If you think you can make 2 = 3 or 1 = 3 or any other weird arrangement, post it up here. I start this thread because this arguement is going all over the place, it needs to be centralized.
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If you think you can make 2 = 3 or 1 = 3 or any other weird arrangement, post it up here. I start this thread because this arguement is going all over the place, it needs to be centralized.
i personnally don't agree with these two arguments:
vsusi:
(-1)^2=1^2
SRBS
((-1)^2)^(1/2)=(1^2)^(1/2)
(-1)^(2*1/2)=1^(2*1/2)
-1^(1)=1^1
-1=1
sql_lall:
only i is defined by its square, namely i^2=-1
but (-i)^2=-1
since the definition of a number squared=-1 is not unique
therefore invalid definition
Don't you just wish all the rejects on this forum who don't know what they're talking about and who are really just here to get attention by arguing would leave?
From what I've read you're one of the few people who understands basic maths in this place...
Well, I do agree with the first argument, largely because it is a circular argument. I can argue that it's wrong (and the same reasoning):
Say (-1) = (1)
Square both sides (valid) to get (-1)^2 = (1)^2. However, it would be wrong to not evaluate these values and still continue. Our next statment would be 1 = 1. It all has to do with one-to-many "functions" and many-to-one (or 1-1) functions. I can't really explain this much better, but we all know that we're applying rules to things where it doesn't apply.
As for the second argument, I kinda like it, if not from the statement that "y = x^2" is not unique, as for the same value of y, x = a or x = -a. To define i as i^2 doesn't really define i uniquely, so is it two things? Does this mean that i = -i, even though the two numbers: a + bi <> a - bi. Or does it?
Taking this a bit further, suppose i^2 = -1.
Does this imply i = -i?
Can you factor this to become 1 = -1?
Somehow, this is wrong - we know 1 <> -1, so what rule(s) does this invalidate, and where?
Destined
Personally, I come here (math forum) because I like to learn. This cannot be done without debate (ever taken any Philosophy?)Quote:
Originally posted by Masterbandit666
Don't you just wish all the rejects on this forum who don't know what they're talking about and who are really just here to get attention by arguing would leave?
From what I've read you're one of the few people who understands basic maths in this place...
However, I learn from people's examples, other people's mistakes, and most importantly, from my own. :)
Destined
i look at things this way:
If there is a word for it, it has it's place in society. Some of these things are good and bad. Personally, I think debate is a good thing. And the beauty of debate is that two people can be saying two different things, and neither one has to be right or wrong(in most cases). Just because people don't have the same views as you doesn't mean you should disregard it. The point I am trying to make is that I would be much happier to help people who don't know what they're talking about than listening to a fool, such as masterbandit, who complains when these people voice their opinions.
PS. I am suggesting that masterbandit, as well as others, change their attitude. If they want to remain as they are, that is certainly their choice and a dissapointing one.
They can in computer memory - you can have two variables assigned the same value, then printed in C:
output:Code:#include <stdio.h>
int main(){
unsigned int i;
int j;
j=0xffff;
i=0xffff;
printf("unsigned: %u signed %d \n",i,j);
return 0;
}
unsigned: 65335 signed: -1
And, no, I'm not asserting that 2=1. ---- People who don't know much about programming get all wrapped around the axle on code like this.
It was my viewpoint that the 2=1 thread, or whatever its called, was just a request for feedback on a certain fallacious proof that 2 = 1. NOT that it was meant to be serious, but more into the investigation of how people use misdirection in math to prove an impossibility.
It was not meant to be taken that the thread starter beleived 2 = 1.
At least, I hope not.
:(
-Lou
An interesting attempt by my grade 11 math teacher to make 0.99999999999999999999999999999 equal to one:
X = 0.99999999999999999999999999999
10X - X = 9
Therefore 9X = 9; X = 1.
Of course, the flaw with this idea (which I pointed out and he ignored, being proud of showing the class how smart he was) is that you get
09.999999999999999999999999999990
-0.999999999999999999999999999999
----------------------------------------------------
08.999999999999999999999999999991
The number of nines is just an arbitrary number.... it is my opinion that even if the number of nines were infinite, the proof would still be incorrect, because we cannot perform multiplications with numbers containing an infinite number of decimal places; it cannot be done.
lol. you'll know the idea of limits when you are taking calculus. it says that when the limit of x approaches c, of f(x) doesn't necessary equal to f(c).
but you can beleive what you want here. you can say its equal, you can say that it isn't. here is an interesting look at the question though:
there should be no real values expressable between 0.(9) (0.9 repeating) and 1. however let x be the greatest real <1. now the average of the two numbers (x+1)/2 is also real, and is greater than x. so there is no lagest real <1. assume 0.(9) is less than one contradicts with the fact that there is no largest real <1 and so 0.(9)=1.
bugz, don't bother.
Alph is in the camp where .9999... <> 1 no matter what you trow at him.
There are a few good threads here about the subject. I might dredge some up.
:)
-Lou
In order to perform said operation, you would require infinite computational power, since x is effectively infinitely difficult to calculate at all, let alone work with. This is like saying that you want to do a math question with all of the decimal places of pi... well, you can't have all of the decimal places of pi, it simply isn't possible. Since it is not possible to do your proposed question, I am forced to conclude that the resulting number (which would be 'larger') is effectively as much larger than x as infinity +1 is larger than infinity:D.Quote:
there should be no real values expressable between 0.(9) (0.9 repeating) and 1. however let x be the greatest real <1. now the average of the two numbers (x+1)/2 is also real, and is greater than x. so there is no lagest real <1. assume 0.(9) is less than one contradicts with the fact that there is no largest real <1 and so 0.(9)=1.
On a side note, I recognize the difference between two numbers being equal, and two numbers being so nearly the same that it may be best to treat them as though they were; the latter is certainly important, and should be done, but is not the same as the former:).
I was suddenly reminded of all of my physics tests. In the first year or two, you're usually looking for an exact solution, but once you hit upper level classes, you often have to make approximations or, as you said, you could be there forever.Quote:
Originally posted by Alphanos
... two numbers being so nearly the same that it may be best to treat them as though they were ...
My favourite was when the prof used (invented?) an "approximately proportional to" symbol. (proportional symbol with a tilde (~) over top.) The equation had gone on for about 2.5 classes, so we weren't bothered for being perfect. ;) I think we were quite skeptical of how accurate the solution was after that. :D hehehe..
Destined
I think some people misunderstood my previous post:
what I was trying to say was that this forum is kind of ruined by people who claim to know this that and the other, and don't. You're right, people come here to learn too, and they're not going to learn anything if someone else comes on here and says something like "2 = 3 and that's right because I said so". I've only been coming here for a few days now, and it seems to me that there are a few individuals who regularly post incorrect maths and claim it to be correct, and it's this that needs to go. People can't learn anything if they have other people aggressively arguing against each other about a topic. Maybe snakeeyes should lighten up, since he immediately assumed I was being derrogatory (spelt right?) whereas the contrary was actually the case.
Here's something along the way this thread is going, to keep people thinking. I brought this up in a lecture recently to provoke thought on how valid our numberline really is. Every positive number on the number line has two square roots right? The square roots of 16 are ±4, the square roots of 0.25 are ±0.5.
So what are the two square roots of '0.9' recurring? (In case someone doesn't know, that means 0.99999999999 and an infinite amount more nines.)
Bear in mind that 0.3 recurring NOT 1/3 (if it was equal then 3 * 0.3 recurring = 1, which it doesn't, 3*0.3 recurring = 0.9 recurring, not 1). That's a flaw in the number line as it is.
So without spending hours typing a massive post on why this that and the other, try defining the square roots of 0.9 recurring.
so what if 0.999 recuring is 1? isn't the problem solved? lol.
I would think that the square roots would be +/- itself, like 1. Saying that a number should be larger than 0.9 recurring and less than 1 is meaningless, so we must effectively treat 0.9 recurring as its own square root; its not completely correct, but the difference between the 'correct' answer and this is meaningless for our purposes.
lol i missed Alphanos' first post.
I think you misunderstood me. i am not trying to calculate x. what i am trying to say is that there isn't a real number that can be considered the largest number less than 1 (proven by contradiction).Quote:
In order to perform said operation, you would require infinite computational power, since x is effectively infinitely difficult to calculate at all, let alone work with.
here is the proof again:
assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1
but since 0.9 repeating is considered the largest real number less than 1 (since its physically not possible to write another another number that's greater than 0.9 repeating), then this belief also contradicts with the fact that there are no largest real thats less than 1. so i am concluding that 0.9 repeating is not less than 1 (proven again by contradiction). -- although this doesn't mean that 0.9 repeating is 1, but at least it's not less than 1
Well I can give you a part answer to my question:
ANY positive real number > 1 will give x2 > x.
ANY positive real number < 1 will give x{sup]2[/sup] < x.
ANY positive real number = 1 will give x2 = x
And since 0.9 recurring is the closest real number to 1, there is no value of x smaller than 1 which will give x2 = 0.9 recurring; also since 12 = 1, there is no value of x larger or the same as 1 which will give x2 = 0.9 recurring.
As such the square roots of 0.9 recurring are undefined on the number line.
So solve x2 = 0.9 recurring, proving your answer is correct (so not just saying "x = blah blah because it does and I'm right). You have to prove it.
You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.Quote:
assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1
Just 1 quick thing first:
As Destined Soul pointed out, this was just saying that if i was defined using "i2 = -1", then this has two values, i and -i.Quote:
sql_lall:
only i is defined by its square, namely i^2=-1
but (-i)^2=-1
since the definition of a number squared=-1 is not unique
therefore invalid definition
Anyway, both proofs to why 0.(9) = 1 are very good ones (both the square and average things.)
We are not really talking about "in any real life situation". Just like in real life Pi cannot be exactly found, but circles exist. And i cannot be found, yet i2 = -1 (:p)Quote:
You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.
You don't have to "make x = 0.9 repeated", it just is.
lol first i was never trying to make x=0.9 repeating. but then again, I don't see that it can't be done. you are saying that 0.9 repeating is not 1. so why you can't make x=0.9 repeating? I can make x=pi can't i? I can work with x=i+7, can't I? so why can't I make x=0.9 repeating? However, just like you know, you can't work with infinity because it is not a number (while i is a complex number, pi is a real number, so is 0.9 repeating)
lol I proved that should be no real number thats closest to 1 that's less than 1. so 0.9 recurring must be equal to 1 :DQuote:
Masterbandit: And since 0.9 recurring is the closest real number to 1
Does a number have to be written down in order to exist?Quote:
Originally posted by Alphanos
You can't do that. I think you misunderstood me; while I understand what you are saying, it cannot be done. How do you make x = 0.9 repeated in any actual real life situation, in any form? This is like saying that you cannot have infinity because infinity +1 is bigger... it doesn't really work.
Even though we cannot ever thoroughly calculate PI, Does not PI
Exist independantly?
Why is an endless series of Nines following a decimal point
impossible?
Certainly, we can conceive of an endless number of nines trailing
a decimal place. We can represent it, with either a properly
defined summation formula, or by .999...
Since there is no such physical thing as a number, can any
number be said to exist in "actual real life"? Of course!
All numbers exist CONCEPTUALLY. So, Numerically, if you
can conceive it with justifiable properties, and it operates on
verifiable rules, then it exists. And I believe that .999... exists,
and is equal to 1.
At least, Thats IMHO,
;)
-Lou
well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1. Even though the error is infitesimly small over infinity it adds up.
therefore if you do it enough (initinely) you could prove that 1 = 0
EDIT: I know that is not possible however with the logic used here it is.
Let a = 1
Let b = 1
Therefor
a = b
a² = ab
a² - b² = ab - b²
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1
2 = 1
lol a-b=0. in math you can't divide by 0 (not that you ever could).
Gandalf:no you are mistaken. we are talking about .9999 repeating. not .99999Quote:
well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1. Even though the error is infitesimly small over infinity it adds up.
therefore if you do it enough (initinely) you could prove that 1 = 0
EDIT: I know that is not possible however with the logic used here it is.
I'm trying to figure out a way in which to make my point clearer, because I can see I was misunderstood.
I am not saying that you have to be able to write down a number to know that it exists, but that you have to be able to write it down in order to use it in an evaluated equation. pi and infinity exist, but they cannot be used in equations that evaluate to a single defined answer. No, we do not use the number pi in equations; we may use 3.1415926 (or more decimal places), but this is only an approximation, because we cannot use the true value of pi because it is too complex. In order to use the true value of pi in an equation, you must write down an infinite number of digits, and it would take an infinite amount of time. Thus we can never perform equations with pi that require a single defined solution; we can either do only algebra with pi, never evaluating what it is, or we can approximate pi. This is the same situation as with 0.999 repeated. We cannot use the true value of 0.999 repeated, because it truly is infinitely close to 1. There can be no larger value than it that is less than 1. This is because its semi-infinite nature excludes it from being used in equations where we need a single exact number as the solution. We can approximate the number, introducing error, or perform algebra with it, like pi. But we cannot solve problems with it. In this case, approximating it introduces so much error relative to the desired precision of the answer (infinite precision) that the equation becomes meaningless. If I asked you to calculate the area of a circle with a radius of 1 centimeter accurate to the 100 trillionth decimal place, you could not do it. The level of precision we would require to evaluate (1+0.9 repeated)/2 is infinitely greater than this.
While you have a theory stating that 0.9 repeated cannot be the largest number less than 1, it cannot be proven, because the necessary procedure to get the result is impossible to do. You may be right, you may be wrong, but you cannot use such an equation as proof, since the amount of error introduced in any attempt is infinite relative to the required accuracy.
If we ignore equations and talk purely in theory, then your proof really makes no sense because there is an infinite number of 9s after the decimal point. It has to be the largest numer less than one, because it is infinitely close to 1.
Conceptual values can only be used when you need algebraic results; otherwise an approximation is necessary.
MartinLiss isn't dividing by zero; its done algebraically until the values are substituted in at the end. This has left me boggled, because I fail to see something wrong with this math.
lol. first, dividing an variable assumes that it is not 0.
here is something:
x(x+1)=5x
if you divide by x, you'll get x+1=5, which can be simiplified to x=4.
however, we divided by x, which assumes x!=0. As you can see, x=0 is another solution to this equation. when you divide by a-b, we also assmue a-b!=0. but as Martin clearly pointed out in the beginning, a=b=1, so we can't deivide by a-b because it has a value of 0.
I proved that ANY REAL NUMBER cannot be the largest number less than 1. It is proven. I've given the proof. What you are talking about a) makes no sense to me and b) does not relate to my proof in any wayQuote:
While you have a theory stating that 0.9 repeated cannot be the largest number less than 1, it cannot be proven, because the necessary procedure to get the result is impossible to do.
my proof is accurate because algebra is used. why you cannot calculate 100 trillionth decimal place of pi, but pi itself (the algebraic symbol) is understood to be EXACT. that's the power of algebra, you don't need to round off in order to find an answer.Quote:
You may be right, you may be wrong, but you cannot use such an equation as proof, since the amount of error introduced in any attempt is infinite relative to the required accuracy.
any math student would tell you that 5pi is not the same as 5*3.1415926, because the former is EXACT, latter is NOT.
[/quote]
If we ignore equations and talk purely in theory, then your proof really makes no sense because there is an infinite number of 9s after the decimal point. It has to be the largest numer less than one, because it is infinitely close to 1.[/quote]
my proof proved that there is no largest number less than one. it is a legal proof. does anyone else think that my proof is wrong, and if so, give me the reason? Alphanos, I don't believe what you are arguing here is of any relavence.
btw Although we can't approximate accurately up to a trillion decimal places of pi, wouldn't you agree that we don't need to? I'd be surprised if the most sophisticated project would need pi up to 500th decimal place (of course other than solely determine a closer approximation of pi). What i am trying to say, the approximation is close enough to be used in almost any situation, unlike what you described of the huge error margins.
Starting with the identity of a = b, then (a-b) is zero. When you divide it from both sides, then you are dividing by zero.Quote:
Originally posted by Alphanos
MartinLiss isn't dividing by zero; its done algebraically until the values are substituted in at the end. This has left me boggled, because I fail to see something wrong with this math.
bugz, please try to read the post more carefully, I don't really think I can explain it in another way more clearly:(.
You have to use ONLY algebra, with NO solving for real numbers if you use conceptual values. If you use an approximation, as is done in almost all uses of conceptual values, the type of proof you're using becomes invalid, because the difference between the approximation and the conceptual value is infinitely more than your desired answer.
Both pi and 0.99 repeated are theoretically real numbers. But you cannot use them as such because their real value is infinitely difficult to work with.
I find this to be completely logical and without fault, although overly cautious.Quote:
Originally posted by bugzpodder
here is the proof again:
assume x is the largest real number less than 1
so the average of x and 1 (1+x)/2 is greater than x but less than 1
this contradictions with the assumption
so there isn't a largest real number less than 1
but since 0.9 repeating is considered the largest real number less than 1 (since its physically not possible to write another another number that's greater than 0.9 repeating), then this belief also contradicts with the fact that there are no largest real thats less than 1. so i am concluding that 0.9 repeating is not less than 1 (proven again by contradiction). -- although this doesn't mean that 0.9 repeating is 1, but at least it's not less than 1
C'mon, jump off the cliff! You'll bounce!
;)
if .999... is not less than 1, and certainly its not greater than 1, then I'm 99.999...% sure its equal to 1, unless it doesn't exist.
:)
1:
Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)Quote:
well with your logic .99999 = 1 so that would mean .9999999 .... 8 = .9999999. So 1 = .99999999 .... 8
if you extrapolate further 1 = .999999 ... 7 = 1.
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1
2:
Ok, answer this. You have circle 1 of diameter D1, and circle 2 of diameter D2, What is the ratio of the area of circle 1 to circle 2??Quote:
Both pi and 0.99 repeated are theoretically real numbers. But you cannot use them as such because their real value is infinitely difficult to work with.
The area of the big circle = Pi*D12
The area of the little circle = Pi*D22
The ratio of the areas = (Pi*D12)/(Pi*D22)
= D12/D22
Now, surely this is using pi in a real situation?? AND using its exact value (in this case, "Pi" stands for the greek letter Pi, which itself stands for the EXACT value of pi)
i am also talking about .999 repeated. If .99 repeated is equal to one then .9999 repeated with an 8 on the infinited decimal point is equal to .99 repeated which is equal to one. If done an infinite number of times you could prove that any number is equal to one with NotLKH's logicQuote:
Originally posted by bugzpodder
No you are mistaken. we are talking about .9999
repeating. not .99999
but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds upQuote:
Originally posted by sql_lall
1:
Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1
bugz, you used pi algebraicly only, with no evaluating in that case. Just the same, if x= the conceptual 0.9 repeated, you COULD say that 2x-2x = 0x = 0 because the number that you use for x is irrelevant. To restate, conceptual values can be used a) algebraically or b) with approximations instead of the real infinitely complex value.Quote:
Ok, answer this. You have circle 1 of diameter D1, and circle 2 of diameter D2, What is the ratio of the area of circle 1 to circle 2??
The area of the big circle = Pi*D12
The area of the little circle = Pi*D22
The ratio of the areas = (Pi*D12)/(Pi*D22)
= D12/D22
Now, surely this is using pi in a real situation?? AND using its exact value (in this case, "Pi" stands for the greek letter Pi, which itself stands for the EXACT value of pi)
You got a ratio of 4/2, which has no relation to the real value of pi, because it was canceled out.
I fail to see how you can look at these two situations differently! If 0.9 repeated is irrational, and there is NO last digit, then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!Quote:
Now, we were talking about 0.(9), which =0.99999999......etc (there is no end to the number)
However, you mention 0.99999...8 This number does have last digit, it is rational, so you can't use the same proof that you did with 0.(9) and 1
lol i didn't post that message. you got your names wrong Alphanos!
but exactly, as you yourself pointed outso what is it equal to? it can't be greater than 1. it can't be less than 0.9 repeating. so its either 1, 0.9 repeating or undefined.Quote:
then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!
undefined won't make any sense.
so if its 1 or 0.9 repeating, it would indicate that the three numbers are the same, like:
if the average of 4 and x is 4, what is x? you would say 4 wouln't you?
the truth is, if there is indeed a difference between two numbers x and y, their average should be in between them. however as indicated, this would happen iff there isn't a difference between x and y, so x=y, or in other words, 0.9 repeating =1
i understand that conceptual numbers can be represented algebraically or can be used as an approximation, but i don't see how it is relevant to anything we've talked about. btw i am representing those conceptual numbers algebraically, not using an approximation, in case thats what u r saying.
Gandalf:Quote:
but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds up
you cannot have .999...8 where there is an infinate number of 9's. you can't just say at the end of infinity I will add a 8, because there is no end to infinity!!! That was what I am trying to say. sorry about the bold, i was just illustrating my point, or in NotLKH's words:IMHO, lol!
Bugz, As you can see, the level of argument when it comes to .999...
lacks here at vbf.
I mean, they misquote you, Gandolf says I have a proof here, when I have None?!?
***?
lol...
-Lou
:p
Gandy,
And, WHAT are you talking about??? My Logic Can you point it out???Quote:
Originally posted by Gandalf_Grey_
i am also talking about .999 repeated. If .99 repeated is equal to one then .9999 repeated with an 8 on the infinited decimal point is equal to .99 repeated which is equal to one. If done an infinite number of times you could prove that any number is equal to one with NotLKH's logic
I apoligize for misquoting you bugz, I wasn't expecting someone else to start aruging, so I thought that it was you:o.
First things first, you should understand that Gandalf is not saying that there can be a number with an infinite number of 9s followed by an 9. He's saying that there can't be!!! His entire argument was to show how stupid it is to say that (0.9 repeated+1) / 2 is greater than 0.9 repeated. If you could pretend to evaluate that, I would expect you to get 0.9 repeated with a 5 at the end (thus the greater part). This makes no rational sense, because the number of 9s is INFINITE.Quote:
lol i didn't post that message. you got your names wrong Alphanos!
but exactly, as you yourself pointed out
quote:
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then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!
--------------------------------------------------------------------------------
so what is it equal to? it can't be greater than 1. it can't be less than 0.9 repeating. so its either 1, 0.9 repeating or undefined.
undefined won't make any sense.
so if its 1 or 0.9 repeating, it would indicate that the three numbers are the same, like:
if the average of 4 and x is 4, what is x? you would say 4 wouln't you?
the truth is, if there is indeed a difference between two numbers x and y, their average should be in between them. however as indicated, this would happen iff there isn't a difference between x and y, so x=y, or in other words, 0.9 repeating =1
i understand that conceptual numbers can be represented algebraically or can be used as an approximation, but i don't see how it is relevant to anything we've talked about. btw i am representing those conceptual numbers algebraically, not using an approximation, in case thats what u r saying.
Gandalf:
quote:
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but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds up
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you cannot have .999...8 where there is an infinate number of 9's. you can't just say at the end of infinity I will add a 8, because there is no end to infinity!!! That was what I am trying to say. sorry about the bold, i was just illustrating my point, or in NotLKH's words:IMHO, lol!
Since you obviously agree with this, this must mean that you agree that (0.9 repeated+1)/2 is not greater than 0.9 repeated. I definately agree that it cannot be less than 0.9 repeated.
Since we have agreed that it can be neither greater than or less than 0.9 repeated, we have two options: either the result IS 0.9 repeated (which clearly does not obey the laws of math), or the question cannot be evaluated. I stated this a while ago.
lol. ok, I understand your position, but i am going to repeat my position again.
.(9) (just gonna stand for .9 repeating to make my life easier) has some "very special" properties, such as it has no square roots, and (.(9)+1)/2 is undefined (not something over 0 -- take the word undefined in the face value).
but it is precisely as I have told you, these two properties does NOT make any sense! for example, no one said you can't divide by 2 (or for that matter, any number that's not 0!) if the number is (1.(9))! or you can't take the sqrt(x) if x is .(9)!
as my proof indicated, x<(x+1)/2<1 if x<1
in terms of an algebraic sense, lets look at two graphs:
y=x, y=(x+1)/2
you have to agree that by solving for the point of intersection,
it will be (1,1).
these two equations are linear functions, which means a line [with no holes].
now how could that be? didn't we say that (.(9)+1)/2 is undefined?
my point is, saying (.(9)+1)/2 is undefined is saying that my original proof that has algebraically proved that x<(x+1)/2<1 if x<1 is wrong!
if you look over my original proof, there is absolutely nothing wrong with what I did or my reasoning, except you claim that 0.(9) does not fit this case.
do you understand of the proof by contradition?
you assume something is true, show that a contradiction exists, and therefore proving your assumption to be false.
what you are doing here however, is claiming something to be true and show that it exempts from properties of real numbers! that is pretty absurd to me.
I did a search for this topic, and came up with every website saying (actually i only visited around 4) .(9)=1
here is their proof:
1/3 = .(3)
2/3 = .(6)
so 1/3+2/3=.(3)+.(6)
so 1=.(9)
in case you claim that 2/3 is not .(6) because you typically see .(6) as .666666...6667 on a calculator. i want to say that as you told me your self, .(6) is a conceptual number, and .666...6667 is a close approxmation to .(6).
in case you are still not convinced,
1/3+1/3+1/3=.(3)+.(3)+.(3)
3*(1/3)=3*0.(3)
1=.(9)
now the only way you are getting out of this proof is to say 1/3 is not .(3) is that what you are saying?
and by the way, you seem to have believe that .(9) is not 1. can you offer any proof of any other source that says the same thing (like a professional website, hopefully university website)
this site was one of the 4-5 sites i visited, it tells a very interesting story ;) :
http://reflectiveeclectic.org/tales_of_truth.htm
this is a sweet proof:
http://mathforum.org/dr.math/faq/faq.0.9999.html
i am not saying that the 8 is at any defined decimal place it is just at infinity. Maybe you aren't getting this specific reference however when i say .999 ... 8 i mean a number infitesimally smaller than .999 .... An infitesimally small number is possible (the difference between .99 ... and 1).Quote:
Originally posted by bugzpodder
you cannot have .999...8 where there is an infinate number of 9's. you can't just say at the end of infinity I will add a 8, because there is no end to infinity!!! That was what I am trying to say. sorry about the bold, i was just illustrating my point, or in NotLKH's words:IMHO, lol!
you can't have a number infinitely small -- when it is actually "infinitely small" it is 0! btw it doesn't make sense to have a 8 at "un-defined" decimal place as you implied. infinity is not a number so you can't have infinity minus 1 nines and a 8. my point is that their is no such thing as .9999....9998 where there is an infinity number of 9s.
the difference between .(9) and 1 is 0. there is no difference. that explains why the number you implied does not exist.
By saying .999..., then the ... symbolically means the established pattern of that which precedes the ... should be understood to continue, with no deviation, forever. Now, the ... in your statement: ".999...8" cannot mean the same thing as the ... in ".999...", because you are requiring a termination in the established pattern. Now, all you have proven, IMHO, isn't that .999... doesn't exist, but that .999...8 doesn't exist.Quote:
Originally posted by Gandalf_Grey_
i am not saying that the 8 is at any defined decimal place it is just at infinity. Maybe you aren't getting this specific reference however when i say .999 ... 8 i mean a number infitesimally smaller than .999 .... An infitesimally small number is possible (the difference between .99 ... and 1).
You are trying to prove or disprove something by creating a
seperate concept, and use it without proving or disproving the
existence of that second concept. You create some association
between the two concepts to build a statement that is
recognizably false, which you then use as verification that the
first concept does not exist. However, what the falsity of that
statement actually proves is that something used to build that
statement does not exist, but it does not point out which concept
is false.
Certainly, if .999... is nonsense, and does not exist, then .999...8
does not exist.
However, .999... can exist, while .999...8 doesn't, which would
return the same results.
So, Prove to me that .999...8 exists upon the condition
that .999... exists, then you can use it as a keystone to build your
arguement. Then, if your arguement produces a statement that is
recognizably false then you will have proven that .999... does not
exist.
-Lou
I'm curious. These proofs are nice and simple, but I wonder if they do hold.
BugZ, you had a link to a page (Dr Math one) that uses a limit in their proof, but does the limit of a function actually ever reach that value? Just because you say that 1/x approaches infinity as x -> 0 from the right, does this mean that AT infinity, x = 0?
I guess my real problem at the moment in completely accepting the fact, is can you say that .(3) + .(6) = .(9) ? Is it an exact answer, or an approximation? No matter how far towards infinity you go (in decimal places), that nth digit will still not allow the number to be exactly one. The limit approaches one, yes, but only that. Approaches.
I really do like the little proof of 1 = 3(1/3) = 3(.(3)) = .(9), but again I'm wondering if math is being applied in the wrong way. If 1 <> .(9) (IF), then either 1/3 <> .(3) or 3(.(3)) <> .(9) (I wonder if, again, there's an approximation occuring that we're overlooking.)
Destined
yes i can assure you they do hold.
Quote:
but does the limit of a function actually ever reach that value?
we are evaluating a sum of a infinite series, namely 9/10+9/100+9/1000+...
because without the x grows out of bounds, it won't be 0.(9)! because there is an infinite number of 9's so you need to have x to grow out bounds thing.
just like e are defined.
e=lim (x->oo) of (1+1/x)^(x)
does that answer your questions (since your questions are all what i've explained)?
here is a post about limits and such: they all came from the same website i gave above
this next story is funny:Quote:
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.
this one makes sense:Quote:
you're _really_ thinking of
0.999...999
which is not the same thing. You're absolutely right that 0.999...999
is a little below 1, but 0.999999... doesn't fall short of 1 _until_
you stop expanding it. But you never stop expanding it, so it never
falls short of 1.
Suppose someone gives you $1000, but says: "Now, don't spend it all,
because I'm going to go off and find the largest integer, and after I
find it I'm going to want you to give me $1 back." How much money has
he really given you?
On the one hand, you might say: "He's given me $999, because he's
going to come back later and get $1."
But on the other hand, you might say: "He's given me $1000, because
he's _never_ going to come back!"
It's only when you realize that in this instance, 'later' is the same
as 'never', that you can see that you get to keep the whole $1000. In
the same way, it's only when you really understand that the expansion
of 0.999999... _never_ ends that you realize that it's not really 'a
little below 1' at all.
Quote:
Another way to approach the question is to subtract.
1.0000000....
- .9999999....
--------------
0.0000000....
Sure looks equal to me. What about the "1" at the end, I hear you ask?
Well, I'll write it as soon as I finish writing infinitely many 0s.
Any decimal place you name (say, the four billion three hundred
twenty-eight million two hundred seven thousand four hundred
ninety-fifth) has a 0 in it. A number with a 0 in every decimal place
is certainly 0.
Hehe.. silly me. :p Yes, now I see what you mean.Quote:
Originally posted by bugzpodder
we are evaluating a sum of a infinite series
I guess the confusion is between a purely-infinite series (0.(9)) and a finite series. The finite series will never reach 1, but the infinite series will. I guess this is another feature of infinity that makes one gasp in its awe. :D
I think the link I was talking about should have the statement that .(9) IS an infinite series.
Knowing this, you can't apply the 0.9999...8 method, as it's a finite number.
Destined
lol, here is an article about 1.999...98. close enough to .9999...98 which is what we were talking about:
http://mathforum.org/library/drmath/view/53387.html
http://www.math.toronto.edu/mathnet/...econd1eq2.html
this is a very interesting proof of 1=2 (of course, it is wrong)
i never caught the error. (i had to try almost every step to find the correct one) y don't you guys take a look.
other such "proofs" are in the middle of this page: http://www.math.toronto.edu/mathnet/falseProofs/
i find it to be rewarding
Is it the first step? I know it's before Step 4, since 1/i = -i, not +i.
For step 1, they say 1/-1 = -1/1. This can be rearranged as (1)(1) = (-1)(-1)
From here on, we get into the arguments that we've been going over for quite some time.
Or, as I am sometimes, am I wrong?
[edit]: Yup... I'm wrong, but not too far off, really. ;)
Destined
well in case you don't know, you can click on the step number and it will tell you on which step an error occured.
*sigh* You don't seem to understand my point of view. I think we'll just have to agree to disagree. I frankly don't care how many websites you can find that agree with you, what I'm interested in is the logic. I fully understand why it is that you think your logic works, but I find it contradicted by my logic regarding the infinite nature of (0.9). You seem to not like the concept of infinity and attempt to define or approximate it, but you really can't do that.
1!=(0.9) because even though there is an infinite number of nines, there is an infinitely small difference between 1 and (0.9). Yes, this is insignificant in 99.999% of the cases where the comparison is made, but this is the case where it does matter. You cannot prove that 1=(0.9) by saying that you don't care about the infintessimally small difference.
On a side note, (0.3)!=1/3; (0.3) is our approximation. Fractions cannot be converted to decimals when they have an infinite number of decimal places.
We cannot use in evaluated math numbers that contain an infinite number of digits. This IS a contradiction in math, and I would think you would know that contradictions exist in math and many other subject areas. Perhaps you've heard of Godel's Incompleteness Theorem. If not, a brief overview can be found here. Godel's Incompleteness Theorem proves that in all areas of knowledge, there are problems that are unsolvable without additional knowledge not contained within the boundaries you are working in. While I have no specific examples of others using this in this particular case, I have not looked since I think logical argument is more important than the number of people who agree with you anyway. I believe that this case of numbers not being usable in math is an application of this theorem.
I will attempt, however erroneously it may be, that the value of .999... doesn't exist at all!
1/9 = .111...
(1/9)*(9/1) = 1
.999...=.111...*9
.999...=(1/9)*(9/9)=1
This is yet another occurence where the English Language fails us. I'm not sure that it can be explained perfectly in any language. We can agree that .999... is definitely .111...*9. But it is actually 1. So .999... does not exist, because infinity is a concept, not a value or measure.
good point Alphanos. Godels incmpleteness theorum has a surprising tie to computer science. In CPU (june 2002) Pete Loshins article "The Nature of Computer Security: An interdisciplinary Approach" outlines Godels theorum relative to computer security. If anyone wants it i could scan it and email the jpegs to them
I agree .999 ... doesn't exist, it [EDIT] Has a has an infinite number of decimal places[/Edit] , and since infinity is just a concept niether exist outside the realm of human thought. No human can comprehend infinity. Alphanos and i were explained the notions of infinity by a teacher who went to a conference on infinity. The infinity which humans think we can comprehend is what is called the countable infinity. That is just a real number that happens to be exessivly large. True infinity though cannot even begin to be comprehended so i guess we just label it as non existant. Sort of like the universal perspective vortex (i think thats the name of it) in Douglas Adams Hitchhikers Trilogy. I must refer to fiction here mainly because it gives a good example about how the human mind cannot grasp the nature of infinity. Anyways the universal perspective vortex would allow a human to grasp how insignificant they were relative to the universe in all of its infinity. In the story it was regarded as the worst possible way to die. :p.Quote:
Originally posted by snakeeyes1000
I will attempt, however erroneously it may be, that the value of .999... doesn't exist at all!
1/9 = .111...
(1/9)*(9/1) = 1
.999...=.111...*9
.999...=(1/9)*(9/9)=1
This is yet another occurence where the English Language fails us. I'm not sure that it can be explained perfectly in any language. We can agree that .999... is definitely .111...*9. But it is actually 1. So .999... does not exist, because infinity is a concept, not a value or measure.
I am not really making a point in this post other than just agreeing with bugzpodder and rembling on about infinity :p
I never said infinity doesn't exist, I said .999... doesn't exist.
I am saying that infinity doesn't exist. You cannot write down infinity because it can only exist inside the realm of human thought. The sideways 8 is just a symbol suxh as pi which details something that no human can fully comprehend. When i say infinity doesn't exist i mean it is not something any person can comprehend to its fullest.
Ok, obviously you disagree with the proofs because you think infinitely small doesn't equal to 0. or in other words:
lim (x->0) x !=0 (now you see, it is x aproaches 0, not x=0 -- in your words, there is an infinitely small difference between x->0 and x=0)
and you also don't believe that 1/3=.333...
because of some wierd reason thats to me is completely could not believe (like 1/3 cannot be translated into a fraction because it is an approximation even though .333... stands for .3 repeating).
and also you believe that infinity exists (which i do so too, but i believe that it is a concept and not a value/number)
now show me a proof of 1 and 2. does anyone ELSE believe you? give me a professional source (hopefully a university site) where it clearly indicates that either lim (x->0) x != 0 or that
1/3=.3333... I want this because I don't believe you are a professional mathematician that specializes in number theory. well you don't have to be one, unless you show me another person's proof or your own proof that may have been published in a professionally recognized publication then I'll believe you.
in short, you offered me no proof while you dismissed my proofs (which has been largely recognized as a valid proof) as rubbish without a proper reason.
By the way the incompleteness theorem just indicated that such a case may exist. as I said, no one has offered me any proper reason as to why .999... != 1, other than .999... is only an approximation even though it clearly is not.