The secret to dividing by 0!

[S=k.logW 12]

You know when you're fooling around on the calculator and you get a strange error. You either tried to find the square root of -1 or divide by 0

 

The square root of negative one is just the letter i if you were wondering , but you see I think that the reason you can't divide by 0 is a simple problem with the way dimensions are perceived. Try representing a line with one point; it doesn't work!

 

Don't understand? Don't worry...it's very complicated. However the solution is simple if you realize what is actually happening.

 

I believe that when you divide by 0 you are essentially adding a dimension! Points become lines, lines become planes, and so on.

 

I probably wrong so don't get your hopes up. :comeatus:

[Knight Hadron 372]

Just as I said earlier on a different thread, I have a response, but it'll have to wait until after I get back from a concert. For starters, think about division by zero as the 3rd dimension of the complex plane. Similar to a quaterion maybe. Also this, by Sasq: http://science.mistu.info/Math/Numbers/Creative_numbers_and_division_by_zero.html/. Just to get you started .

[Thorn O Discord 118]

Just as I said earlier on a different thread, I have a response, but it'll have to wait until after I get back from a concert. For starters, think about division by zero as the 3rd dimension of the complex plane. Similar to a quaterion maybe. Also this, by Sasq: http://science.mistu.info/Math/Numbers/Creative_numbers_and_division_by_zero.html/. Just to get you started .

 

I really like that guy's idea of creative numbers.

[Sigma 768]

I tried this once. Allow me to crush your hopes and dreams.

 

Let's say that the definition of this non real number division by zero dividend is defined as g=1/0.

 

From this, we can tell that 2g=2/0, 3g=3/0, and so on, each having their special place on a line perpendicular to the real number line.

 

However, there will be a point of intersection where a number will be both a part of the real number line and a part of the new line we created. This number will be 0g=0/0=0.

 

Now, another thing we can derive from this definition is that 0g=1, since by multiplying both sides of g=1/0, we simply multiply by zero on one side, and cancel out the division by zero on the other.

 

So, if 0g=0 and 0g=1, then 0=1.

 

What about making the point of intersection 0/0=1? Well, multiplying by any non-one number on both sides will lead to another contradiction. For instance:

 

0/0=1

(0*0)/0=1*0

0/0=0

0=1

 

Sorry, but you can still break math with this method.

Edited April 4, 2015 by Sigma

[Brony Number 42 10,069]

How do I insert symbols, preferable Latex? 

 

Consider the formal definition of a limit

lim (x to c) f(x) = L means

For each e > 0 there exists a d > 0 such that if

0 < |x-c| < d

then

| f(x) -L | < e

 

Does there exist L = lim (x to 0) 1/x ?

For every e

|1/x - L| < e there exists |x| < d. So we are taking a region of size 2d around x=0, and we want |1/x| to be within e of L. But |1/x| becomes arbitrarily large as x gets arbitrarily close to 0, thus |1/x| cannot be with e of any limit.

 

Also, sqrt(-1) = + or - i.  And a good way to represent complex numbers is to use polar notation

 

i = exp(i pi/2) 

[Nerdy Luigi 2,065]

I think you're thoughts have done goofed. Dividing by zero is an impossibility for the logical reason that zero is nothing. You can't divide something among nothing.

 

#I<3math

Edited April 5, 2015 by TwillyFSniper

[Snuggle Fiend 135]

Wow. This forum has everything 0.0

[IcyHaze 594]

Well, dividing by zero is undefined.

If you twll a man to enter a room and take 0 items from the room, and.told him to do that as much as he liked, potentially he could enter 0 times or an infinite number of times - it wouldn't matter, as nothing would change.

I think that's as simplistic as this gets.

[Knight Hadron 372]

@@IcyHaze, then, how do you explain sqrt(-1)? It's "illogical" by your thinking. Sometimes mathematics is just like that.

 

The rest of you: I'll write up responses in the morning.

[GenderIsAnIllusion 2,177]

Mathematicians have argued over the answer to dividing by zero since its implementation as a number.

 

If you were to graph "1/x" on a coordinate plane you would find that at x=0 it's either infinity, or negative infinity, depending on where you are approaching from on the plane. We call this undefined, because we cannot define will any certainty which one is correct, or if both are equally correct, or neither.

[IcyHaze 594]

@@IcyHaze, then, how do you explain sqrt(-1)? It's "illogical" by your thinking. Sometimes mathematics is just like that.

 

The rest of you: I'll write up responses in the morning.

I never mentioned the word "Illogical" at all :3

 

Besides, sqrt(-1) is an imaginary unit which is simply used to solve equations - but it cannot be a solution itself. It can be manipulated, but it itself is not truly a real number.

[Brony Number 42 10,069]

 

 

Besides, sqrt(-1) is an imaginary unit which is simply used to solve equations - but it cannot be a solution itself. It can be manipulated, but it itself is not truly a real number.

It is a solution: sqrt(-1) = +,- i

It is no more or less "real" than any other number. It is used in sinusoidal functions. Does the number 1 exist in the real world any more than i? It is in the Schrodinger operator, so quantum mechanics uses it. You cannot describe AC electricity without it, because sin(t) implicitly uses i. You ask me to show you 1 apple. I ask you to show me 1 alternating current. Thus i exists just as much as 1.

[Knight Hadron 372]

@@BronyNumber42, Do you think that division by zero could be the 3rd dimension on the complex plane?

 

 

I tried this once. Allow me to crush your hopes and dreams.

 

Let's say that the definition of this non real number division by zero dividend is defined as g=1/0.

 

From this, we can tell that 2g=2/0, 3g=3/0, and so on, each having their special place on a line perpendicular to the real number line.

 

However, there will be a point of intersection where a number will be both a part of the real number line and a part of the new line we created. This number will be 0g=0/0=0.

 

Now, another thing we can derive from this definition is that 0g=1, since by multiplying both sides of g=1/0, we simply multiply by zero on one side, and cancel out the division by zero on the other.

 

So, if 0g=0 and 0g=1, then 0=1.

 

What about making the point of intersection 0/0=1? Well, multiplying by any non-one number on both sides will lead to another contradiction. For instance:

 

0/0=1

(0*0)/0=1*0

0/0=0

0=1

 

Sorry, but you can still break math with this method.

Let me disprove your argument. Let's say that, like you say, g=1/0. Now, 0g does not equal 1. g=1/0, 0g=0*(1/0). But 1/0 is a solid number, a constant. It is in its simplest form. Just like there are wierd rules for exponets and i, there are rules for division, multiplication, and 0. New ones. 1/0=some constant (or infinity depending on what you want). So 0g is simply 0 1/0's, or 0. 0 * g = 0.

 

So, assuming 0/0=1, which I don't know how you got there but whatever.

0/0=1

0*1=0*(0/0)

0=0, because 0*g is 0, as I have shown.

 

You can't break math, you just have to extend math.

How do I insert symbols, preferable Latex? 

 

Consider the formal definition of a limit

lim (x to c) f(x) = L means

For each e > 0 there exists a d > 0 such that if

0 < |x-c| < d

then

| f(x) -L | < e

 

Does there exist L = lim (x to 0) 1/x ?

For every e

|1/x - L| < e there exists |x| < d. So we are taking a region of size 2d around x=0, and we want |1/x| to be within e of L. But |1/x| becomes arbitrarily large as x gets arbitrarily close to 0, thus |1/x| cannot be with e of any limit.

 

Also, sqrt(-1) = + or - i.  And a good way to represent complex numbers is to use polar notation

 

i = exp(i pi/2) 

Sorry, haven't studied limits yet. Still stuck in Geometry with my head past calc. So, I can't disprove your argument. Yet.

Edited April 5, 2015 by Knight Hadron

[Brony Number 42 10,069]

 

 

So, assuming 0/0=1, which I don't know how you got there but whatever.

Right there your math fails. You can't make statements that you can't prove. Division by 0 does not create new dimensions. Infinity is an interesting topic. There are different sizes of infinity. The smallest is the countably infinite. An example of this are the number of integers. There are just as many odd integers as there are total integers (odd and eve) because they are both countably infinite. Then there is uncountably infinite. An example of this is a continuum, or the number of points in the interval (0,1) which is the same as the number of points in (0,\infinity). There are other infinities found by taking super sets of a previous set of infinite size.

[Knight Hadron 372]

Right there your math fails. You can't make statements that you can't prove.

That, is why I was weary about assuming it. The post that I was quoting and disproving assumed it, and I was showing the error in the maths that he did, not his underlying assumtions.

[Sigma 768]

Let me disprove your argument. Let's say that, like you say, g=1/0. Now, 0g does not equal 1. g=1/0, 0g=0*(1/0). But 1/0 is a solid number, a constant. It is in its simplest form. Just like there are wierd rules for exponets and i, there are rules for division, multiplication, and 0. New ones. 1/0=some constant (or infinity depending on what you want). So 0g is simply 0 1/0's, or 0. 0 * g = 0.

 

So, assuming 0/0=1, which I don't know how you got there but whatever.

0/0=1

0*1=0*(0/0)

0=0, because 0*g is 0, as I have shown.

 

You can't break math, you just have to extend math.

 

So, you're saying that it's impossible to get 0g=0(1/0)=0/0? If that's the case, the it's impossible to get 2g=2(1/0)=2/0 as well. The ability to treat the division like a fraction, and then multiply by the numerator is an essential part of the system that would require quite the replacement should you remove it.

 

There are two ways to get 0/0=1. Either make it a defining characteristic in the system in an attempt to remove the paradoxes, or derive it like so:

 

0g=0(1/0)=0/0, because of what I said above in red.

0g=0(1/0)=1, because division, being the inverse of multiplication, should get cancelled out by multiplication.

If 0g=0/0, and 0g=1, then 0/0=1 because of the transitive property.

 

Multiply by any nonzero number in this situation, and you'll reach a paradox:

0/0=1

(0/0)*1=1*1

0=1

 

Agreed (Except for the statement claiming that you can't break math. Math can be broken if you use a system that leads to paradoxes). I'm not saying division by zero is impossible; just that more work needs to be done. Making division by zero an exception to certain rules is fine as long as those rules aren't something of foundational importance such as multiplication and division being inverses of each other. Aversion when representing i in radical form is a thing, yet complex numbers are still a widely accepted system.

Edited April 6, 2015 by Sigma

[Twilight Dirac 721]

The reason you can't divide by zero is simple, there exist no number that satisfies the definition.  1/0 can alternatively be written as n such that n*0 = 1.  But because multiplying zero by anything gives zero, there are no numbers n that will satisfy the equation.

 

At best you can talk about the limit of 1/x as x approaches zero, an even this is unbounded and taken as infinity.

[DATA EXPUNGED 324]

The reason you can't divide by Zero is because of the relation between Multiplication and Division.

Take, for instance, the following example. 12 / 6 = 2 because 2 (6) = 12

This is the same for all division and multiplication problems.

However, trying to do this with 0 gets you this:

12 / 0 = x because x(0) = 12

0 multiplied by any value will always be 0.

No value for x will satisfy this equation. 

 

 

 

You know when you're fooling around on the calculator and you get a strange error. You either tried to find the square root of -1 or divide by 0

 

 

To the OP, you really shouldn't put an exclamation mark beside the zero in your title, because I initially thought you were saying 0 factorial, which is 1. 

Edited April 6, 2015 by DATA EXPUNGED

[Admiral Regulus 2,769]

In simplest terms possible, x/0 is known as a point of singularity. You can read all about that here.

 

http://en.m.wikipedia.org/wiki/Singularity_(mathematics)

[Knight Hadron 372]

The reason you can't divide by Zero is because of the relation between Multiplication and Division.

Take, for instance, the following example. 12 / 6 = 2 because 2 (6) = 12

This is the same for all division and multiplication problems.

However, trying to do this with 0 gets you this:

12 / 0 = x because x(0) = 12

0 multiplied by any value will always be 0.

No value for x will satisfy this equation. 

 

So, you're saying that it's impossible to get 0g=0(1/0)=0/0? If that's the case, the it's impossible to get 2g=2(1/0)=2/0 as well. The ability to treat the division like a fraction, and then multiply by the numerator is an essential part of the system that would require quite the replacement should you remove it.

 

There are two ways to get 0/0=1. Either make it a defining characteristic in the system in an attempt to remove the paradoxes, or derive it like so:

 

0g=0(1/0)=0/0, because of what I said above in red.

0g=0(1/0)=1, because division, being the inverse of multiplication, should get cancelled out by multiplication.

If 0g=0/0, and 0g=1, then 0/0=1 because of the transitive property.

 

Multiply by any nonzero number in this situation, and you'll reach a paradox:

0/0=1

(0/0)*1=1*1

0=1

 

Agreed (Except for the statement claiming that you can't break math. Math can be broken if you use a system that leads to paradoxes). I'm not saying division by zero is impossible; just that more work needs to be done. Making division by zero an exception to certain rules is fine as long as those rules aren't something of foundational importance such as multiplication and division being inverses of each other. Aversion when representing i in radical form is a thing, yet complex numbers are still a widely accepted system.

Both of you don't think of division by zero as another number system, but rather as an extension of our integers. Like the complex numbers, i=sqrt(-1), and i is constant. You can still say sqrt(-2), but we just say 2i for simplicity. Same here. g=1/0. g is a constant, it is not an integer. No integer value is equal to 1/0 -- you have to invent a new value. Your proof that 0/0=1 is wrong because you assume that 0(1/0)=1. 0*(1/0)=0*g=0, not 1. 1/0 is the simplest form of a division-by-zero number. I'm still struggling to get solid rules with this, but I can at least say that 0*(1/0) can not cancel out.

 

12/0=x, 12=0*x, 0=12, is what you say. But, using g: 12/0=12*(1/0)=12g=x, and x has g in it. g is a new type of number, so your saying that "no number satifies x" is false, at least in my construction, because I made a new number to solve this paradox.

[Brony Number 42 10,069]

 

 

12/0=x, 12=0*x, 0=12, is what you say. But, using g: 12/0=12*(1/0)=12g=x, and x has g in it. g is a new type of number, so your saying that "no number satifies x" is false, at least in my construction, because I made a new number to solve this paradox.

This is wrong. You are trying to define a nonsense number, g = 1/0 is the same as saying g = (rainbow dash)/(pinkie pie), it's not defined. 

 

12/0=12*(1/0)=12g=x

 

You cannot make the substitution 12g = x because, by your own definition, (1/0) is a self contained object. You cannot operate on just the 1 and thus you cannot make the substitution 12*1/0 = 12/0. It is like saying 

 

sin x / n = six = 6

 

The symbology 1/0, as you said yourself, is not an operation but a constant. You can't have it both ways. You can't say it is an object then say it is an operation.

 

 

 

but I can at least say that 0*(1/0) can not cancel out.

Yet this is exactly what you do with 12*1/0 = 12/0

[Knight Hadron 372]

This is wrong. You are trying to define a nonsense number, g = 1/0 is the same as saying g = (rainbow dash)/(pinkie pie), it's not defined. 

 

12/0=12*(1/0)=12g=x

 

You cannot make the substitution 12g = x because, by your own definition, (1/0) is a self contained object. You cannot operate on just the 1 and thus you cannot make the substitution 12*1/0 = 12/0. It is like saying 

 

sin x / n = six = 6

 

The symbology 1/0, as you said yourself, is not an operation but a constant. You can't have it both ways. You can't say it is an object then say it is an operation.

 

 

 

Yet this is exactly what you do with 12*1/0 = 12/0

I understand what you are saying, but why can't I do 12*1/0=12/0? Some operations are prohibited with division by zero, but others are standard. 12*1/0=12/0 is not the same rules that someone might use to make 0*1/0=1, because 0's cancel in the latter, 12*(1/0)=12/1 * 1/0=12*1 / 1*0=12/0, but even then 1/0 is not a distinct unit. But I think I can do that, the same way sqrt(-4)=2*sqrt(-1)=2i.

 

I'm defining a nonsense number to make it make sense. I'm trying to define a set of rules for a new number system. I'm not advocating one way or the other, but rather playing around with mathematics.

[Brony Number 42 10,069]

but why can't I do 12*1/0=12/0?

Go read my formal proof of why this limit does not exist. The 12*1 part is not the problem. It's the 12/0 part.

I'm defining a nonsense number to make it make sense.

You cannot define something that is in contradiction to something else that is established. I cannot define 1+1 = 5. Likewise, you cannot define g=1/0 and then act like it behaves like a normal operation of a/b. You could have some unknown object g, and put it through operations or functions. But you cannot at the end say, "Ah ha! It's actually 1/0" like you snuck it in. If the object itself is undefined, then anything you do with it is meaningless.

 

Here is an example:

g(x) = x

f(x) = 1/x

The domain of f is: (- infinity, 0 ) union (0, infinity), so all real numbers except 0. The domain of g is all real numbers including 0

Now the composite function

f(f(x)) = x

But

g(x) does not equal f(f(x)). This is because in the composition f(f(x)), you must carry through the domain of f(x) into f(x). The domain excludes 0. So a plot of f(f(x)) will look like a plot of g(x) except that f(f(0)) does not exist, so there is a hole at that point.

 

Basically, f(x) cannot be defined at x=0, and no matter what kind of manipulation you do to it, that domain does not change. You can't sneak that singularity in.

Edited April 6, 2015 by BronyNumber42

[Sigma 768]

I'm still struggling to get solid rules with this, but I can at least say that 0*(1/0) can not cancel out.

 

To me, that would be like saying you can get i from sqrt(-1), yet you can't get -1 from i². Even if 'g' remains a constant, you should still be able to use multiplication by zero to turn it into a nonzero real number.

 

Also, this isn't an argument, but rather an interesting question. What would be g/0?

[Brony Number 42 10,069]

 

 

What would be g/0?

It's meaningless. g is divergent. Any further analysis on it is meaningless.