$$a,b\in\Bbb Z$$ $$n,k\in\Bbb N^+$$ $$a\equiv b\bmod n$$ $$\text{Use induction to prove}$$ $$a^k\equiv b^k\bmod n$$
Asked
Active
Viewed 981 times
-1
-
2what has been tried ? is this homework ? – Jul 16 '17 at 12:35
-
Hint $ $ This Congruence Power Rule follows by inductively applying the Congruence Product Rule. – Bill Dubuque Jul 16 '17 at 12:50
-
Are you allowed to use the fact that if $a\equiv b\mod n$ and $c\equiv d\mod n$ then $ac\equiv bd\mod n$? – Dave Jul 16 '17 at 12:57
-
Yes, this was proven in an earlier question but i'm stumped with this last bit – Kane Jul 16 '17 at 13:11
-
@Kane you have $a\equiv b$ and $a^k\equiv b^k$ – Shuri2060 Jul 16 '17 at 13:13
2 Answers
3
The base case is a given.
The successor step follows from
$$x \equiv y \pmod{n} \land x' \equiv y' \pmod{n} \implies xx' \equiv yy' \pmod{n}$$.
Shuri2060
- 4,353
Henno Brandsma
- 242,131
2
Use the induction hypothesis to suppose $a^k\equiv b^k\mod n$. Using the fact that: if $a\equiv b\mod n$ and $c\equiv d\mod n$ then $ac\equiv bd\mod n$ we have: $$\begin{align}aa^k&\equiv bb^k\mod n\qquad\text{since $a\equiv b\mod n$}\\\implies a^{k+1}&\equiv b^{k+1}\mod n\end{align}$$
Dave
- 13,568