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It is not difficult to prove that $0/0$ is not equal to any constant $c \ | \ c/c = 1 \ \land \ c \in \mathbb{R} \cup \mathbb{C}$.

First assume that $0/0 = c$. Consider the true equation $2\times0=0$. Based on our initial assumption, dividing the equation by zero gives $2\times0/0=0/0 \ \Rightarrow \ 2c=c$. Based on our definition of $c$, dividing the equation by $c$ gives $2=1$. However, $2 \ne 1$. $\ \therefore\ $ $0/0 \ne c.\quad\blacksquare$

Sorry for the incomprehensible spacing. Anyways, whenever I attempt to prove that $0/0 \ne 0$, I always end up arriving at $0=0$. Can someone find (or make) a proof? I couldn't find one online either.

Just some clarifications: In in the argument that ends with a box, I am attempting to prove that $0/0 \ne 0$.

Extra clarifications: This is not a duplicate. I am not asking for an intuitive proof that division by zero is nonsensical. I am asking for a concrete mathematical proof that $0/0 \ne 0$. And sorry about my previous edit, that was incorrect but I fixed it.

Shawhin Layeghi
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  • Can you explain what you are trying to prove (in the argument that ends with a box) and also what you mean by "Based on our definition of $c$, dividing the equation by $c$..." Are you assuming $c$ is a nonzero real number? What properties are you alluding to in the phrase "based on our definition of $c$"? – Michael Mar 12 '18 at 17:10
  • you do not need a proof to let you know that 0/0 $\neq$ 0. The lefthand side is completely meaningless. Division by zero is just not defined for numbers. – Perry Bleiberg Mar 12 '18 at 17:13
  • It looks like the poster is trying to prove that it is not possible to assign a numerical value to $0/0$. The first paragraph shows that if $0/0$ is a nonzero real number, then a specific contradiction appears.

    OP's question is how to show that $0/0 = 0$ doesn't make sense either.

     – Umberto P. Mar 12 '18 at 17:13

2 Answers2

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The quotient $a/b$ is the unique number $c$ such that $bc=a.$

So the quotient $0/0$ should be the unique number $c$ such that $0c=0.$

The problem is non-uniqueness.

If one considers $5/0$ rather than $0/0$ then the problem is non-existence of any number $c$ for
which $0c=5.$

Either non-existence of non-uniqueness makes division by $0$ undefined.

Michael Hardy
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Division is defined by repeated subtraction. How many time can you subtract zero from zero? or, a better question might be -- how many answers are there to the last question? What does this mean about the value $\frac{0}{0}$?

dan post
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