Why are the value of -a-0 nbsp- - 1-Why is not value of -a-0 nbsp- is 2-3 or any real number-

Answer : We know nbsp; : nbsp; ( nbsp; Anything )0 = nbsp; nbsp; 1 nbsp; In case of a0 , we can write it , As nbsp; : a0 = a1 1 = nbsp; a1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VI

Mathematics

Closure Property

Why are the v...

Question

Why are the value of (a)⁰= 1?
Why is not value of (a)⁰is 2,3 or any real number?

Open in App

Solution

Answer :

We know : ( Anything )⁰ = 1

In case of a⁰ , we can write it , As :

a⁰ = a^{1 - 1} = a¹ $\times$ a^{- 1} = $\frac{a}{a}$ = 1

So,

Value of a⁰ is always 1 not 2 , 3 or any other real number .

Suggest Corrections

Similar questions

Q. Consider

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

, where

a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}

are real numbers and n is a positive integer
The value of

a_{4 n - 1}

Q. Consider

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

, where

a_{0}, a_{1}, a_{2}, . . ., a_{4 n}

are real numbers and

n

is a positive integer. The value of

a_{2}

Let $(1 + x^{2})^{2} (1 + x)^{n} = A_{0} + A_{1} x + A_{2} x^{2} + . . . . . . . . . . . . . . I f A_{0}, A_{1}, A_{2}$ are in A.P., then the value of n is

Q. Let

{(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} X^{2} + \cdot \cdot \cdot

A_{0}, A_{1}, A_{2}

are in A.P. then the value of n is

Q. Consider

(1 + x + x^{2})^{2 n} = 4 n \sum r = 0 a_{r} x^{r}

, where

a_{0}, a_{1}, a_{2} . . . . . . a_{4 n}

are real numbers and

n

is a positive integer
The value of

n - 1 \sum r = 0 a_{2 r}

Join BYJU'S Learning Program

Why are the value of (a)0 = 1? Why is not value of (a)0 is 2,3 or any real number?

Answer : We know : ( Anything )0 = 1 In case of a0 , we can write it , As : a0 = a1 - 1 = a1 × a- 1 = aa = 1 So, Value of a0 is always 1 not 2 , 3 or any other real number .

Why are the value of (a)⁰= 1?
Why is not value of (a)⁰is 2,3 or any real number?

Answer :

We know : ( Anything )⁰ = 1

In case of a⁰ , we can write it , As :

a⁰ = a^{1 - 1} = a¹ $\times$ a^{- 1} = $\frac{a}{a}$ = 1

So,

Value of a⁰ is always 1 not 2 , 3 or any other real number .