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# raise to power zero?

Why is it that anything raise to power 0 is 1?

Note by Raideep Singh
3 years, 1 month ago

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$$0 = k-k$$ , 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....
;) · 3 years, 1 month ago

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.) · 3 years, 1 month ago

Take example as 2^0. It can be expessed as 2^(x-x) = (2^x)/(2^x)=1 · 3 years, 1 month ago

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,

$$a^{x-x} = 1$$

$$\Rightarrow \boxed{a^{0} = 1}$$. · 3 years, 1 month ago