如题，看不出哪里不对劲

$\forall a = kb, k\in Z, 都有\frac{a}{b}\equiv ax \pmod m$

$\Rightarrow$

$\forall a = kb, k\in Z, 都有a \equiv abx \pmod m$

$\Rightarrow$

$\forall a = kb, k\in Z,a与m互质, 都有a \equiv abx \pmod m$

$\Rightarrow$

$1 \equiv bx \pmod m$

又

$\forall a = kb, k\in Z, \frac{a}{b} \equiv \frac{a}{b} \pmod m$

$\Rightarrow$

$\forall a = kb, k\in Z, \frac{a}{b} \equiv ax \pmod m$