Principle of Mathematical Induction (ab)^n = a^n*b^n Proof

[@TheMathSorcerer]

[@angelmendez-rivera351]

One should be careful to note that this is not true for arbitrary n, only for natural n, and only if a, b are elements that commute.

[@syesjohnny4119]

(a^n)(b^n)=ab^2 cuz a*a...a(len n)*b*b...b(len n)=ab*ab...ab(len n)

[@YoungLink51423]

ty

[@TheMathSorcerer]

No problem 😃

[@24pete7]

At 3:00, after expanding (ab)^k+1, why did you use the part of the hypothesis along with what we are suppose to be proving?

[@darkorys8012]

because when you use the induction method, you make a suposition that a general case n=k is true, and using the assumption you manipulate the case n=k+1 to equal the k+1 conclusion, he needs to use the hipothesis otherwise you can't demonstrate by induction.

[@anushkatyagi123]

Ye for abelian group k lye true h na only ??

[@angelmendez-rivera351]

(a·b)^n = a^n·b^n does assume that a, b are elements of G, with (G, ·) being an Abelian power-associative magma.

[@siyadsiyadplmk5011]

A^m×b^n=? Sir pleas