\(0 = k-k\) , Thus \(n^0 = n^{k-k}\) (say k is non-zero as we want to prove for 0) = \(\frac{n^k}{n^k} = 1\)

But this won't be true for \(n=0\) as the denominator becomes \(0^k = 0\) for non zero k , and hence the value is not defined (this is why we say that \(0^0\) is not defined . )
Thus ANYTHING raised to 0 is not 1 .... any "non-zero" thing raised to 0 is 1....
;)

## Comments

Sort by:

TopNewest\(0 = k-k\) , Thus \(n^0 = n^{k-k}\) (say k is non-zero as we want to prove for 0) = \(\frac{n^k}{n^k} = 1\)

But this won't be true for \(n=0\) as the denominator becomes \(0^k = 0\) for non zero k , and hence the value is not defined (this is why we say that \(0^0\) is not defined . ) Thus ANYTHING raised to 0 is not 1 .... any "non-zero" thing raised to 0 is 1....

;)

Log in to reply

as u said that , 0 power 0 is not defined.... bt as far i know.. 0 to the 0 is e whose value is 2.17828(approx.)

Log in to reply

Take example as 2^0. It can be expessed as 2^(x-x) = (2^x)/(2^x)=1

Log in to reply

Let a number \(a\) be raised to \(x\).

We have \(a^x\).

Let's divide it by itself-

\(\frac{a^x}{a^x}\).

Surely, these two quantities are the same-

\(\frac{a^x}{a^x} = \frac{a^x}{a^x}\)

Let's apply the law of indices to L.H.S & normal division to the R.H.S. Now, we have,

\(a^{x-x} = 1\)

\(\Rightarrow \boxed{a^{0} = 1}\).

Log in to reply