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Why 0/0= undefined? We always write $0/x\in\mathbb R$, where $x \in\mathbb{R}\setminus\{0\}$, and this $0/x = 0$. Again $x/0$ is always undefined. Now when it's about $0/0$ why we don't follow the first rule. And why I don't support is because if 0$/0 = §$, then $0*§=0$. But it's not the case. $0*§=§$. Then why $0/0=§$?

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marked as duplicate by Clement C., MJD, Joey Zou, user1551, Behrouz Maleki Aug 24 '16 at 19:50

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  • $\begingroup$ As a general rule, we define something if it useful to define it. Here, there is no reason to define $0/0$. Allowing $0/0$ as an expression can lead to lots of errors. $\endgroup$ – Thomas Andrews Aug 24 '16 at 13:24
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The definition of $\frac{a}{b}$ is $a\cdot b^{-1}$, where $b^{-1}$ (the multiplicative inverse of $b$) is the unique real number such that $b^{-1}\cdot b = b\cdot b^{-1} = 1$. $0$ has no such multiplicative inverse, since $0\cdot x = 0 \neq 1$ for any real number $x$. Thus, division by $0$ can never be defined.

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  • $\begingroup$ Well think inversely sir. 0*0=0 , 0/a =0 then why we give emphasis on its division by 0 not on 0 divided by any number? I meant since 0 is the numerator then why this 0/0 can't be equal to 0 ? $\endgroup$ – ffahim Aug 24 '16 at 13:27
  • $\begingroup$ This isn't a matter of what we choose to give emphasis to. In order to define division by a number, you multiply by the multiplicative inverse of that number, which is not defined for $0$. It's as simple as that. $\endgroup$ – florence Aug 24 '16 at 13:33
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The main reason is that $a/b$ is defined as the unique number $x$ such that $a=xb$. In the case of $0/0$, there are a lot of different $x$ such that $0=0x$. Following that idea, $0/0$ has the value of all real numbers simultaneously. That possibility breaks so many rules that it's better to leave it undefined.

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