.9999 ... = 1
there seems to be general consensus about this.
However i don't know if that proof in OP is valid.
Name:
Anonymous2007-02-18 17:39
So does this mean .7 repeating = .infinite 7s and then one 8?
Name:
Anonymous2007-02-18 18:04
Rather than saying "giving infinity a value," it's perhaps a bit
clearer to say, "giving the concept of a limit of an infinite sequence
of numbers a value."
.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.
You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:
THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.
Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.
There is nothing special about .999... The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.
And it doesn't limit itself to single repeated decimals. When we say:
1/7 = .142857142857142857...
none of the finite parts of the decimal is equal to 1/7; it's just
that the more you add, the closer you get to 1/7, and in addition, 1/7
is the UNIQUE number that they all get closer to.
Finally, you can show for all such examples that doing the arithmetic
on the series produces "reasonable" results:
Since:
1/3 = .333333...
2/3 = .666666...
1/3 + 2/3 = .999999... = 1.
By the way, there is nothing special about 1 as being a non-unique
decimal expansion. Here are a couple of others:
This is _the_ classic math troll. Just let the children play.
Name:
Anonymous2007-02-20 22:12 ID:SPnDISQ2
>>19
Please explain how the concept of an infinite geometric series is misusing infinity.
And you can take your shitty hotel metaphors with you when you GTFO.
Name:
Anonymous2007-02-20 23:38 ID:TfbhBCm1
Infinity is not a real number.
Name:
Anonymous2007-02-21 3:24 ID:D4Pcez+l
.99999 repeating is one there are proof's of it and even a wiki
my proof 1 is the same as 3/3 correct?
and 1/3 = .33333333333
and if we ad another 1/3 to get 2/3 that =.6666666 correct
so if we ad 1/3 or .3333repeating to 2/3 or .6666repeating we get both 1 and .99999 repeating
Name:
Anonymous2007-02-21 3:29 ID:D4Pcez+l
you think your so smart don't you, you dirty fucks with all you "fancy" words like "Calculus" "Maths" and "2"
QED. You guys fail to see that 0.999...∞ still misses the infinitely tiny fraction ...∞ to make it 1.
More evidence:
I have 0.999...∞, let me take off an infinitely small fraction ...∞
Now I have 0.999...∞ - ...∞ = 0.999...∞
As you can see, I have taken a number with an infinitely small fraction less than 1, took off an infinitely small fraction and still have the same number with an infinitely small fraction smaller than 1! Amazing, isn't it!
Like Hotel Infinity should make clear to you, you can't use infinity in regular arithmetic. 0.999...∞ repeating is a CONCEPT and the only thing you can derive from it is that it is smaller than 1. An infinitely small fraction smaller than 1, which is why we can only use it as a concept, and not in mathematical formulas.
>>30
Why should >>29 bother posting an argument? People have already presented dozens of arguments (which, unlike yours, are flawless and actually written by someone who has finished junior high) and you ignored all of them.
Name:
Anonymous2007-02-21 14:34 ID:tl5dVrYu
>>31
Flawless how? I pointed out exactly what flaw was made and provided proof for this reasoning. You disprove it, if you can. If this simply is not enough to explain things to you, I don't think you are capable of understanding the concept matter of infinity.
The problem is you are trying to put a finite point in your mind to an infinitely tiny fraction, which makes it not infinitely tiny anymore. As I explained, infinity is a concept, and if you don't understand that concept well you will make mistakes like "0.999... = 1"
>>32
You didn't point out a flaw, you pointed out that you have ABSOLUTELY NO FORMAL TRAINING IN MATH.
Name:
Anonymous2007-02-21 15:56 ID:yXM2Rzjs
>>No sir, it is you who does not understand the concept of infinity.
>Now I have 0.999...∞ - ...∞ = 0.999...∞
That is a flawed argument. You can't take sigma(9/10^n) from {n = 0 to ∞} and subtract 9/10^∞. hell, 9/10^∞ isn't even a number. What you can do though is take the limit as n approaches infinity of 9/10^n and subtract that.
sigma(9/10^n) from {n = 0 to ∞} - limit(9/10^n) as {n approaches ∞}
Unfortunately for you the limit as n approaches infinity of 9/10^n is ZERO, ZEROZEROZEROZEROZEROZEROZEROZEROZEROZERO!
In Effect:
sigma(9/10^n) from {n = 0 to ∞} - 0 = sigma(9/10^n) from {n = 0 to ∞}
Nice try dumbshit.
Name:
Anonymous2007-02-21 17:10 ID:C+qsu6k9
people should seriously start sdudying maths, calculus and analisys before demonstrating shitty things
Name:
Anonymous2007-02-21 17:40 ID:5QdPT5h7
okay
it's like with the graph of a limit
it touches one, but it isn't one. it's so close it doesn't matter and it's not worth arguing about. If you physically produced something with that accuracy, chances are it would get quantum particles touching or something, and theyd jump around and switch places as particles will do, and it would be the same fucking thing anyhow.
>>37
It's not "close to one", it IS 1. End of story. If you don't like it, go post in a fucking philosophy board you useless asshole. The real numbers are defined as the set of limits of cauchy sequences of rational numbers, and the cauchy sequence of rational numbers {0.9, 0.99, 0.999, ... } converges both to 0.999... and 1. Hence, they are equal.
thats just plain wrong.. it clearly is 0.999.....
3+6 = 9
this repeats infinitely.. its not even hard to understand. Youre talking as if 0.333.. repeating infinitely is 0.333... + some fraction.
You have totally misunderstood infinity and math, even though you claim that youre the one who claims to understand it. Just gtfo.
1 - 0.99... = 0.000000000000...
1 = 0.99...
Name:
Anonymous2007-02-21 19:25 ID:xvBnJwcw
Does 0.000... = 0? 1 - 0.999... = 0.000..., and since 0.999... ≠ 1, 0.000... must be infinitesimally greater than 0. Probably represented as 0.000...1.
0.00... means 0 WITH INFINITELY MANY ZEROS AFTER IT, SO ITS EQUAL TO 0
YOU DUMB FUCKS ARE DRIVING ME CRAZY. IF YOU STILL DONT GET IT JUST GO LEARN MATH FOR A COUPLE OF YEARS AND COME BACK
Name:
Anonymous2007-02-21 23:27 ID:f1AaW16/
>>44
Wrong. 0.000... means 0 + 0.0 + 0.00 + ... + 0.0001 which is not equal to 0, dumb fuck. lern2math plz
Name:
Anonymous2007-02-22 1:22 ID:kjwB4oM4
>>40
0*10^n will always be zero, so you can add as many zeroes as you want to a number without changing its value (and therefore without changing it).
As for 0.000...1, that is a malformed number and has no meaning. Consider .123...4, where the ellipsis still represents an infinite number of digits. Each digit is multiplied by 10 to some n proportional to its distance and direction from the decimal point, so .123... can be represented as (1*10^-1)+(2*10^-2)+(3*10^-3)+.... What is n for the 1 in .000...1? The correct application of that hotel thing from earlier tells us that it can't be infinity.
Alternatively, I challenge you to write this number - write an infinitely long string of zeroes, and then put a 1 at the end of it. If you manage to do so, well, that wasn't a very infinite string, was it?
>>43
Troll. That crap where the teacher supposedly tried to create a number for the reciprocal of zero does nothing to resolve the conflict, since 0*(1/0) would equal 0*(nullity), which would still be required to equal both 0 and 1 by 0x=0 and n/n=1.
>>48
0x=0 for all x ∈ ℝ (ℂ, too, unless there's an exception I am missing.)
Name:
Anonymous2007-02-22 18:03 ID:f+L8Wtr8
This thread proves that math sucks.
Name:
Anonymous2007-02-22 23:03 ID:kjwB4oM4
>>50
This thread proves that the average person is not qualified to speak about math.
Name:
Anonymous2007-02-23 14:34 ID:qwl1UYde
0.9999recurring is just a different string for the same value. it just so happens that the decimal number system is crude when it comes to representing such values
Name:
Anonymous2007-02-23 15:50 ID:cjWS/YOi
>>52
There's nothing wrong with decimal. If you think there is something difficult about .999..., you just fail at math.
Name:
Anonymous2007-02-23 16:36 ID:+STHOy9U
>>53
I'm not >>52
Although it should be noted that decimals are just estimations. The rational numbers can be expressed as a fraction, and the irrationals can generally be expressed by some infinite series and then estimated to the digits necessary.
Name:
Anonymous2007-02-23 17:29 ID:PxOGa7sl
>>54
A decimal is not at all an estimation. Each one specifies a specific value. You can have a decimal specifying a value which ITSELF is an approximation (ie, 3.14159 ~= pi), but as I said, it is the value which is an approximation here, not the decimal - the decimal is specifying one single real number.
Name:
Anonymous2007-02-23 19:05 ID:cjWS/YOi
>>55
Amen. All real numbers unambiguously represent a single quantity.
Not uniquely, though. That's the whole point of this topic.
The fact that this is the classic math troll has already been established. Kindly go fuck yourself.
Name:
Anonymous2007-02-23 23:26 ID:MQuf7/CZ
>>55
A decimal is what thou dost experience with thine eye, but the number is not what thou dost experience with thine eye, but experience with thine mind.
BUT THE PROBLEM OCCURS CAUSE
1.000r = 0.999r + (1.000r-0.999r)
and 1.000r = 0.999r + (1.000r-0.999r) + (1.000r-0.999r) etc, onto infinity.
ALL NUMBERS ON THE REAL NUMBER LINE WELL END UP EQUAL TO EACH OTHER! SINCE EVERY REAL NUMBER IS EQUAL TO IT'S
INFINITELY CLOSE NEXT NUMBER, ETC.
I can't explain this well as I am not a mathematician, but real number line is epic fail!
Name:
Anonymous2007-02-26 7:09 ID:beEuLrzo
Oh and to clarify my post, I know that 1.000r-0.999r is infinitely small, but what happens when you add an infintely small number an infinite amount of times....that's right every number on the real number line is equal to each other....cats and dogs living together!
Name:
Anonymous2007-02-26 7:35 ID:beEuLrzo
1.000r = 0.999r so therfore 1.000r - 0.999r = 0
thus 1.000r = 0.999r + (1.000r-0.999r)
and 1.000r = 0.999r + 2 x (1.000r-0.999r), x3, x4, etc. onto eternity
BUT 1.000r = 0.999r + (1.000r-0.999r) x infinity
IF 1 / inifinty = 0, then 1.000r - 0.999r = 1 / infinity
SO 1.000r = 0.999r + infinity x (1 / infinity)
infinity x (1 / infinity), the infinities cancel out so you are left with 1 / 1
1.000r = 0.999r + 1 / 1
1 = 1.999r
1 = 2 WTF!!!!!!!!!!!
EVERY NUMBER ON THE REAL NUMBER LINE IS EQUAL TO EACH OTHER FAGGOTS!!!!!
Why the fuck is this thread still alive?
0.999.. = 1
End of fucking story. If you don't understand, go get a fucking education.
Name:
Anonymous2007-02-26 21:56 ID:e7Jw2Rtp
infinity divided by infinity is a degenerate form and can equal anything because guess what, INFINITY IS NOT A NUMBER. It is a concept that is used to understand things. Here's a short and informal proof of why infinity is not a natural number:
For inifinity equals the largest natural number,
Let w equal the very largest number of the natural numbers. Add one to w. w < w + 1. Therefore w is not the largest number. Therefore, any "number" that could actually be larger than itself is not a natural number.
For your information: 0/0, infinity/infinity, 1^infinity, 0^0, infinity - infinity and a whole host of others are degenerate forms and can literally be ANY number.
Name:
Anonymous2007-02-26 23:31 ID:Z64rdbNq
'Undefined' and 'can be any number' are two different things. 0^0, for example, is undefined because it has two different values by definition (0^x = 0, x^0 = 1). Infinity isn't a number, so things like infinity/infinity just simply aren't defined, though there isn't much difficulty evaluating 1^infinity - = 1*1*1*1*... = 1.
Proof:
some number "S" = lim(x->infinity) of (1+1/x)^x
(natural log) lnS = lim(x->infinity) ln(1+1/x)^x
(property of logs) lnS = lim(x->infinity) x*ln(1+1/x)
(x = 1/(1/x))lnS = lim(x->infinity) (ln(1+1/x))/(1/x)
(L'Hopital's rule)lnS = lim(x->infinity)((1/(1+1/x))*(-x^-2))/(-x^-2)
(reduce)lnS = lim(x->infinity)(1/(1+1/x))
(take limit)lnS = 1/(1+1/infinity)= 1/(1+0)=1
(exponentiate both sides) e^lnS = e^1
S = e
Degenerate forms can be many things even if it doesn't appear so. (Note: this does not prove the existence or value of e, it proves that the lim(x->infinity) of (1+1/x)^x = e)
Name:
Anonymous2007-02-28 3:21 ID:dOHNHbmO
>>83
You cannot raise 1 to the power of infinity, because infinity is not a number. Instead, you would ask what is the limit as n->infinity of 1^n. Clearly, 1^n = 1 for any integer n, so
lim(1^n,n->infinity)=lim(1,n->infinity)=1
FOr your response, 83, you never once demonstrated that 1^infinity = e. You seem to be assuming that you can freely interchange (1^infinity) and (1+1/infinity)^infinity, which you clearly cannot. As I discussed above, 1^infinity is simply impossible to calculate because infinity is not a number. Instead, you have to take the limit as x->infinity of 1^x, and I proved at the beginning that lim(1^x,x->infinity)=1.
It's not even a question of "can you do certain steps to get around the math." It's simply that you were wrong in your statement that "1^infinity can = e." It cannot. Ever. Ever. Ever.
Take some real analysis before saying any more. And if you have taken real analysis, either you did not pay attention or your professor sucked.
>>83
You idiot, "can equal" doesn't make any fucking sense. Equality is unique because it's transitive. It either IS EQUAL or IS NOT EQUAL. Take your "can equal" and shove it up your ass.
Name:
Anonymous2007-02-28 6:28 ID:T4uWJSOO
>>85
Oh what? I only just removed it from that very same place!
Name:
Anonymous2007-02-28 23:53 ID:gBFGW6L4
(direct substitution)lim(x->infinity)(1+1/x)^x = (1+1/infinity)^infinity
=(1+0)^infinity = 1^infinity
Because one (or for that matter any positive constant) over the "biggest number possible" (infinity) yields an infinitesimally small number that at the "point" (or limit) at infinity becomes zero for lim(n->infinity)(1/n). If you don't believe me, try to plug in really large number for the function 1/x. You will not achieve 0 until the "point" (or limit) at infinity. Plug in all the positive numbers you want, it will ALWAYS be a little more than 0 until the "point" (or limit) at infinity. Granted, you can not find this "point" on the real number line but the 1/x function has a horizontal asymptote that is only touched when you "reach" "infinity" (more commonly known as taking the limit as x approaches infinity) (lim(x->infinity)(1/x)) = 0 not something close to 0 but instead equals 0).
And yes it is "can equal" because 1^infinity is not a specific number just in the same manner that the variable "x" "can equal" (have a value equivalent to) 1, 2 or any other real number.
For instance 1^infinity = 169
some number "S" = lim(x->infinity) of (1+ln(169)/x)^x
(natural log) lnS = lim(x->infinity) ln(1+ln(169)/x)^x
(property of logs) lnS = lim(x->infinity) x*ln(1+ln(169)/x)
(x = 1/(1/x))lnS = lim(x->infinity) (ln(1+ln(169)/x))/(1/x)
(L'Hopital's rule)lnS = lim(x->infinity)(-ln(169)/(x^2+x*ln(169)))/(-1/x^2)
(reduce, 1st step)lnS = lim(x->infinity)(x^2*ln(169))/(x^2+x*ln(169))
(reduce, 2nd step)lnS = lim(x->infinity)(x*ln(169)/(x+ln(169))
(add and subtract the same quantity, 3rd step)lnS = lim(x->infinity)(x*2*ln(13)+4*(ln(13))^2-4*(ln(13))^2)/(x+2*ln(13))
(seperate)lnS = lim(x->infinity)((x*2*ln(13)+4*(ln(13))^2)/(x+2*ln(13))-(4*(ln(13)^2)/(x+2*ln(13))
(factor)lnS = lim(x->infinity)(2*ln(13)*(x+2*ln(13))/(x+2*ln(13))-(4*(ln(13)^2)/(x+2*ln(13)))
(reduce, 3rd step)lnS = lim(x->infinity)(2*ln(13)-(4*(ln(13)^2))/(x+2*ln(13)))
(take limit)lnS = 2*ln(13)-4*(ln(13))^2/(infinity+2*ln(13))=2*ln(13)-0=ln(169)
(exponentiate both sides) e^lnS = e^ln(169)
S = 169
As you can see, the degenerate form 1^infinity can be ANYTHING. It's just a little bit harder to do those that aren't common exponents of e.
Name:
Anonymous2007-03-01 2:04 ID:0yudiVxE
some number "S" = lim(x->infinity) of (1+ln(169)/x)^x
(natural log) lnS = lim(x->infinity) ln(1+ln(169)/x)^x
fail
\ln S = \ln \lim_{x \tendsto \inf} (1+\frac{\ln 169}{x})^x
Name:
Anonymous2007-03-01 6:13 ID:N/NM6Clu
actually you CAN use infinitesimals in math, but there are none in the real number set.
[9*(sigma from 1 to inf) 10^-x converges to 1]. This may seem like an approx, but it is not. When an infinite series converges to a number, it IS that exact number. This concept is used in calculus all the time for finding area under curves. By saying that 0.999~ is not = 1, you are basically saying that integral calculus is useless (because integration is based on summing up an inf number of rectangles, kind of like the inf series shown above).
>>98
The limit of the sum that produces the decimal expansion that is .999... is 1, and therefore they are one and the same. High school calculus teaches this simple concept, and calc trumps Alg II any day.