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To the power of zero

  1. May 12, 2003 #1
    Now.

    2^4 = 16
    2^3 = 8
    2^2 = 4
    2^1 = 2
    2^0 = 1

    i can see how:
    2^3 = 2*2*2
    2^2 = 2*2
    2^1 = 2

    but why does anything to the power of 0 equal 1.

    i only know the rule that enything ot the power of 0 = 1 but why is it like that. Tell me, mathmaticly why it is.
     
  2. jcsd
  3. May 12, 2003 #2
    That's a damn good question. I would like to predict that people will answer that algebra alone provides this answer, rather than anything "logical". In the sense that if manipulating an equation that's large and has this X to the power of 0 in it, when manipulated the answer is only correct if we assume x^0 = 1

    Let's see if I'm close!
     
  4. May 12, 2003 #3
    23 / 23 =(2*2*2)/(2*2*2) =1

    2a / 2b = 2a-b

    got it?

    now

    23 / 23 =(2*2*2)/(2*2*2)= 23-3 = 20 = 1

    in general

    ab / ab
    = (a*a*a*....*a) / (a*a*a*...*a)
    [there are b a's in both the numerator and denominator]
    =1

    got it ?

    ab / ab = ab-b = 1

    therefore
    20 = 1
    30 = 1
    40 = 1
    a0 = 1

    where b is an interger, a is a real number
     
  5. May 12, 2003 #4

    Hurkyl

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    Like much of arithmetic, the meaning of a zero exponent (and negative, fractional, and irrational exponents) is chosen because it forms an extension of previous mathematics that obeys most of the original rules.

    Historically and logically, arithmetic starts with only the positive whole numbers and the increment operation, and everything beyond that is just layers of extensions.

    The relevant case here is that exponentiation is first recursively defined for positive integral exponents by:

    a1 = a
    a * an = an+1

    So the natural (ha ha!) thing to do is to flip the recursive step around and go backwards:

    an = an+1 / a

    and for a [x=] 0, this provides us with the definition for nonpositive integral exponents. In particular, for nonzero a:

    a0 = a1 / a = a / a = 1


    This isn't the only way, nor the only motivation, to arrive at nonpositive integral exponents. For instance, KL Kam gave reasoning based on using exponent arithmetic to extend the definition.


    And this discussion wouldn't be complete without touching on 00. Notice that the above recursive definition fails when a is zero, because we can't divide by zero! KL Kam's fails for the same reason. Mathematically, 00 is left undefined because there isn't a value you can choose which will work with all common manipulations where that may arise. If a value is assigned to it at all, it is context defined and is often explicitly stated... most commonly because it can simplify notation when you write a polynomial like:

    a + bx + cx2

    as

    ax0 + bx1 + cx2

    (Yes, I know it doesn't simplify this notation, but this rewriting is useful for more complicated expressions, like Taylor series)

    So because you're using x0 as a marker for the constant term it makes sense to define x0:=1 for x=0.
     
  6. May 12, 2003 #5

    HallsofIvy

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    Just look at the list you gave:
    2^4 = 16
    2^3 = 8
    2^2 = 4
    2^1 = 2

    To go from 2^4 to 2^3 you have to divide by 2: 16/2= 8.
    To go from 2^3 to 2^4 you have to divide by 2: 8/2= 4.
    To go from 2^2 to 2^1 you have to divide by 2: 4/2= 2.
    To go from 2^1 to 2^0 what do YOU think you should do?
    (Hint: 2/2= 1)

    That also why 2^(-1)= 1/2 and 2^(-2)= (1/2)/2= 1/4, etc.
     
  7. May 12, 2003 #6
    Yes...specialist math teacher told me that it works by

    a^1 / a^1
    =
    A^1-1
    =
    a^0

    and so as a^1 / a^1 = 1

    a^0 = 1

    and in the other ways you showed me..
     
  8. May 12, 2003 #7
    KL - Brilliantly said. You answers a question I'd bet most people wouldnt never know because it's overlooked. And you did it perfectly! Well done!
     
  9. May 13, 2003 #8
    On a related note, is there a reason 0! = 1?

    I've only heard one explination for that, but it was not algebraic.
     
  10. May 14, 2003 #9
    Here is an explanation that I've heard.
    4! = 4*3!
    3! = 3*2!
    2! = 2*1!
    1! = 1*0!

    If we work backward, we get
    3! = 4!/4
    2! = 3!/3
    1! = 2!/2
    0! = 1!/1 = 1

    Thanks. Yeah, I think lots of people overlook it but I'm sure there are lots of members here who can answer this question.
     
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