Why is dividing by zero impossible in math?

[mathdad]

We know division by zero is not possible but what is the math reason why it is impossible to divide by zero?

 

[Prove It]

Let's say you wanted to do 5/0, then you're asking "how many 0's are there in 5?"

Well, try adding up 0's until you get to 5...

Hang on, 0 + 0 = 0... If I keep adding 0 we still stay at 0...

How can we ever possibly get to 5?

 

[HallsofIvy]

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

- - - Updated - - -

Saying that "$$\frac{a}{0}= c$$" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.

In fact, some texts make a distinction between $$\frac{a}{0}$$, for a non-zero, and [tex]\frac{0}{0}[tex]. If $$a\ne 0$$ then $$\frac{a}{0}= c$$, for any number, c, is equivalent to $$a= 0*c= 0$$ which is false. There is NO number c that satisfies that so we say it is "undefined" or simply impossible.

On the other hand, $$\frac{0}{0}= c$$ is equivalent to $$0= 0*c= 0$$ which is true- but is true for any number c. There is no unique number c such that this is true so we say that it is "undetermined".

The difference is really only important in "limits". If I am trying to find $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ and find that g, separately, goes to 0 while f goes to a non-zero number, I have the case $$\frac{a}{0}$$ for a non-zero: there is no such limit. But If both f and g go to 0 then there still might be a limit. For example, if $$f(x)= x^2- 9$$ and $$g(x)= x- 3$$ both f(3)= 0 and g(3)= 0. But for x not equal to 0, $$\frac{x^2- 9}{x- 3}= \frac{(x- 3)(x+ 3)}{x- 3}= x+ 3$$ so $$\lim_{x\to 3}\frac{x^2- 9}{x- 3}= \lim_{x\to 3} x+ 3= 6$$.

 

[mathdad]

Great information.

- - - Updated - - -

Say there are 5 people in a particular classroom. They came to class without a pencil. If the principal walks into the classroom and tells me to distribute one pencil per student but I have no pencils, no one will get a pencil. How can I divide a number by nothing? So, number ÷ nothing = undefined.