1
$\begingroup$

In 1914 Albert Bennett suggested the following operation:

$$a * b=a^0_2b=\exp(\ln a \ln b)$$

Now, given this function, addition and multiplication, and their properties, can one express exponentiation and power function?

The operation $a^0_2b$ has the following properties:

$$a*b =b*a$$ $$(ab) * c = (a * c)(b * c)$$ $$a * e = e * a = a$$

I also wonder whether a function $a^0_{-1}b=\log(\exp(a)+\exp(b))$ can help in this endeavor.

$\endgroup$
  • $\begingroup$ According to his 1915's paper, it seems that $a \# b = \exp(\log a + \log b)$, which actually denotes the usual multiplication. $\endgroup$ – Sangchul Lee Nov 6 '12 at 0:11
  • $\begingroup$ By noting that the mapping $\log : (\Bbb{R}^{+},\#,*) \to (\Bbb{R},+,\cdot)$ gives an isomorphism, it seems natural to define the exponentiation via $$\log(a^{*b}) = (\log a)^{\log b}.$$ $\endgroup$ – Sangchul Lee Nov 6 '12 at 0:18
  • $\begingroup$ @sos440 I just used my own notation. If I had access to his paper I would change it. $\endgroup$ – Anixx Nov 6 '12 at 0:19
  • $\begingroup$ The two operations, as @Anixx has quoted them, do not satisfy distributivity. So @sos440’s definition seems far more plausible. p.s. I met old Bennett during my first years at Brown. $\endgroup$ – Lubin Nov 6 '12 at 0:25
  • $\begingroup$ It is not a matter of notation. Given definition of $\#$ and $*$, we cannot form an analogy of addition and multiplication since the distribution law $a * (b \# c) = (a*b)\#(a*c)$ fails. This is what I pointed out. And maybe you can refer to Note on an operation of the third grade. $\endgroup$ – Sangchul Lee Nov 6 '12 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.