3
$\begingroup$

Considering Peano axioms we'll define addition, multiplication and exponentiation operations.

We can then prove that addition and multiplication operations are commutative and associative. The proof for addition operation is pretty straightforward but for for multiplication it a bit more difficult. For example, to prove associativity, we have to prove distributivity first.

Moreover, for the definition itself it is not obvious that commutativity should take place, because the definition is not symmetric.

$a \cdot 0 := 0$

$a \cdot S(b) := a \cdot b + a$

Here $S(b)$ is the "next" function from Peano axioms

This definition is not symmetric with respect to $a$ and $b$, so commutivity is not obvious.

For exponentiation operation neither commutivity nor associativity take place. And I am very curious about the logical explanation why that happens

NOTE: The explanation I am looking for is somehow similar to a Abel–Ruffini theorem which explains WHY there is no an algebraic solution for a general polynomial equation

$\endgroup$
  • $\begingroup$ @MauroALLEGRANZA Yeah, but that's not the real explanation. Basically you said that it happens because it happens :) $\endgroup$ – mnaoumov Jul 1 '16 at 8:47
  • $\begingroup$ Your second equation should read $a \cdot S(b) := a + a \cdot b$ $\endgroup$ – gammatester Jul 1 '16 at 8:48
  • $\begingroup$ @gammatester thanks, it was a typo $\endgroup$ – mnaoumov Jul 1 '16 at 8:50
  • $\begingroup$ @MauroALLEGRANZA yeah, but the question is not why the multiplication is commutive. The question is why exponentiation is not! $\endgroup$ – mnaoumov Jul 1 '16 at 9:00
4
$\begingroup$

I'm not sure how to explain it using the Peano axioms other than that it's not, that this follows from the axioms and the definition and there's nothing more to it.

On the other hand, if you need some intuitions, you can find them by interpreting number $n$ as $\big|\{0,1,2,...\ldots,n-1\}\big|$ and observing that $a = |A|$ and $b = |B|$ implies $a \cdot b = |A \times B|$. No wonder multiplication is commutative, because the roles of sets $A$ and $B$ are symmetric. Similarly for $(a\cdot b)\cdot c = a \cdot (b \cdot c)$ which follows from $(A \times B) \times C \equiv A \times (B \times C)$ (I'm assuming you have the intuition why the last one is true).

Yet, for exponentation we have $a^b = |B \to A|$, i.e., the size of the set of functions from $B$ to $A$ (it is sometimes also written as $A^B$). However, note that function of the form $B \to A$ is much different than a function $A \to B$, for example if $B$ is empty and $A$ is not, then the latter cannot exist. The two operands on both sides of the arrow have different roles, so no wonder exponentiation is not commutative.

Also $(A \to B) \to C$ differs a lot from $A \to (B \to C)$. To see this, observe that the first one expects as input a rather huge amount of information compared to the second which is equivalent to $(A \times B) \to C$ (the formula $(a^b)^c = a^{b\cdot c}$ is not a coincidence). Thus, it's not surprising that the exponent is not associative.

I hope this helps $\ddot\smile$

$\endgroup$
  • $\begingroup$ Thanks! That's the most intuitive explanation I could think of! $\endgroup$ – mnaoumov Jul 1 '16 at 9:04
2
$\begingroup$

Of course the question is somewhat philosophical, hence hard to answer. But here is my take on it:

Addition (more precisely disjoint union) is the coproduct in the category of finite sets (basically natural numbers) and coproducts are always associative.

Because of that, the dual of coproduct, a.k.a. products are always associative too and for finite sets the product is of course multiplication (more precisely the cartesian product).

But the dual of a product is again a coproduct, so while $+$ and $\cdot$ are dual to each other (and hence share a lot of properties), exponentiation is not dual to any of them.

$\endgroup$
  • $\begingroup$ thanks, that's also a good explanation $\endgroup$ – mnaoumov Jul 1 '16 at 9:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.