A power automorphism maps every subgroup of a group to within itself, with equality if the group is finite. More specifically, for a subgroup $H$ of $G$, a power automorphism $f$ has $f(H) \subseteq H$. If $G$ is finite then $f(H) = H$.
There are lots of these of the form $f:G \to G$, $f(x) = x^...
Suppose you have an integer $X$ written in a base $b>1$ as:
$$X=d_n b^{n}+\ldots+d_1 b^1+d_0.$$
Dividing both sides by $b$ we get:
$$\frac{X}{b}=d_n b^{n-1}+\ldots+ d_1+ \frac{d_0}{b}.$$ Why does this imply that $d_0$ is the remainder of the division of $X$ by $b$?
Thanks.
Suppose you have an integer $X$ written in a base $b>1$ as:
$$X=d_n b^{n}+\ldots+d_1 b^1+d_0.$$
Dividing both sides by $b$ we get:
$$\frac{X}{b}=d_n b^{n-1}+\ldots+ d_1+ \frac{d_0}{b}.$$ Why does this imply that $d_0$ is the remainder of the division of $X$ by $b$?
Thanks.
