∀a=kb,k∈Z,都有ab≡ax(modm)\forall a = kb, k\in Z, 都有\frac{a}{b}\equiv ax \pmod m∀a=kb,k∈Z,都有ba≡ax(modm)
⇒\Rightarrow⇒
∀a=kb,k∈Z,都有a≡abx(modm)\forall a = kb, k\in Z, 都有a \equiv abx \pmod m∀a=kb,k∈Z,都有a≡abx(modm)
∀a=kb,k∈Z,a与m互质,都有a≡abx(modm)\forall a = kb, k\in Z,a与m互质, 都有a \equiv abx \pmod m∀a=kb,k∈Z,a与m互质,都有a≡abx(modm)
1≡bx(modm)1 \equiv bx \pmod m1≡bx(modm)
又
∀a=kb,k∈Z,ab≡ab(modm)\forall a = kb, k\in Z, \frac{a}{b} \equiv \frac{a}{b} \pmod m∀a=kb,k∈Z,ba≡ba(modm)
∀a=kb,k∈Z,ab≡ax(modm)\forall a = kb, k\in Z, \frac{a}{b} \equiv ax \pmod m∀a=kb,k∈Z,ba≡ax(modm)