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studyopen101

  • 4 years ago

When a number has an exponent of zero, why does it equal 1? Shouldn't it be 0?

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  1. kevinkeegan
    • 4 years ago
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    anything to the power of 0 = 1

  2. integralsabiti
    • 4 years ago
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    anything but 0.

  3. integralsabiti
    • 4 years ago
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    \[0^0\neq0\]

  4. kevinkeegan
    • 4 years ago
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    if all fails use the calculator

  5. studyopen101
    • 4 years ago
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    I know it equals 1, but I want to know why.

  6. BMann
    • 4 years ago
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    no because the rule is 0 exponents always equal 1

  7. kevinkeegan
    • 4 years ago
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    why do you want to know why

  8. studyopen101
    • 4 years ago
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    math teacher told our class to find out

  9. m_charron2
    • 4 years ago
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    I guess you can prove it by stating this : x^y = x^a * x^b (provided that y = a+b) If b = 0, x^b needs to =1, else you'd have x^a * 0 = 0 =/= x^y

  10. m_charron2
    • 4 years ago
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    so by having x^0 = 1, x^a * 1 = x^a = x^(a+0) = x^y. Makes sense?

  11. studyopen101
    • 4 years ago
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    can u plug in numbers to that as an example please?

  12. integralsabiti
    • 4 years ago
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    consider a^1 ÷ a^1 = a / a = 1 but as per laws of indices, a / a = a^(1 – 1) = a^0 = 1 consider a^2 ÷ a^2 = a × a / a × a = 1 but as per laws of indices, a^2 / a^2 = a^(2 – 2) = a^0 = 1 or 5^3 ÷ 5^3 = 125 / 125 = 1 and 5^3 ÷ 5^3 = 5^(3 – 3) = 5^0 = 1 this shows that any number with exponent 0 is equal to 1

  13. m_charron2
    • 4 years ago
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    ah, yes, I much prefer intedralsabiti's proof than mine, a lot simpler

  14. m_charron2
    • 4 years ago
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    his proof also proves that 0^0 doesn't work because you'd need to divide 0 by 0. 0^2/0^2 is impossible, so 0^(2-2) also is, and finally, 0^0 is impossible as well.

  15. integralsabiti
    • 4 years ago
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    i dont deserve that medal.take it back. It is copy-paste

  16. m_charron2
    • 4 years ago
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    bah, then it's a medal for google-searching skills ;-) whatever works really

  17. integralsabiti
    • 4 years ago
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    thx :D

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