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For instance, in split-complex numbers we definitely have $2\cdot(j/2+1/2)=j+1$, which is absolutely valid. Can we then say that $\frac{j+1}{j/2+1/2}=2$? If not, why we cannot define it this way?

Anixx
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3 Answers3

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You can define whatever you want

You can define $\frac{j+1}{j/2+1/2}$ to mean whatever you want.

But the notation suggests we are working in a field where $j+1$ has an inverse. This is not true for the split complex numbers so will confuse the reader.

For example if you write $\frac{j+1}{j/2+1/2}=2$ then your fraction bar does not work the same way the normal fraction bar works. If it did we could multiply both sides by $j-1$ to get

$$\frac{j+1}{j/2+1/2} =2$$

$$\implies \frac{j+1}{j/2+1/2} (j-1)=2(j-1)$$

$$\implies \frac{(j+1)(j-1)}{j/2+1/2} =2(j-1)$$

$$\implies \frac{0}{j/2+1/2} =2(j-1)$$

$$\implies 0 =2(j-1)$$

which is untrue.

Daron
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  • Hmm, but $j-1$ is another zero divisor and it is not evident that we can multiply the both sides by it even if the fraction is valid. – Anixx Jun 11 '22 at 16:16
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    @Anixx That's even worse. In every other context, we can perform the same operation on both sides of an equality sign. If you have to break that rule to introduce your new notation that means it is really bad notation. – Daron Jun 11 '22 at 16:18
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Division by a zero divisor can never give a unique quotient. Suppose that $a$ is a zero divisor, so $\,\color{#c00}{a\hat a=0},\,\ a,\hat a\neq 0.\,$ Then $\, a\:\!\:\!x = b\Rightarrow \color{#c00}a(x+c\color{#c00}{\hat a}) = b,\,$ so there is no unique solution $\,x := b/a.\,$ OP is case $\,a = r(j+1),\ \hat a = j-1,\ r = \frac{1}2,\,$ where $\,a\hat a = r(j+1)(j-1)=0,\,$ by $\,j^2=1.$

Bill Dubuque
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  • But the equation $x(j/2+1/2)=j+1$ has unique solution $x=2$ does not it? – Anixx Jun 11 '22 at 16:30
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    @Anixx No, since the answer proves that $,x = 2+c\color{#c00}{(j-1)}$ is a solution for all $,c.\ $ The map $,x \to zx,$ has nonzero kernal so it cannot be $1$-$1\ \ $ – Bill Dubuque Jun 11 '22 at 16:32
  • Oh I see, $x=3/2+j/2$ also satisfies it... – Anixx Jun 11 '22 at 16:33
  • But then, can we talk about a kind of main value of such division? – Anixx Jun 11 '22 at 16:35
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    @Anixx What do you mean by "main value"? – Bill Dubuque Jun 11 '22 at 16:42
  • Something along the lines of $0^0=1$ even though $\log_0 x=0$ for any non-zero $x$ – Anixx Jun 11 '22 at 16:51
  • @Anixx It's unsolvable - just scale by $1+j,$ (or eval at $j = 1).,$ You deleted the comment but I'll leave this till you see it. – Bill Dubuque Jun 11 '22 at 17:04
  • Yes, I see it, thanks. I realize it's insolveble because in $\mathbb R^2$ there is no solution for $(x,y)(1,0)=(0,1)$. – Anixx Jun 11 '22 at 17:37
  • Well, given that the main value (and the only algebraically sane choice) for $0/0$ is $0$, we should be able to divide $(0,2)/(0,1)=2$, and so $(j+1)/(j/2+1/2)$. Even though there are many solutions to the equation. – Anixx Jun 12 '22 at 08:26
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    Again, what is the "main value"? How do you propose to canonically choose a solution from an infinite set of solutions? (and why try to do so?) – Bill Dubuque Jun 12 '22 at 08:32
  • Cauchy main value at complex plane of $0/0$ is $0$. also, it is the only non-contradicting choice. – Anixx Jun 12 '22 at 08:36
  • E.g. in $,\Bbb Z_{15}!:,\ \color{#c00}{3\cdot 5 = 0},$ so $,3x = 6,\Rightarrow, \color{#c00}3(x!+!\color{#c00}5c)= 6,,$ which yields the three solutions $,x=2,,2!+!5,,2!+!2(5) = 2,,7,,12.,$ Therefore the "fraction" $,x = \frac{6}2 ,=, 2,7,12,$ is multi-valued. $ $ In some contexts it proves convenient to extend fractional arithmetic to such solution sets, e.g. see the Remark here on their use in the extended Euclidean algorithm in fractional form. – Bill Dubuque Jul 06 '22 at 20:01
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Given a commutative ring. Division is an operation derived from multiplication, i.e., $c=a:b$ is defined as $c=a\cdot b^{-1}$, where the divisor $b$ must be invertible. Note that invertible elements (units) are not zero-divisors. Similar situation for subtraction which is derived from addition.

Wuestenfux
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