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raise to power zero?

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

Note by Raideep Singh
2 years, 12 months 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....
;) Aditya Raut · 2 years, 12 months ago

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@Aditya Raut 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.) Aman Agrawal · 2 years, 12 months ago

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Take example as 2^0. It can be expessed as 2^(x-x) = (2^x)/(2^x)=1 Kushagra Jaiswal · 2 years, 12 months ago

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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,

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

\(\Rightarrow \boxed{a^{0} = 1}\). Ameya Salankar · 2 years, 12 months ago

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