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Thread: any no raised to zero

  1. #1
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    Question any no raised to zero

    Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

    Regards,
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  2. #2
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    Consider Numerical Approach:

    $\displaystyle 5^2 = 25$
    $\displaystyle 5^1 = 5$
    $\displaystyle 5^0= 1$
    $\displaystyle 5^{-1}= \frac{1}{5}$

    See, We are dividing by 5 each of the power values to get the one below. At any number to the power of zero, we will get one. It's like that for any number.

    For example, 9:
    $\displaystyle 9^2 = 81$
    $\displaystyle 9^1 = 9$
    $\displaystyle 9^0= 1$
    $\displaystyle 9^{-1}= \frac{1}{9}$
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  3. #3
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    Quote Originally Posted by avinash View Post
    Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

    Regards,
    Well $\displaystyle 0^0$ is undefined, but other than that it is so that $\displaystyle a^xa^y=a^{x+y}$ works for any real $\displaystyle x$ and $\displaystyle y$ and $\displaystyle a >0$ . Also for other consistency reasons.

    RonL
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  4. #4
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    Quote Originally Posted by avinash View Post
    Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

    Well that is not always true $\displaystyle 0^0$ is not defined
    .
    However, if $\displaystyle x \ne 0 \Rightarrow \quad x^0 = 1$.
    Every nonzero number has a multiplicative inverse denoted by $\displaystyle x^{-1}$.
    Use the law of exponents: $\displaystyle 1 = \left( x \right)\left( {x^{ - 1} } \right) = x^{1 - 1} = x^0 $

    Alternatively: $\displaystyle 1 = \frac{{x^k }}{{x^k }} = x^{k - k} = x^0 $
    Thanks from ishandixit
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