a) “Exponentiation on reals does not distribute over multiplication to the right.”
Hello thehollow89If there are two binary operations defined on a set $\displaystyle S$, that are denoted by (say) $\displaystyle \circ$ and $\displaystyle *$, then $\displaystyle \circ$ is right-distributive over $\displaystyle *$ if $\displaystyle (x *y)\circ z = (x\circ z) *(y\circ z), \forall x,y,z \in S$. So if $\displaystyle \circ$ represents exponentiation - in other words, $\displaystyle x\circ y = x^y$ - and $\displaystyle *$ represents multiplication - so $\displaystyle x*y = xy$ - then
$\displaystyle (x*y) \circ$$\displaystyle z = (xy)^z = x^zy^z = (x\circ z)*(y\circ z)$
So exponentiation is right-distributive over multiplication.
If, however, to the right (which is a phrase I haven't come across in this context) means that the multiplication is on the right, then:
$\displaystyle (x \circ $$\displaystyle y) * z = x^yz$
but $\displaystyle (x*z) \circ$$\displaystyle (y*z) = (xz)^{yz}$
And these two are, of course, unequal. So this, if my interpretation is correct, is what the sentence 'Exponentiation does not distribute over multiplication to the right' means. (See PS.)
Note that this is similar to addition and multiplication, where
$\displaystyle (a+b)\times c = (a\times c) + (b\times c)$
but $\displaystyle (a \times b) + c \ne (a+c) \times (a+b)$.
So multiplication is right-distributive over addition, but (if my assumption about the use of the phrase 'to the right' is correct) multiplication 'does not distribute' over addition 'to the right'.
Maybe someone else on this Forum can confirm that my interpretation is correct?
Grandad
PS Alternatively, it could mean $\displaystyle x \circ (y*z) = x^{yz}$, which when compared to $\displaystyle (x \circ y) * (x \circ z) = x^yx^z = x^{y+z}$ is again, not equal. So it is not distributive using this interpretation either. On reflection, I think this is the more likely meaning of 'exponentiation does not distribute over multiplication to the right'.