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sasogeek

  • 2 years ago

Prove that any number \(\huge n \) to the power 0 is equal to one. \(\huge : n^0=1 \)

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  1. KingGeorge
    • 2 years ago
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    What about \(n=0\)? That's undefined.

  2. sasogeek
    • 2 years ago
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    make that exception, then prove for any other number

  3. ParthKohli
    • 2 years ago
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    Do you know that \(\Large \color{purple}{\rightarrow x^n \div x^1 = x^{n - 1} }\)

  4. KingGeorge
    • 2 years ago
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    We know that \(n^1=n\) and \(n^{-1}={1 \over n}\) for all \(n\). If we multiply together,\[n^1 \cdot n^{-1}=n^{1-1}=n^0\]\[n\cdot{1 \over n}={n \over n}=1\]So \(n^0=1\)

  5. ParthKohli
    • 2 years ago
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    \(\Large \color{purple}{\rightarrow x^1 = x }\) \(\Large \color{purple}{\rightarrow x^1 \div x^1 = 1 }\) as they cancel out. Also, x^1 over x^1 = x^0 \(\Large \color{purple}{\rightarrow x^0 = 1 = x^1 \div x^1 }\)

  6. UnkleRhaukus
    • 2 years ago
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    nice answers

  7. across
    • 2 years ago
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    @KingGeorge, I would say that for the case where \(n=0\) is more subjective than it is undefined nowadays. ;)

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