0

I came across a version of maths on a site (it requires you to sign up to read the article), where addition is defined as the reciprocal of the sum of the reciprocals. In the below article, the author describes the basic operations in that variant (addition and subtraction) and proceeds to use those to get analogies of the derivative operator (and product, chain and quotient and sum rules), integral, Taylor series in that variant. I would like to learn more about this variant of maths, which the author calls "parallel maths" because it looks similar to the parallel addition of resistances. Also, have you seen this referred to with another name anywhere? If so, please do let me know.

https://www.cantorsparadise.com/a-fifth-fundamental-operation-of-arithmetic-and-the-beauty-of-parallel-calculus-93a2dfe28dda

If you would like to read the article, I have uploaded screenshots of the article on imgur. https://imgur.com/a/1NfJ8Xr

Entity 903
  • 31
  • 3
  • 1
    So $a\oplus b = \frac{1}{\frac{1}{a}+\frac{1}{b}}$? It seems to me that they still need to define and use the usual addition to build up to this, so its not like they are working with completely different operations, just more operations. $\oplus$ also does not appear to be a group operation, it doesn't have an identity for instance... and as such "(parallel) subtraction" wouldn't be able to be defined in the usual way (adding by the additive inverse (the value that when added gives the identity element)). – JMoravitz Feb 02 '24 at 15:13
  • 1
    This appears to be motivated by the equivalent resistance of two parallel resistors in an electronic circuit. Are you familiar with the basics of electric circuits? – John Douma Feb 02 '24 at 15:20
  • How does the author define multiplication? – TurlocTheRed Feb 02 '24 at 17:39
  • @JohnDouma I am kind of familiar with them. – Entity 903 Feb 02 '24 at 18:03

1 Answers1

1

I think what you need is Group Theory or Field Theory. It allows for the systematic analysis of binary operations.

So what we have is $x,y \in \mathbb {R}$ and $x\oplus y=\frac{1}{1/x+1/y}=\frac{xy}{x+y}$. Though with the addition of $\pm \infty$.

To have a group, you need a closed, associative binary operation with inverses and an identity element. This new operation is closed under addition and it's associative. To get the other properties, you need to take some liberties.

Inverses are tricky. We need $e\in G$ so that $\frac{xe}{e+x}=x\implies x^2+ex=ex\implies x^2=0$.So no finite $e$ exists. $e=-\infty$ gets you there.

What about inverses? $\frac{xy}{x+y}=-\infty\implies y=-x $.

To do calculus, you need invertible multiplication.

Because of the paywall, I don't know how the article defines multiplication. Repeated addition seems feasible.

$x\oplus x\oplus ...\oplus x$ n times is $x/n$. So $x \otimes n= x/n$. This suggests we use $n=1$ as the multiplicative identity.

$(x \otimes y)\otimes z = x/yz. x\otimes( y \otimes z) = x \otimes y/z=zx/y$ so it's not associative. So under this definition of multiplication we do not have a field.

An attempt at differentiation, incomplete.

$\frac{[(x\oplus h) \ominus x]}{h}=\frac{\frac{xh}{h+x}-x}{h}= \frac{-x^2}{h+x}\otimes (1/h)=? \frac{-x^2h}{h+x}$

In usual calculus we have $h\to 0$. Do we need $h\to 0$ or $h \to -\infty$ since the latter is the additive identity? In the first case, we get $0$, in the latter, $-x^2$.

So this is one example of analyzing the "new maths" you mention. It's a special case of Group/Field Theory.

TurlocTheRed
  • 5,683
  • The author has mentioned that "parallel addition" of x n times would be x/n, so yeah what you have done. I'll add a link to an imgur with the article. – Entity 903 Feb 02 '24 at 17:54
  • I have done the above. I think reading the article would help you answer my question. – Entity 903 Feb 02 '24 at 18:06