MHB Why is dividing by zero impossible in math?

AI Thread Summary
Dividing by zero is impossible in mathematics because it leads to contradictions. For example, attempting to express a division like 5/0 asks how many zeros fit into 5, which is impossible since adding zeros never reaches 5. When a non-zero number is divided by zero, it results in an undefined value because there is no number that satisfies the equation. In contrast, dividing zero by zero is considered indeterminate since it can equal any number, but lacks a unique solution. Understanding these concepts is crucial, especially in the context of limits in calculus, where different scenarios can yield different outcomes.
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We know division by zero is not possible but what is the math reason why it is impossible to divide by zero?
 
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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?
 
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 \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.<br /> <br /> - - - Updated - - -<br /> <br /> Saying that "\frac{a}{0}= c" is equivalent to "a= 0*c". But 0 times anything is 0 so that would say a= 0.<br /> <br /> In fact, some texts make a distinction between \frac{a}{0}, for a non-zero, and \frac{0}{0}. 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.<br /> <br /> On the other hand, \frac{0}{0}= c is equivalent to 0= 0*c= 0 which <b>is</b> true- but is true for <b>any</b> number c. There is no unique number c such that this is true so we say that it is "undetermined".<br /> <br /> 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 <b>not</b> 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.
 
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