I am an eighth grader with an interest in abstract math. I was just wondering since multiplication is distributive over addition and exponents were distributive over multiplication, shouldn’t addition be distributive over successor function? The only problem is, to my knowledge, you can’t have one term successing another term. Is there a case where you could define the successor function so addition would be distributive over it? Thanks
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- $\begingroup$ In addition to TomKern's answer, addition also distributes over the operation $a * b = \ln(e^a + e^b)$ -- see en.wikipedia.org/wiki/…. $\endgroup$Mathemagician314– Mathemagician3142025-05-17 11:53:17 +00:00Commented May 17 at 11:53
- 1$\begingroup$ the main problem is that while add, mul, exp are all binary, S(.) is a unary operation. Also note that Addition-Multiplication have both of these nice properties: $a \times (b + c) = a \times b + a \times c$; $(b + c) \times a = b \times a + c \times a$ ; but Exponentiation-Multiplication is not symmetric like that: $(b \times c) ^ a = b ^ a \times c^a$, but $a^{b \times c} \neq a^b \times a^c$ $\endgroup$whoisit– whoisit2025-05-17 16:39:31 +00:00Commented May 17 at 16:39
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1 Answer
$\begingroup$ $\endgroup$
Wikipedia https://en.wikipedia.org/wiki/Hyperoperation defines the hyperoperation before addition as $a \;[0]\; b = b+1$
We then check that:
$a + (b\;[0]\;c) = a+c+1$
And:
$(a+b) \;[0]\; (a+c) = a+c+1$
So we have left distributivity.
Also:
$(b \;[0]\; c)+a = c+1+a$
And:
$(b+a) \;[0]\; (c+a) = c+a+1$
So we have right distributivity.
