Congruence of Powers

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Theorem

Let $a, b \in \R$ and $m \in \Z$.

Let $a$ be congruent to $b$ modulo $m$, i.e. $a \equiv b \pmod m$.


Then:

$\forall n \in \Z_{\ge 0}: a^n \equiv b^n \pmod m$


Proof

Proof by induction:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$a \equiv b \implies a^k \equiv b^k \pmod m$


$P(0)$ is trivially true, as $a^0 = b^0 = 1$.


$P(1)$ is true, as this just says:

$a \equiv b \pmod m$


Basis for the Induction

$P(2)$ is the case: : $a^2 \equiv b^2 \pmod m$ which follows directly from the fact that Modulo Multiplication is Well-Defined.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$a \equiv b \implies a^k \equiv b^k \pmod m$.


Then we need to show:

$a \equiv b \implies a^{k+1} \equiv b^{k+1} \pmod m$.


Induction Step

This is our induction step:

Suppose $a^k \equiv b^k \pmod m$.

Then $a^k a \equiv b^k b \pmod m$ by definition of modulo multiplication.

Thus $a^{k+1} \equiv b^{k+1} \pmod m$.


So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_+: a \equiv b \implies a^n \equiv b^n \pmod m$.

$\blacksquare$


Sources