Question

Prove that $\tt{{x}^{0}=1}$. (x^0 = 1)

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Answer 1

$\huge \sf \pink{\boxed{x0 = 1}}$

The power of zero is 1 by dividing indices i.e. (x^n/x^n)=x^(n-n) =x^0 and this is equal to 1 because any number divided by the same number is 1. this is true for all value of x. except for the special case where x is 0 in which case 0^0 is undefined.

Answer 2

$\mathcal{\huge{\underline{\underline{\orange{Solution:-}}}}}$

Method 1 :-

To Prove:- x^0=1

Proof:-

We know that x^0=x×x (0 Times)

we all know 1−1=0

so we can say that x0=x^1−1

Now We Got x^0=(x¹)×(x-¹)

x^0=x¹×1/x¹where x1 and x1 gets cancelled

Now ,we got x^0=1

Hence Proved

Method 2 :-

As we know that;

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

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

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

we have to find the value of x^0=

x, x^2 , x^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 ,

x^0= 1

x^1=x

x^2=x×x

x^3=x×x^2

Let;

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

As we know that,

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

Where as,

A = first term and d= common difference

Therefore, A= 1 Or x^0= 1

Hence proved

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