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If we introduce $j$ for $\frac{1}{0}$, then $\frac{2}{0}$ will be $2j$ and $\frac{3}{0}$ will be $3j$ ,etc.

So, I'm asking to call the solution of $x\cdot 0=1$ $j$ and discover properties of $j$ just like we did with $i$.

But, if we say $j\cdot 0=1$ for some $j$, then, since for any $j$, $j\cdot 0=0$, so it will give the contradiction $0=1$.

But, we can mend the rules for $j$. I mean, we follow the rule $\sqrt{a}\cdot \sqrt{b}=-\sqrt{ab}$ for negative $a$ and $b$. If we don't follow this rule, then this contradiction happens:

Let's suppose $i$ exists.

Then $i^2=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$, which is a contradiction, because we've assumed $i^2=-1$. So, $i$ does not exist.

But we changed the rules for $i$ to avoid this contradiction.

Then, why can't we change the rules for $j$ to avoid contradictions? I mean, we can say $x\cdot 0=0$ only if $x\neq j$.

EDIT: As a side note, why can't we define $\frac{1}{0}$ to be $\infty$? It seems soo.. intuitive to say that $\frac{1}{0}=\infty$ and $\frac{1}{\infty}=0$,

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    I'm only vaguely knowledgable about nonstandard analysis, but you may want to look it up. The thing is, though, that we want to do as much math as possible in the set of real numbers. You are going to have to come up with a really strong argument about why we should care about $j$, because most mathematicians, I suppose, are perfectly happy with $1 \over 0$ being undefined. – Steven Alexis Gregory Mar 12 '17 at 03:43
  • In order to realize such $j$, you need to make amend of ordinary arithmetic rules to the point that the resulting system no longer looks like the usual algebra. I can hardly imagine how this will make us happy at the expense of giving up all that familiarity. – Sangchul Lee Mar 12 '17 at 03:48
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    Have you tried looking up Riemann sphere, or extended real numbers? – Hrhm Mar 12 '17 at 03:49
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    @stevengregory But imaginary numbers were also considered useless in the start and mathematicians were happy with it being left undefined. –  Mar 12 '17 at 03:50
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    So... $j\cdot (2\cdot 0)=j\cdot 0=1$ but $(j\cdot 2)\cdot 0=(2j)\cdot 0=2$? Associativity of multiplication fails then, which means that such a structure wouldn't even be a ring... – JMoravitz Mar 12 '17 at 04:19
  • Sorry to break the news, but you don't understand what the $\sqrt{\phantom{1}}$ symbol means outside of real numbers... – zipirovich Mar 14 '17 at 05:12
  • @zipirovich Please explain what I don't understand. Are you trying to say that we didn't 'make up' the rule $\sqrt{a}\cdot \sqrt{b}=-\sqrt{ab}$ for negative $a$ and $b$? Funny that no one else pointed that out. –  Mar 14 '17 at 05:53
  • Short version: the square root is a multivalued function, so identities such as "$\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$" don't make much sense anymore. So we neither have that rule nor did we make up a new one with a negative sign. Yes, we can choose a principal value of a multivalued function -- but that wouldn't yield a well-defined property for "$\sqrt{a}\cdot \sqrt{b}=??$" either, at least for two reasons: because that can be done in different ways (although there is often the more "standard" one), and more importantly, because $a$, $b$, and $ab$ can be on different branches. – zipirovich Mar 14 '17 at 06:02
  • That's why a "definition" of $i$ as "$i=\sqrt{-1}$" always makes me cringe. I believe the definition should be that $i$ is such a number that $i^2=-1$. And it's not the same statement! Because with it immediately comes another such number, viz. $-i$ also satisfying $(-i)^2=-1$. And then $\sqrt{-1}={-i,i}$, a set of two numbers. That's why your "proof" of a "contradiction" that "$i^2=1$" was incorrect: it's not true that "$i^2=\sqrt{-1}\sqrt{-1}$" because it's not true that "$i=\sqrt{-1}$". – zipirovich Mar 14 '17 at 06:04
  • @zipirovich Different people will have different opinions about it. If you ask me, I don't have much problem with writing $i=\sqrt{-1}$. It's the rule of exponents that, if $a^b=c$ then, $a=c^{\frac{1}{b}}$. I'd be happy to have that rule being applied to all complex numbers. –  Mar 14 '17 at 06:22

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