either way, it's true. The
limit is of the 1^infinity form, but in this case it's e, not 1. Try
repeating the work with (2/x) in the exponent, or with (1/x^2), or with
1/(sqrt(x)), and see how that changes the answer.
That's why we call it indeterminate - all those different versions of
the limit approach 1^infinity, but the final answer could be any
number, such as 1, or infinity, or undefined.
Some of these are indeed indeterminate FORMS, they are not numbers. So they do not have a defined position on the number line and are not represented by a geometric series. Since this it not our case it does not apply and is an interesting fact but meaningless in this case. I think you added #5 to the list for your own convinence, It is not indetriminat. 1^inifity =1 because the lim of 1^n as n tends to infinity is 1.Originally posted by: sao123
How have you never heard of indeterminate numbers???
All indeterminate numbers take on 1 of 7 forms... all numbers of these forms are indeterminate...
1) oo / oo
2) 0 / 0
3) 0 * oo
4) oo - oo
5) 1 ^ oo
6) 0 ^ 0
7) oo ^ 0
indeterminant numbers have no set definition, they could = 0 ,1 , 2, 5, 10,000, oo, -oo
Your notation sux, I think I know what you are saying.take the most simplest example....#2.
0 / 0... = 0 or 1?
Rule 1 -> N/N for any N, = 1
Rule 2 -> 0/N for any N, = 0
or....
0 * N = 0 for all N.
N = 0/0 for all the same N. Therefore 0/0 can equal any number N.
0 = N for all possible N???
Example B...Rule 4... what is oo - oo?
1 + (infinity - infinity) = 1 + 0 = 1, but
(1 + infinity) - infinity = infinity - infinity = 0.
0 = 1???
Example C...Rule 6... what is 0 ^ 0?
Rule 1... 0 * 0 * 0 * 0...... = 0
Rule 2... N ^ 0 (for all N) = 1.
0 = 1???
Now. Using the original equation again...
9 * E( .1 ^ N) [from 1 to N] {N > 0} + (.1^N) = 1
here you lose me intirely, I have no clue what this means or where you got it. Therefore am unable to say if it is right or wrong. I suspect that you are badly mishandling the terms of the sum. All that follows is suspect..99999999.... repeating forever is the case N = oo (countable infinity)
I can then rewrite the equation to be...
9 * (10^N) * E(1^N) + (10^N) * (1^N) = 1
And in the case that N = oo, both terms contain 1 ^ oo (See rule 5), an indeterminate number that can take on any value...
Both terms contain an indeterminate number. Rule: Adding or Multiplying an indeterminate number with a determinate number is still indeterminate.
Therefore the most general equation is adding 2 indeterminate numbers.
The sum of 2 indeterminant numbers must also be indeterminant. IE it will never always equal a determinant number (1).
Therefore... .9999999.... = 1 some of the time, but not all of the time.
Pure speculation and wishfull thinking on your part. This is pure garbage.The idea is that you can make this equal different values by
approaching 1^infinity in different ways. First, we know that 1 to any
finite power is 1. But on the other hand, we know that any number
other than one, raised to an infinite power, would be infinite if the
base is greater than 1, and zero if the base were less than 1. It may
seem natural to you to define
1^infinity = lim[x->infinity] 1^x
which would be 1, since 1^x is always 1; but in order to have a solid
definition, it turns out that we have to allow both numbers to vary,
and define it as
1^infinity = lim[x->1, y->infinity] x^y
But this is defined only if we get the same limiting value regardless
of how we approach 1 and infinity. If we hold x constant at 1, we get
1 as before; but if we hold y constant at infinity (which isn't really
legal, but it gives us a good extreme case to picture), while x
approaches 1 from above or from below, we get infinity or zero. If we
let x and y simultaneously vary, we can get any answer we like,
depending on which moves faster toward its destination.
Therefore It could equal any value whatsoever.
if you have "1^infinity" what you really have is some kind of limit:
the base isn't really 1, but is getting closer and closer to 1 perhaps
while the exponent is getting bigger and bigger, like maybe (x+1)^(1/x)
as x->0+.
Your methods are very bad!The question is, which is happening faster, the base getting close to
1 or the exponent getting big? To find out, let's call:
L = lim x->0 of (x+1)^(1/x)
Then:
ln L = lim x->0 of (1/x) ln (x+1) = lim x->0 of ln(x+1) / x
So what's that? As x->0 it's of 0/0 form, so take the derivative of the
top and bottom. Then we get lim x->0 of 1/(x+1) / 1, which = 1.
So ln L = 1, and L = e. Cool!
[/quote]Is it really true? Try plugging in a big value of x. Or recognize this
limit as a variation of the definition of e. Either way, it's true. The
limit is of the 1^infinity form, but in this case it's e, not 1. Try
repeating the work with (2/x) in the exponent, or with (1/x^2), or with
1/(sqrt(x)), and see how that changes the answer.
That's why we call it indeterminate - all those different versions of
the limit approach 1^infinity, but the final answer could be any
number, such as 1, or infinity, or undefined. You need to do more work
to determine the answer, so 1^infinity by itself is not determined yet.
In other words, 1 is just one of the answers of 1^infinity.
9 * (10^N) * E(1^N) + (10^N) * (1^N) = 1
I'll second that. I'm glad you took the time to try to figure out what the hell SAO was trying to say because I sure didn't have the patience to do so.Originally posted by: RossGr
Your notation sux.
Originally posted by: ZeroNine8
It seems like what was proven is that the limit of 0.999... = 1 as the number of repeated digits goes to infinity. I have no problem believing that 1 is the limit of the convergent series 0.999..., however just because it is the limit does not mean it is equal. Because infinity isn't a number, merely a concept, you can't claim that 0.999... with infinite digits actually equals 1, only that it approaches it as the limit with more and more 9's.
In plenty of cases, the limit of a convergent series is never actually reached by that series.
Originally posted by: RossGr
9 * (10^N) * E(1^N) + (10^N) * (1^N) = 1
I have been trying to interpret this expression to figure out what you are trying to say.
I may have it.
I am going to assume that your E is the Sigma for summnation. Are you attempting to factor out a .1^n ?
If so sorry that does not work.
Consider this
2+4+8+16 = 2^1 + 2^2 + 2^3 +2^4
Can you factor 2^3 out of this expression? Is that what you are doing? If so it is not correct. All that can be factored out is a 2.
In our case that would mean you could factor out a .1 not .1^n if you could factor out a .1^n you would have a 1 left not a 1^n.
You need to review some basic algebra.
Originally posted by: sao123
I myself am checking into the validity of the equation.
I do believe i made an algebraic error, however, the principle still applies.
.1^N can be factored into 2 terms, one of which can be 1^N.
Originally posted by: nrg99
what about 1=0?
Originally posted by: uart
Originally posted by: nrg99
what about 1=0?
Sorry, the closest I can get to that is 0^0 = 1. Is that whacky enough for you?
lets define the difference bettween 0.9999.. and 1 as Q and let Q=1*10^-N where N tends to infinity (no one can argue that that isnt the case)