PDA

View Full Version : 1=0.



Pages : [1] 2

The_Hermit
08-05-2008, 09:15 PM
assuming that x and y represent the same nonzero number:

x=y
x^2=xy
x^2 - y^2=xy - y^2
(x+y)(x-y)=y(x-y)
x+y=y
2y=y
2=1
1=0

what just happened here?
I honestly don't know...

Thantos
08-05-2008, 09:19 PM
if x=y then x-y = 0
so in the third step you are dividing by 0

The_Hermit
08-05-2008, 09:37 PM
oh, wow. good catch.
Thanks, man.

indigo0086
08-06-2008, 06:11 AM
My friend showed me this in high school, he was good at math so he thought he could trick us and I found it out fast. I do remember his being different, though. It still had the divide by zero thing.

Thantos
08-06-2008, 07:35 AM
Yeah I've seen this one enough to immediately know it is dividing by 0.

QuantumPete
08-06-2008, 07:57 AM
Yeah I've seen this one enough to immediately know it is dividing by 0.

I think that's pretty much the only way to get 1=0. Undefined behaviour ;)

QuantumPete

abachler
08-06-2008, 11:36 AM
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy.


x=y

this is a statement of equality which makes x and y dependant variables.

x^2=xy

this is a conjecture of equivelancy which only holds true for the case of x == y

x^2 - y^2=xy - y^2

this further assumes the case that x == y. If x and y are independant this statement is false.


(x+y)(x-y)=y(x-y)

You are mixing general solutions with special case solutions and presenting the special case solution as a general solution which is erroneous.
y(x-y) = xy - y^2
(x+y)(x-y) = x^2 - xy + xy -y^2 = x^2 - y^2 not xy - y^2




x+y=y
this holds true only for the case of x = 0


2y=y
this is true only for the case of y = 0

the rest is nonsense as a result of faulty logic. makign a statement which holds true only for a single case and then applying it as if it held for all cases (false generalization,).
2=1
1=0

for 0<x<9 all odd numbers are prime. TRUE

since odd numbers are prime and 9 is an odd number 9 is prime. FALSE.

that is what you are doing, assuming that because a statement (xy = y^2) holds true under restricted conditons (x = y), that it holds true without those restrictions.

Sang-drax
08-07-2008, 07:39 AM
I think that's pretty much the only way to get 1=0. Undefined behaviour ;)
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1
sqrt(-1)*sqrt(-1)= 1
i * i = 1
-1 = 1
-1/2 = 1/2
0 = 1

Sang-drax
08-07-2008, 07:41 AM
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy.No, every step was correct except the step where he was dividing by zero.

indigo0086
08-07-2008, 07:49 AM
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1
sqrt(-1)*sqrt(-1)= 1
i * i = 1
-1 = 1
-1/2 = 1/2
0 = 1



still, it's clear to see that you're mixing equality with equivalence. Also that last step seems to just be the magic trick in the sea of misdirection.

QuantumPete
08-07-2008, 08:47 AM
sqrt(1)= 1

That's still cheating. :P But yes, you can do it without a divide by 0.

QuantumPete

Stonehambey
08-09-2008, 12:18 PM
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy....

Well of course, why do you think that most mathematical theorems and proofs start with "Given that...", "assume that..." or "Let us define...". In a mathematical proof, it is ok to define a set of initial conditions and then proceed logically assuming these are true, so long as you say so. So the fallacy in this classic 1=2 proof, is not where you suggest.

For example, Pythagoras' theorem says something along the lines of "Given a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the two remaining sides." What you appear to be arguing is that this would not be true if it was not a right angled triangle, well of course! That's why we define it to be one in the first place.

Similarly, it's perfectly ok to say that x^2 - y^2 = xy - y^2 in this case, since the first thing we did in this proof was say that x=y.

vart
08-09-2008, 10:09 PM
When you get to this step:

sqrt(-1)*
you get to the complex numbers - talking about complex numbers - you have 2 possibilities for the expression instead of one
you choose the wrong one to preserve the equility

abachler
08-13-2008, 05:43 PM
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1 wrong
sqrt(-1)*sqrt(-1)= 1 wrong
i * i = 1 wrong
-1 = 1
-1/2 = 1/2
0 = 1

sqrt(-1) = i
i * i = -1

cpjust
08-13-2008, 09:48 PM
Originally Posted by Sang-drax
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1 wrong
sqrt(-1)*sqrt(-1)= 1 wrong
i * i = 1 wrong
-1 = 1
-1/2 = 1/2
0 = 1
sqrt(-1) = i
i * i = -1

Actually, the first one is right (look at the parenthesis):
sqrt( (-1) * (-1) ) = sqrt( 1 ) = 1

The second one is where reality flies out the window since it uses the Product Rule for Radicals (http://algebra-tutoring.com/product-rule-radicals-1.htm), but the product rule for radicals says:

The nth root of a product is equal to the product of the nth roots. provided that all of the expressions represent real numbers.

Blackroot
08-13-2008, 10:46 PM
#define 1 0


Haha... HAHAHAHAHA! Take that LOGIC! /dance.

misterMatt
08-14-2008, 12:12 AM
What about the whole .999... = 1 'proof'?

http://en.wikipedia.org/wiki/0.999...

vart
08-14-2008, 12:30 AM
sqrt(-1) = i

sqrt(-1) = +i
and
sqrt(-1) = -i
as well

cpjust
08-14-2008, 08:20 AM
sqrt(-1) = +i
and
sqrt(-1) = -i
as well

Oh that's right! So in that case:

sqrt( 1 ) = sqrt( -1 ) * sqrt( -1 ) = (+i) * (-i) = 1
Problem fixed. :)

Stonehambey
08-14-2008, 10:00 AM
What about the whole .999... = 1 'proof'?

http://en.wikipedia.org/wiki/0.999...

There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.

As for the sqrt one. Remember that "square root" is not a function, since it yields to possibilities (maybe if you do a masters degree in maths you might come across a type of function which gives two answers, but for most people, it is sufficient to define a function as a mapping blah blah which has ONE element in the co-domain ;)). So we in fact start having trouble on the second line where it says

sqrt(1) = 1

what it should say is

sqrt(1) = 1 <-- AND -1 AS WELL!!


All the square roots do a good job at confusing you ;)

Has anyone seen the differential one? It's one of my favourites

x^2 = x + x + x + x +... (x times)

(d/dx)(x^2) = (d/dx)(x + x +...)

2x = 1 + 1 + 1 +... (x time)

2x = x

2 = 1

indigo0086
08-14-2008, 10:27 AM
cute. I'm glad I'm not in bed with mathematics.

vart
08-14-2008, 10:54 AM
you forgot to d/dx of the (x times) part

Stonehambey
08-14-2008, 05:21 PM
you forgot to d/dx of the (x times) part

Sorry, I just meant the (x times) bracket to mean that x squared is x added to itself x times, it's not actually part of the maths :)

@nthony
08-14-2008, 10:32 PM
There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.
I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

lim[n->inf] Σ9/10^i =
9 * lim[n->inf] Σ[i=1..n](1/10)^i =
9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
9 * ( 0 - 0.1 ) / -0.9 = 1

Therefore it's safe to say that the [i]limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).

vart
08-14-2008, 10:58 PM
it's not actually part of the maths
yes it is - you have


sum_0^x 1


so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x

Stonehambey
08-15-2008, 03:34 AM
I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

lim[n->inf] Σ9/10^i =
9 * lim[n->inf] Σ[i=1..n](1/10)^i =
9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
9 * ( 0 - 0.1 ) / -0.9 = 1

Therefore it's safe to say that the [i]limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).

lol, I told you people have a hard time accepting it ;)

x = 0.9999...

10x = 9.9999...

9x = 10x - x = 9.999... - 0.999... = 9

9x = 9

x = 1



yes it is - you have
Code:

sum_0^x 1

so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x

It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )

Kernel Sanders
08-15-2008, 04:01 AM
lol, I told you people have a hard time accepting it ;)

x = 0.9999...

10x = 9.9999...

9x = 10x - x = 9.999... - 0.999... = 9

9x = 9

x = 1

I never liked that one, because 9.999... has one more '9' than 0.999... has. It's confusing because we're dealing with infinity here, but you're still manufacturing an extra decimal place. I'm with @nthony that the limit of 0.999... = 1, but that does not mean that .999... = 1

**EDIT**
From the wiki link

There are many proofs that 0.999… = 1, of varying degrees of mathematical rigour. A short sketch of one rigorous proof can be simply stated as follows. Consider that two real numbers are identical if and only if their difference is equal to zero. Most people would agree that the difference between 0.999… and 1, if it exists at all, must be very small. By considering the convergence of the sequence above, we can show that the magnitude of this difference must be smaller than any positive quantity, and it can be shown (see Archimedean property for details) that the only real number with this property is 0. Since the difference is 0 it follows that the numbers 1 and 0.999… are identical. The same argument also explains why 0.333… = 1⁄3, 0.111… = 1⁄9, etc.

That does it for me better than anything ever has, but I still don't like it (I never liked dealing with infinity). The difference is 1/inf, which is defined as zero, but to me it's just another limit thing. The limit of 1/x is 0, but it still never gets there



It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )

Is it because that only holds true for positive numbers?

Stonehambey
08-15-2008, 05:17 AM
Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.

0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point :)

Kernel Sanders
08-15-2008, 05:27 AM
Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.

Well, 1/9 isn't actually equal to .111... because 1/9 is impossible to represent using base 10 if you get right down to it


0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point :)

Yeah - I can't say I'm convinced, but that doesn't mean I don't believe it's true. A lot of people who are a hell of a lot better at math than I have spent a hell of a lot more time reasoning about it and come to the conclusion that they are equal. I never was very comfortable with infinity and I guess this is just an artifact of that

cpjust
08-15-2008, 08:35 AM
In that case, what is 0.999...8 equal to? i.e. an infinite number of 9's with an 8 at the end. That probably doesn't make much sense mathematically and maybe even logically, but then again, if 0.999... = 1, then that means 1.000...1 = 1 also right? and if you subtract 1 from that it leaves 0.000...1. If you then subtract that from 0.999... you get 0.999...8.
Just because it can be "proven" mathematically, doesn't necessarily mean it's true. It's just true as far as we can currently prove. ;)

Shakti
08-15-2008, 09:13 AM
You asume there is an end, as shown in this statement/question: "In that case, what is 0.999...8 equal to? i.e. an infinite number of 9's with an 8 at the end."

However when you deal with infinity this is not the case.
0.999...8 equals 0.999...8 because it ends, thus does not continue infinitly.

Stonehambey
08-15-2008, 09:15 AM
In that case, what is 0.999...8 equal to? i.e. an infinite number of 9's with an 8 at the end. That probably doesn't make much sense mathematically and maybe even logically, but then again, if 0.999... = 1, then that means 1.000...1 = 1 also right? and if you subtract 1 from that it leaves 0.000...1. If you then subtract that from 0.999... you get 0.999...8.
Just because it can be "proven" mathematically, doesn't necessarily mean it's true. It's just true as far as we can currently prove. ;)

This makes absolutely no sense whatsoever. Think about what you're saying, "an infinite number of 9's, with an 8 at the end"

But thanks for the laugh :)

rudyman
08-15-2008, 09:26 AM
The number 1 can be written as 0.999... just like it can be written as 1.000...
With that kind of mentality, it's almost like shorthand for a limit:

The difference between 1 and 0.99 is 0.01.
The difference between 1 and 0.99999 is 0.00001
So as you get 0.999999 gets infinitely close to 1, the difference gets infinitely close to 0.

The same applies as 1.00000001 gets infinitely close to 1.

All three representations (1, 0.999..., and 1.000...) are mathematically equal, they're just different ways that humans have expressed the number 1.


Well, 1/9 isn't actually equal to .111... because 1/9 is impossible to represent using base 10 if you get right down to it

It's impossible to represent 1/9 as a number in base 10 unless you have an infinite amount of 1's, which is usually what's meant by the ellipsis.

cpjust
08-15-2008, 11:01 AM
This makes absolutely no sense whatsoever. Think about what you're saying, "an infinite number of 9's, with an 8 at the end"

But thanks for the laugh :)

Well I did say it doesn't make much sense didn't I? :D
I just invented (0.000...1) by subtracting (0.999...) from 1

Shakti
08-15-2008, 11:44 AM
but you still base your logic on the fact that 0.000...1 is a limited number, but it contains unlimited number of 0's so you would not be able to write a 1 at "the end" because there is no end, there are an unlimited amount of 0's.

cpjust
08-15-2008, 12:57 PM
but you still base your logic on the fact that 0.000...1 is a limited number, but it contains unlimited number of 0's so you would not be able to write a 1 at "the end" because there is no end, there are an unlimited amount of 0's.

I know you can't write a 1 on the end on paper, but I'm just doing it in my head (and using ...1 to represent it) -- basically the smallest possible non-zero number.

laserlight
08-15-2008, 01:00 PM
basically the smallest possible non-zero number.
But if there was a smallest positive real number, then the set of real numbers is countable.

cpjust
08-15-2008, 01:42 PM
But if there was a smallest positive real number, then the set of real numbers is countable.

There are an infinite number of numbers, regardless of whether they're natural, whole, integer, rational or irrational numbers, so I'm not sure what your point is?
If you could count forever, then it's countable. :D

laserlight
08-15-2008, 01:50 PM
There are an infinite number of numbers, regardless of whether they're natural, whole, integer, rational or irrational numbers, so I'm not sure what your point is?
My point is that if you give me the "smallest possible non-zero number", I can give you a smaller non-zero number, simply by dividing it by a number greater than 1. Clearly, there is no smallest non-zero number, or more accurately, there is no smallest positive real number. I assumed that you were aware of the concept of countable and uncountable sets, so I expressed my point indirectly.


If you could count forever, then it's countable.
If you can map a set (as in a one to one mapping, possibly with some natural numbers left unmapped as in the case of a finite set) to the set of natural numbers (or some other countable set), then that set is countable.

Blackroot
08-17-2008, 07:52 PM
You aren't dealing with a technical infinity (.99999999999->) VS 1.(000000->). Although a large technicaly 9 != 10 therfore .9 != 1. You can just say a number is *.(infity zeros)1* larger then it is defined to be. .999(infinity 9's) is completely seperate from 1.(zero is defined in mathematics as no value, if you have zero objects you have no objects)

-- Subtracting these two numbers does not equal zero nor does adding them equal 2 * either number. No matter what you say to be true about these, you cant disreguard the fact that .9(infinity) does not equal one it equals very close to one, but not 1. Much in the same way 1.1 != 1.

There's nothing to accept, the statement that both number are the same is false, prehaps in the ideogy of infinity it could be said to have merit, but in such a case you shoulden't be comparing infinity with a real number anyway, as infinity is a hypothetical for "A point of undefined value ~ Something with no end" VS a whole number, which ends immediately after it's decimal (1.) You cant subtract infinity from this number and get a logical result, it merely becomes (1. - infinity) So why would you get a logical result when you substract (1 - .9(infinity 9s)) You're still dealing with infinity, you'll still get an illogical result.

Slightly offtopic though, neither 1 = 0 nor .999 = 1 are true in the realm of reality. (.999 apples != 1 apple!) (1 apple != 0 apples!) So there, my long winded rant is over. Enjoy :P

cpjust
08-17-2008, 08:04 PM
My point is that if you give me the "smallest possible non-zero number", I can give you a smaller non-zero number, simply by dividing it by a number greater than 1. Clearly, there is no smallest non-zero number, or more accurately, there is no smallest positive real number. I assumed that you were aware of the concept of countable and uncountable sets, so I expressed my point indirectly.


If you can map a set (as in a one to one mapping, possibly with some natural numbers left unmapped as in the case of a finite set) to the set of natural numbers (or some other countable set), then that set is countable.

OK, let me start over... The number I was talking about with (0.000...1) is not a quantifiable amount. Maybe if I said (1/infinity) my point would be more clear, although after all this I think I forgot what my original point was anyways. :)

laserlight
08-17-2008, 11:19 PM
No matter what you say to be true about these, you cant disreguard the fact that .9(infinity) does not equal one it equals very close to one, but not 1.
That is rubbish: 0.999... and 1 are symbols. Proofs have shown that the numbers represented by these symbols are the same.


prehaps in the ideogy of infinity it could be said to have merit, but in such a case you shoulden't be comparing infinity with a real number anyway
This is a strawman argument since the proofs in question do not involve treating the concept of infinity as a real number.


Slightly offtopic though, neither 1 = 0 nor .999 = 1 are true in the realm of reality. (.999 apples != 1 apple!) (1 apple != 0 apples!)
By that reasoning 0.5 + 0.5 = 1 is false is the "realm of reality" as well, since adding half an apple and another half an apple does not give you an apple. In any case, note that we are talking about 0.999..., not 0.999. Part of your problem is inaccuracy of notation, which also leads you to postulate that strawman argument as mentioned.

abachler
08-17-2008, 11:52 PM
Well part of the problem comes from differeing definitions of infinity. In mathematical terms infinity is a fixed quantity that exists outside the set of real numbers. Most laymen think of it as a really large number'. Its not really large it is in fact unquantifiably large. It cannot be treated as so large that it simply obliterates all other values that are entangled with it, it must be treated as a unique quantity and kept serpate from the entangled terms, just as the imaginary number i must be treated as a seperate quantity, otherwise the maths break down.

5i * 5i yields 25i^2 = -25, not 25i

5 * infinity * 5 * infinity = 25 * infinity ^2, not 25 infinity or inifinty

5/0 = 5 * infinity not infinity

5/0 * 0 = 5

5*infinity * 0 = 5

If you do NOT treat infinity and hence division by zero as special unique quanities then the maths break down and in that illogical realm yes, you can prove any number to equal any other number, which is itself the proof that you must keep them seperate.

mike_g
08-18-2008, 07:03 AM
Most laymen think of it as a really large number'. Its not really large it is in fact unquantifiably large.
Lol, I would be a just a little bit surprised if I met anyone older than 10 that beleived that.

cpjust
08-18-2008, 08:02 AM
5/0 * 0 = 5

5*infinity * 0 = 5

Shouldn't those both equal 0? Anything times 0 equals 0 right?

matsp
08-18-2008, 08:03 AM
Shouldn't those both equal 0? Anything times 0 equals 0 right?

But isn't infinity large enough to change that ;)

--
Mats

@nthony
08-18-2008, 07:39 PM
infinity * zero is undefined as is 5/0. the limit of the latter is another story (inf).

cpjust
08-18-2008, 07:55 PM
infinity * zero is undefined as is 5/0. the limit of the latter is another story (inf).

Yeah, I remember my teachers always saying that x/0 is undefined, not infinity; but why is infinity * 0 undefined?

tabstop
08-18-2008, 08:06 PM
Yeah, I remember my teachers always saying that x/0 is undefined, not infinity; but why is infinity * 0 undefined?

Because infinity * 0 = infinity / (1/0) = infinity / infinity, which I think we already did.

cpjust
08-18-2008, 10:18 PM
Because infinity * 0 = infinity / (1/0) = infinity / infinity, which I think we already did.

Well using that logic: if x/0 = infinity and 0/x = 0 then


x * 0 = 0
0 x
because then the x's and 0's would cancel each other out and you'd be left with 0.