Question

how a^0 = 1?

Answer 1

by defult

Answer 2

I am going to show you.

\(\LARGE\color{black}{ \bf a^0 = a^{b-b}=a^b/a^b=1 }\) number divided by itself equals 1.

Answer 3

Need more clarification ?

Answer 4

a^0=a^(b-b) (because b-b=0 )
a^(b-b)=a^b / a^b (because when you divide you subtract exponents. we went beackwards here.)

a^b / a^b =1

just like anytime when you have a number divided by itself it is equal to 1, "a^b / a^b " is also 1 for the same reason.

Answer 5

Mathematicians defined \(a^0 = 0\), so that the rules of exponents work.

Also, if you observe the following, that definition will make sense.

An integer exponent greater than or equal to two is the number of equal factors multiplied together.
For example:
\(3^5 = 3 \times 3 \times 3 \times 3 \times 3\)
\(4^2 = 4 \times 4\)

Now look at this pattern:

\(a^5 = a \times a \times a \times a \times a\)
\(a^4 = a \times a \times a \times a\)
\(a^3 = a \times a \times a\)
\(a^2 = a \times a\)

That same definition of an integer exponent cannot work for an exponent of 1, since there is no multiplication with one single factor.
Definition: \(a^1 = a\), so we can add this line to out pattern.

\(a^5 = a \times a \times a \times a \times a\)
\(a^4 = a \times a \times a \times a\)
\(a^3 = a \times a \times a\)
\(a^2 = a \times a\)
\(a^1 = a\)

Notice that every time you subtract 1 from the exponent on the left side, you are dividing the right side by a. We can use this fact to continue the pattern:

\(a^5 = a \times a \times a \times a \times a\)
\(a^4 = a \times a \times a \times a\)
\(a^3 = a \times a \times a\)
\(a^2 = a \times a\)
\(a^1 = a\)
\(a^0 = 1\)
\(a^{-1} = \dfrac{1}{a}\)
\(a^{-2} = \dfrac{1}{a^2}\)
etc.

With the above pattern in mind, it makes sense to add these two definitions:

\(a^0 = 1\)

\(a^{-n} = \dfrac{1}{a^n}\)

In addition to the definitions making sense with the pattern, these definitions also work with the rules of operations with powers.

\(a^m \times a^n = a ^{m + n} \)

\(\dfrac{a^m}{a^n} = a^{m - n} \)

\((a^m)^n = a^{mn} \)