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Math Help - Exponent Rules: (x)^1 and (x)^0

  1. #1
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    Exponent Rules: (x)^1 and (x)^0

    We know that anything raised to the first power equals itself.
    For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?
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  2. #2
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    Quote Originally Posted by RTC1996 View Post
    We know that anything raised to the first power equals itself.
    For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?
    Well, x^1 = x makes sense, does it not? If x^2 = x*x it also follows that x^1 = x. One x multiplied together (so to speak) is just x.

    Similar logic applies to x^0. Zero x's multiplied together is obviously 0. [EDIT: Wow, I've handled x^0 enough times you'd think I wouldn't mess this up! x^0 = 1 as explained below. Let that be a lesson to me!]

    Do note that 0^0 is a little tougher to settle. Why? An explanation can be found here:
    sci.math FAQ: What is 0^0?
    Last edited by Grep; December 1st 2010 at 05:54 PM. Reason: Fixed inaccuracy for 0^0
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  3. #3
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    \displaystyle x^0=x^{a-a}=\frac{x^a}{x^a}=1
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  4. #4
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    Quote Originally Posted by Grep View Post
    Well, x^1 = x makes sense, does it not? If x^2 = x*x it also follows that x^1 = x. One x multiplied together (so to speak) is just x.

    Similar logic applies to x^0. Zero x's multiplied together is obviously 0.

    Do note that 0^0 is undefined, though. Why? An explanation can be found here:
    sci.math FAQ: What is 0^0?
    \displaystyle x^0\neq 0
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  5. #5
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    \frac{a^x}{a^x} = 1...we know this.

    \frac{a^x}{a^y} = a^{x-y}

    \frac{a^x}{a^x} = 1 = a^{x-x} = a^0

    We have shown one statement.

    \frac{a^{x+1}}{a^x} = a = a^1

    This is the other statement.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    \displaystyle x^0\neq 0
    Sorry, wrote too quickly there, didn't I? My apologies.
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  7. #7
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    Quote Originally Posted by RTC1996 View Post
    We know that anything raised to the first power equals itself.
    For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?
    \displaystyle\frac{x^5}{x^3}=\frac{(x)(x)(x)(x)(x)  }{(x)(x)(x)}=\frac{(x)(x)(x)}{(x)(x)(x)}\;(x)(x)=(  1)x^2

    We can get this result by forming the new index from 5-3.
    Basically 3 of the x's in the denominator "cancel" 3 of the x's in the numerator to give 1.

    Hence we can write the "general" way as shown in earlier posts.

    Therefore, in index form... \displaystyle\frac{x^5}{x^5}=\frac{x^n}{x^n}=1=x^0 if we subtract the indices.

    x^0 is 1 because it is an "index" way to write a number divided by itself.

    \displaystyle\frac{x^5}{x^4}=\frac{x^3}{x^2}=\frac  {x^{n+1}}{x^n}=x^1=x

    Following on from this, you will see why negative indices represent a power of a number divided by a higher power of the same number.
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