# Principle of mathematical induction

• Oct 19th 2009, 09:23 AM
flower3
Principle of mathematical induction
Use the principle of mathematical induction to prove that if $\displaystyle a \in R$ then $\displaystyle ( a^{m} ) ^n = a^{mn} , \forall m,n \in N$

TRUE OR NOT ?? :
1) WANT TRUE FOR n=1 AND m=1 :
$\displaystyle (a^{1 )^ 1 } = a^{1.1} \to a=a$
2) SUPPOSE TRUE FOR n=k AND m=v :
$\displaystyle (a^ {v ) ^k }= a^{vk}$
3) WANT TRUE FOR n=k+1 AND m=v+1 :
$\displaystyle (a^ {v+1 ) ^k+1 }= a^{(v+1)(k+1)}$ HOW I PROVE IT ???
• Oct 19th 2009, 02:03 PM
redsoxfan325
Quote:

Originally Posted by flower3
Use the principle of mathematical induction to prove that if $\displaystyle a \in R$ then $\displaystyle ( a^{m} ) ^n = a^{mn} , \forall m,n \in N$

TRUE OR NOT ?? :
1) WANT TRUE FOR n=1 AND m=1 :
$\displaystyle (a^{1 )^ 1 } = a^{1.1} \to a=a$
2) SUPPOSE TRUE FOR n=k AND m=v :
$\displaystyle (a^ {v ) ^k }= a^{vk}$
3) WANT TRUE FOR n=k+1 AND m=v+1 :
$\displaystyle (a^ {v+1 ) ^k+1 }= a^{(v+1)(k+1)}$ HOW I PROVE IT ???

$\displaystyle (a^{v+1})^{k+1}=(a^v\cdot a)^{k+1}=(a^v)^{k+1}\cdot a^{k+1}=(a^v)^k\cdot a^v\cdot a^{k+1}$

Spoiler:
Since from the inductive step $\displaystyle (a^v)^k=a^{vk}$,

$\displaystyle (a^v)^k\cdot a^v\cdot a^{k+1}=a^{vk}\cdot a^v\cdot a^{k+1}=a^{vk+v+k+1}=a^{(v+1)(k+1)}$

QED