0*infinity=(0+0+0+0...)=0 (x+0=x x=0)
1=0.999... (agreed) 1-0.999...=0
(type 1) 1-0.999...=(1/10)*(1/10)*(1/10)... (from division by 10, 1-0.9 etc)
ab=0 if a or b=0; 1/10=0 ;10*1/10=10*0 ;1=0 ;x*1=x*0 ;x=0
(type 2) (1/10)*(1/10)*(1/10) ...=(1/10)^infinity=1/infinity=0
1/infinity=0 then 0*infinity=1 bu 0*infinity=0 ;1=0 ;x*1=x*0; x=0
Every number equals zero .(by 1=0.999...)
>>6
We're all talking about real numbers. If you want to take another system, be my guest. It's like arguing that 1 + 1 = 0 because it's true in the field with two elements.
Name:
FrozenVoid2007-03-01 12:18 ID:fPuNCwa3
>>14
Only in binary.I'm always using real number of the decimal system.
If you like you can substitute infinity by N,in any of the lines and they still
hold.
>>15
Infinity isn't a real number. If you're using the real number system you can't just go "LOL MULTIPLY BY INFINITY" because Wikipedia states some OTHER SYSTEM allows multiplication by infinity, any more than you could say "1 + 1 = 0" just because some OTHER SYSTEM allows this.
Furthermore, 1+1 = 10 in binary you uneducated inbred mouthbreather.
Name:
FrozenVoid2007-03-01 14:17 ID:fPuNCwa3
>Infinity isn't a real number.
If i can use it as variable,its fine.Else variable N(infinitely large)
. If you're using the real number system you can't just go "LOL >MULTIPLY BY INFINITY" because Wikipedia states some OTHER >SYSTEM allows multiplication by infinity, any more than you could say >"1 + 1 = 0" just because some OTHER SYSTEM allows this.
Of course i can,wikipedia can.So do i.
>Furthermore, 1+1 = 10 in binary you uneducated inbred mouthbreather.
1 carries over.see the next digit.
Name:
FrozenVoid2007-03-01 14:22 ID:fPuNCwa3
If you read post >>4
You can clearly see the section title.here i'll help you. http://en.wikipedia.org/wiki/Infinity#Operations_involving_infinity_and_a_real_number_x
Operations (equations)
Involving (containing numbers)
Infinity (concept of arbitraly large number)
and (also containing)
a (noun prefix)
Real (reals as in fracitonal decimal,with dot.)
Number (abstract representation of quantity)
X (a variable)
>>18 >>17
"Infinity is not a real number but the extended real number line adds two elements called infinity"
Extended real number line != real number line. GTFO Troll
Name:
FrozenVoid2007-03-02 1:35 ID:+sP42Nst
Whats the matter,you can't handle infinity being in equations with real number X?
Something wrong with you?
Wikipedia uses infinity with real numbers.
I use them with real numbers.
Only sad idiots can't use it because their religion doesn't allow it.
"I can't use infinity in calculations because God forbids it"
>>20
Listen, I'm explaining this one last time: The real numbers and the extended real numbers are TWO DIFFERENT THINGS. You can't say "wow look, A is true in the extended real numbers so it must be true in the real numbers too!" which is EXACTLY WHAT YOU ARE DOING. Once again, it's like saying 1 + 1 = 0 in the real numbers because 1 + 1 = 0 in F2. Of course, you probably don't even know what a field is because you're a useless uneducated cunt. FOAD.
I'm really surprised that no one has called him on being unable to argue without Wikipedia's cock in his mouth.
Name:
FrozenVoid2007-03-02 5:01 ID:+sP42Nst
The problem is that mathfags assume they can forbid equations with infinity.While they are commonly used.
Its like forbidding to use powers of 2.Really stupid.
Using the same logic of this thread here's the proof of 3=5
3*0=5*0 => 3=5
Name:
Anonymous2007-03-03 13:47 ID:EPaNpJoA
>>1
1/10=/0 (lim(n->infinity)1/(10^n)=0)
1=/0
let x = 1
x = x^2
x-1 = x^2-1
(x-1) = (x-1)*(x+1)
1 = x+1
0 = x = 1 But
1-1 = (1-1)*(1+1)
0 = 2+0
0/0 = (2+0)/0
1 = 2
0 = 1
you pulling the same fail math as above.
>>31 >>32
These examples are not disproving my argument,they are
completely absurd and only purpose is equate the argument to non-sense.
Please grow up and drop your stupid analogies that make no sense whatsoever.There is no single line of my argument so far,you didn't
disprove.Let me refresh the course.
0*infinity=(0+0+0+0...)=0 (x+0=x x=0)
Adding infinite zeros is equivalent to adding zero.Case closed.
1=0.999... (agreed) 1-0.999...=0
Equation Assumed to be true,be we show the absurdity of it in the following transformation. Difference between equals is zero
(type 1) 1-0.999...=(1/10)*(1/10)*(1/10)... (from division by 10, 1-0.9 etc)
The difference between "equals" is also a infinite series
ab=0 if a or b=0; 1/10=0 ;10*1/10=10*0 ;1=0 ;x*1=x*0 ;x=0
The multiplication results is zero,only is of one of the members is zero.
The only member is 1/10 so it equates zero.Following shows we proof now every number equals zero. Absurd? As 0.999...=1
(type 2) (1/10)*(1/10)*(1/10) ...=(1/10)^infinity=1/infinity=0
The series can be represented by (1/10)^n (n which grows to infinity)
converted to 1/infinity And assumed to be equal to zero)by 1-0.999...).
1/infinity=0 then 0*infinity=1 bu 0*infinity=0 ;1=0 ;x*1=x*0; x=0
This result is in contradiction with first (see line with 0*infinity=0) bringing
1 in euality with 0,proving that:
Every number equals zero .(by 1=0.999...)
Name:
Anonymous2007-03-03 15:34 ID:EPaNpJoA
For your type one argument:
Correct:
1-.099999... = 0
Incorrect:
lim(x->infinity)(1/(10^x))=something not 0
Correct:
lim(x->infinity)(1/(10^x))= 0
Informal proof:
plug in numbers for f(x)=1/(10^x)
x f(x)
1 .1
2 .01
3 .001
4 .0001
5 .00001
... ...
"infinity" 0
As x approaches infinity, f(x) approaches the x-axis and when x "meets" infinity, f(x) "meets" the x-axis and because the x axis is the line y = 0 and f(x) = y then f(x) = y = 0 = x axis and therefore f(x) = 0. Keep in mind that no organism (even if they had billions of years or more to do it) can count to infinity by integral values; it's that far away. Yet, by 5, f(x) is already one hundred thousandth and by 100, f(x) is one googleth. By the time one would "reach" infinity, f(x) ceases to exist and is zero.
As for second type:
Incorrect: infinity/infinity must equal 1.
Correct: infinity/infinity can equal any real number.
proof:
lim(x->infinity)(3*x/x)
direct substitution yields: infinity/infinity (n multiplied by infinity is infinity) which if x/x = 1, then lim(x-infinity)(3*x/x) should equal one.
reduce: lim(x->infinity)(3)(allowable since 3x/x = 3 except for x = 0, but we're heading away from zero, not towards it).
take limit:lim(x->infinity)(3)= 3
3 =/ 1, the limit of the function 3x/x appears to have two answers yet f(x) is a function and cannot have two y-values defined for any x-value ("infinity" in this case). Which limit is it? Well f(x)=3x/x equals f(x)=3 for all numbers except x=0 and plugging x's into 3x/x yields
x f(x)
1 3
2 3
3 3
4 3
5 3
6 3
... ...
I wonder if it'll be 3 at infinity or 1?
infinity/infinity = 3 in this case.
Name:
FrozenVoid2007-03-03 15:45 ID:hCm1D25G
lim(x->infinity)(1/(10^x))=something not 0
Where did you find those limits? I didn't mention them.
Where do you reach zero?its infinitely away."As x approaches infinity"
DId you just say limit=result?
Incorrect: infinity/infinity must equal 1.
Where did you find this? I only mention 1/infinity=0 (becuase of 1-0.999..=0) and 0*infinity=0 which both prove (by a/b=c then cb=a)
that 1=0.
Name:
Anonymous2007-03-03 16:37 ID:EPaNpJoA
>>35
when you said: >>1
1-0.999...=(1/10)*(1/10)*(1/10)... (from division by 10, 1-0.9 etc)
I intepreted this to mean that:
1-.9 = .1
1-.99 = .01
1-.999 = .001
1-.9999 = .0001
1-99999... = .0000...
such that the pattern is 1-.9999... = lim(x->infinity)(1/(10^x))
Becuase if you meant that:
1-.9 = 1/10
1-.99 = 1/10
1-.999 = 1/10
1-.9999 = 1/10
etc.
Then you need to use a calculator or something because 1-.99 =1/100.
Also, you "reach" zero at the same point at which you "reach" infinity: the infinitly far away. Any positive real number put into 1/(10^x) will return a real number (albiet one very small real number) but when infinity goes into 1/(10^x), you get 0 because the infinitly small is considered to be nonexistant (take the smallest thing possible, cut it in two, what's left? another "smallest thing possible" or nothing? if the former, then you didn't start with the "smallest thing possible").
>>35
"Incorrect: infinity/infinity must equal 1.
Where did you find this?"
right here:
"1/infinity=0 then 0*infinity=1"
because these are the steps you have to take:
1/infinity = 0
(1/infinity)*infinity = 0*infinity
1*(infinity/infinity) = 0*infinity
(next step is not always true)
1*(1)=0*infinity
1=0*infinity
Name:
Anonymous2007-03-03 18:42 ID:/6tL6/pY
What number exists between 0.999... and 1 on the real number line?
>>33
Stop being fucking absurd
"0*infinity=(0+0+0+0...)=0 "
Again, INFINITY IS NOT A FUCKING NUMBER, even if we consider the extension of reals, 0*infinity is indetermined. Your loved wikipedia will show it to you http://en.wikipedia.org/wiki/Infinity
1=0.999... (agreed) 1-0.999...=0
Equation Assumed to be true,be we show the absurdity of it in the following transformation. Difference between equals is zero
"ab=0 if a or b=0; 1/10=0 ;10*1/10=10*0 ;1=0 ;x*1=x*0 ;x=0
The multiplication results is zero,only is of one of the members is zero."
Total failure, ab are just 2 elements, and you have infinite of them, and no, you cant group infinite elements..."
"1/infinity=0 then 0*infinity=1 bu 0*infinity=0 ;1=0 ;x*1=x*0; x=0"
RIIIIIIIIIGHT, again, infinity is not a number, and, even if we consider a extension of reals, infinity/infinity is indetermined.
>>37
if we take .999... to mean a decimal point followed by an "infinite" amount (perhaps it's better to say, uncountable amount) of nines afterwards then the answer to the question:
"what number is between .999... and 1?" is none. Or rather there exists no distinct number between .999... and 1 because they are the same number. Just like there are no numbers between 0 and .000... because they become the same number.
>>38
that number would actually be less than both 1 and .999... since both .999...0 and .999... have the "same" amount of digits (infinitly so for both). If one was subtract the former from the latter, one would get .000...9 which is greater than than the subtraction of 1.000...0 and .999...9 which yields .000... and therefore .999...0 is farther from one than .999... is. You would somehow have to create a number that has more digits than infinity in order to create a number between 1 and .999...
Name:
Anonymous2007-03-04 21:02 ID:McstwcYw
>>41
"(perhaps it's better to say, uncountable amount)"
No, it's not better, because uncountable means something different.
>>41
>If one was subtract the former from the latter, one would get .000...9
.000...9 is not a number. That expression makes no fucking sense. You can't write something *after* an infinite number of digits, there is no after, THERE IS NO END.
If one was to subtract the former from the latter, one would get ZERO, ZIP, NADA, SAGE THIS FUCKING THREAD DIE DIE DIE
Name:
Anonymous2007-03-05 21:14 ID:KqA9OQxe
In my very first class, on my very first day in university, the professor had everyone in the entire lecture hall chant in unison, "Infinity is not a number."
Best thing he ever taught us.
Also, I just noticed the Wikipedia page "Proof that 0.999... equals 1" has been merged into the 0.999... page. That's kindof boring. In any case, what with FrozenVoid liking Wikipedia so much, how has no one linked this yet?
http://en.wikipedia.org/wiki/0.999...
>In mathematics, the recurring decimal 0.999… , which is also written as 0.\bar{9} , 0.\dot{9} or \ 0.(9), denotes a real number. Notably, this number is equal to 1. In other words, "0.999…" represents the same number as the symbol "1".
Name:
GMontag2007-03-07 2:43 ID:tGho8tNT
>>42
>No, it's not better, because uncountable means something different.
Would 0.999... with aleph_null 9s be a different number than 0.999... with aleph_one 9s?
Name:
Anonymous2007-03-07 4:23 ID:ObWbNNpv
>>47
.999... with aleph-one 9s doesn't really make sense. Aleph-zero is the cardinality of N; it seems obvious that the number of digits in .999... is of similar size.
>>47 Would 0.999... with aleph_null 9s be a different number than 0.999... with aleph_one 9s?
As >>48 said, there aren't aleph_one 9s. They're countable, look:
0.9 9 9 9 9 9 9 ...
---1 2 3 4 5 6 7 ...
There isn't a different number with aleph_one 9s or a way of writing it with aleph_one 9s; the decimal expansion is by definition just a sequence of numbers, hence any decimal expansion has countably many digits.
Name:
Anonymous2007-03-09 11:52 ID:4Rr10EF4
Let x=.99999..
Then 10x=9.999..
So 10x-x=(9.999..)-(.999..)
Then 9x=9
Thus, x=1.
GG.
Name:
Anonymous2007-03-11 0:43 ID:eKtBbXdh
1/1 = 1.
1/.999... = 1.000...
10/9.999... = 1.000...
Therefore, 1 = .999...
There is no .000...0001 at the end of .999... because the nines go on infinitely.
Name:
Anonymous2007-03-14 4:17 ID:MsSYv7X+
you assume at the top that .999 = 1 its the only way your formula will work
>>58
How do you chart a single number? Because that's what 0.999... is. It isn't a fucking sequence you useless shithead.
Name:
Anonymous2007-03-15 22:28 ID:WPqjQ+gn
.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 to infinity will eventually (lol) reach the number 1.
It's the same sequence as .999... (decimal expansion), but with a0 = 3, a2 = 1, and all other a's as zero. The limit of that sequence is 3.01000..., which is what you're looking for.
>>64
Tell me, do the sequences {1/2, 3/4, 7/8, 15/16, ... } and {2/3, 8/9, 26/27, ... } approach the same value? If so, why do you think that the sequences {0.9, 0.99, 0.999, ... } and {1.0, 1.00, 1.000, ...} do not? The article you linked specifically states that a number is equal to an infinite sum of the form Sum[i=0,+inf](a_i/10^i). And, ironically, it specifically links the article about proving 0.999... = 1!
>>68 >>69
K, so I reply to >>64 and get two responses from obviously different people. Please try to determine which of you morons I'm addressing before rushing to hit the reply button next time.
>>71
Rereading >>64 I can see you weren't really disagreeing with me, sorry. Nevertheless, you mostly missed my point. Although you can take the sequence of successive one-digit approximations to a number and graph it as a sequence of course, I was objecting to people saying "graph 0.999... and see that it approaches 1 but never equals it", primarily because this kind of statement is just a fancy way of hiding behind the "omg numbers are like, processes, dude!" that so many deniers use. Also because it's silly to say "graph a number as a sequence" when there are infinitely many sequences which converge to any number.
Note that 1/Infinity is not equivalent to Infinity*0 . If the second were true, it would have to be true for every x, and, by transitivity of the equals relation, all numbers would be equal. This is what is meant by being undefined, or indeterminate.
Straight from the same Wikipedia page. Way to go, OP.