Question

Question- Prove that a^0 =1.

Answer 1

PROOF:-

As we know that;

a^1= a :- (LET:-1st series)

a^2= a^2 :-(LET:-2nd series) and,

a^3= a^3:- (LET:-3rd series)

And we have to find the value of a^0=?

Examining the above three series you can see that the product of all the three series is differ by a or you can say they have the common difference of the value a for your better understanding I am elaborating the above series:-

a , a^2 , a^3(This are all the values of the above series and you can see that it differ by 'a' this means that the given series is in A.P)

Therefore, the value just before a will be 1 because ,

a^0= 1

a^1=a

a^2=a×a

a^3=a×a^2

Let;

…(X)….. , a , a^2 , a^3(X= first term)

As we know that,

No. Of terms = X + (No. Of terms-1)d

Where as,

X= first term and d= common difference

Therefore, X= 1 Or a^0= 1

Hence proved!

Answer 2

$\large \:answer$

In short, the multiplicative identity is the number 1, because for any other number x, 1*x = x. So, the reason that any number to the zero power is one is because any number to the zero power is just the product of no numbers at all, which is the multiplicative identity, 1.