Exponentiation is nonassociative $ a^{b^c}\neq (a^b)^c$

Question

I am trying to determine whether the definition of * gives a binary operation on the set.

On $\mathbb{Z^+}$, define * by letting $a*b = a^b$

I think that the binary operation is not commutative because if $a,b \in \mathbb{Z^+}$, if $a = 1$ and $b = 2$, we see that $a^b \neq b^a$.

I am not sure how to determine whether the binary operation is associative because I am not sure how to set the binary operation up.

Would it look something like this? $(a^b)^c = (b^c)^a$ ?

Answer 1

The systematic way is to use the definition $A\ast B=A^B$ for positive integers $A,B$. Now just rewrite $A=a$ and $B=b\ast c$. Then the definition says that $$ a\ast (b\ast c)=a^{(b\ast c)}=a^{b^c}. $$ On the other hand writing $A=a\ast b$ and $B=c$ gives $$ (a\ast b)\ast c=(a\ast b)^c=(a^b)^c. $$ Obviously this is different for some $a,b,c$.

Answer 2

In general, $a^{b^c}\neq (a^b)^c$. You could take $a=2,b=2$ and $c=3$. You get $2^{2^3}=2^8\neq2^6=(2^2)^3$.