Tnuts lab)" - a"b" by using principle of mathema12. Prove the rule of

1.	Tnuts lab)" - a"b" by using principle of mathema12. Prove the rule of exponents (ab)" = ananinduction for every natural number.hot
Answer» LetP(n)P(n)be the given statement i.e.,P(n):(ab)n=anbnP(n):(ab)n=anbn We note thatP(n)P(n)is true forn=1n=1since(ab)1=a1b1(ab)1=a1b1. LetP(k)P(k)be true, i.e., (ab)k=akbk(ab)k=akbk----------(1) We shall now prove thatP(k+1)P(k+1)is true wheneverP(k)P(k)is true. Now, we have =(ab)k+1=(ab)k(ab)=(ab)k+1=(ab)k(ab) =(akbk)(ab)=(akbk)(ab)[ by (1) ] =(ak.a1)(bk.b1)=ak+1.bk+1=(ak.a1)(bk.b1)=ak+1.bk+1 Therefore,P(k+1)P(k+1)is also true wheneverP(k)P(k)is true. Hence by principle of mathematical induction,P(n)P(n)is true for alln∈Nn∈N.

Tnuts lab)" - a"b" by using principle of mathema12. Prove the rule of exponents (ab)" = ananinduction for every natural number.hot

Answer»

LetP(n)P(n)be the given statement

i.e.,P(n):(ab)n=anbnP(n):(ab)n=anbn

We note thatP(n)P(n)is true forn=1n=1since(ab)1=a1b1(ab)1=a1b1.

LetP(k)P(k)be true, i.e.,

(ab)k=akbk(ab)k=akbk----------(1)

We shall now prove thatP(k+1)P(k+1)is true wheneverP(k)P(k)is true.

Now, we have

=(ab)k+1=(ab)k(ab)=(ab)k+1=(ab)k(ab)

=(akbk)(ab)=(akbk)(ab)[ by (1) ]

=(ak.a1)(bk.b1)=ak+1.bk+1=(ak.a1)(bk.b1)=ak+1.bk+1

Therefore,P(k+1)P(k+1)is also true wheneverP(k)P(k)is true. Hence by principle of mathematical induction,P(n)P(n)is true for alln∈Nn∈N.