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Prove, using only the field axioms of real numbers, that $0/0$ is undefined.
I have thought about it for a while and come up with an idea how to solve this. First, I would like to prove (using field axioms - again) that $x \cdot \frac{1}{y} = \frac{x}{y}$, (but I am still working on this - any hints would be most appreciated),therefore $\frac{0}{0} = 0 * \frac{1}{0}$ and then, say that - by virtue of one of the axioms there exists only one number $b$ satisfying the equation $ab = 1$ for all $a \ne 0$. Since in our case we want to take the reciprocal of 0 - the axiom clearly states that it does not exist.
Is this a good way to prove that the expression in question is undefined?

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  • $\begingroup$ What does $\frac xy$ mean if it is not an abbreviation for $xy^{-1}$? $\endgroup$ – Mees de Vries Nov 3 '17 at 14:19
  • $\begingroup$ @MeesdeVries This is obvious, but as long as we are limited to the axioms, I like to think thoroughly through every step in my proof. $\endgroup$ – Aemilius Nov 3 '17 at 14:20
  • $\begingroup$ According to filed axioms $\dfrac 1 a$ is defined only for $a \ne 0$. $\endgroup$ – Mauro ALLEGRANZA Nov 3 '17 at 14:20
  • $\begingroup$ I am speaking on an "every step of the proof" level. Typically, the field operations are $+, \times, 0, 1$. Thus $\frac xy$ is not something you can formulate directly in the language of fields. Hence it must be defined. How is it defined? $\endgroup$ – Mees de Vries Nov 3 '17 at 14:21
  • $\begingroup$ Thus, the issue is that $\dfrac 0 0 = 0 \times \dfrac 1 0$ is undefined because $\dfrac 1 0$ is undefined. $\endgroup$ – Mauro ALLEGRANZA Nov 3 '17 at 14:23
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Let $R$ be any ring with $1$. Suppose that $0^{-1}$ makes sense. Then that means $0 \cdot 0^{-1}=1 = 0^{-1}\cdot 0$. Let $a \in R$ be arbitrary. Then $$ a = 1a = 0^{-1}0a = 0^{-1}0=0. $$ Hence $R=\{0\}$.

Thus, if you want your ring to have more than one element, you better not let $0$ have an inverse. So, not only should $\frac{0}{0}$ not make sense, no "fraction" of the form $\frac{a}{0}$ should make sense.

In case you do not believe that $a0=0$, we have $$ a0 = a(0+0) = a0+a0. $$ Now cancel an $a0$ from each side. If you do not believe in subtraction/cancellation someone else will have to help you.

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  • $\begingroup$ Your equations rely on $a \times 0 = 0$, but if you already know that then you're done (since $0 \neq 1$ in a field). $\endgroup$ – Mees de Vries Nov 3 '17 at 14:26
  • $\begingroup$ I was avoiding assuming $1 \neq 0$ until the end. Some people allow this, but either way, the argument above forces $R=\{0\}$ no matter what. $\endgroup$ – Randall Nov 3 '17 at 14:28
  • $\begingroup$ Fine, but your argument is still missing the proof that $a \times 0 = 0$. $\endgroup$ – Mees de Vries Nov 3 '17 at 14:29
  • $\begingroup$ I'm assuming that's not a dealbreaker/showstopper. The OP can insert it. $\endgroup$ – Randall Nov 3 '17 at 14:29
  • $\begingroup$ I disagree; I think it is the "difficult" part of the proof. But I will provide my own answer then. $\endgroup$ – Mees de Vries Nov 3 '17 at 14:30
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The expression $\frac xy$ is not immediately in the language of fields, which only has the symbols $+,\times,-,0,1$. Thus we must define what $\frac xy$ means, and the fastest thing to do is as follows: if the equation $ya = x$ has a unique solution $a$, then $a = \frac xy$. (Typically you would define $y^{-1}$ first, and then set $\frac xy = xy^{-1}$, and we would have to rule out the existence of $0^{-1}$, but this route also works.)

Thus, let us show that $0 \times a = 0$ does not have a unique solution. In fact, for any $a$ we have $$ 0 \times a = (0 + 0) \times a = 0 \times a + 0 \times a. $$ Now subtract $0 \times a$ from each side to obtain $0 = 0 \times a$. This means that the equation defining $\frac 00$ does not have a unique solution -- it has at least the solutions $0,1$ -- and thus $\frac 00$ is not well-defined.

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  • $\begingroup$ How do you infer that because of the fact that $0 = 0 a$, the expression $0/0$ has more than one value? $\endgroup$ – Aemilius Nov 3 '17 at 15:38
  • $\begingroup$ Like I mentioned before, $\frac xy$ does not a priori mean anything. We have to give it meaning. The standard way to do this (without going via $y^{-1}$) would be to say: $\frac xy$ is defined if and only if $ya = x$ has a unique solution in $a$, and then that solution is $a = \frac xy$ by definition. Since $0 = 0a$ for at least two distinct $a$, we do not have a unique value for $\frac 00$. $\endgroup$ – Mees de Vries Nov 3 '17 at 15:42
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Suppose $\frac{0}{0}$ is defined.

Then, by the Existence of a Multiplicative Inverse, $\frac{0}{0}=1$

By the Existence of an Additive Inverse, we have $a+(-a)=0$. Using this with $a=1$, we can write:

$1-1=0 \Leftrightarrow \frac{0}{0} - \frac{0}{0} =0 \Leftrightarrow \frac{0-0}{0}=0 \Leftrightarrow \frac{0}{0}=0$

which leads to a contradiction by uniqueness of your multiplicative inverse.

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    $\begingroup$ Thanks, your answer is clear and hits the spot. However, I've got one question - you wrote that $0/0 - 0/0 = (0-0)/0$. But none of the axioms tells anything about fraction addition. Can I use it? $\endgroup$ – Aemilius Nov 3 '17 at 15:33
  • $\begingroup$ Consider the Distributivity of Multiplication over Addition that tells you $a\times(b+c)=a\times b+a\times c$ and let $a$ be the inverse of $0$. $\endgroup$ – Zachary Nov 3 '17 at 22:44

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