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Why is -1^0 = -1 while (-1)^0 = 1

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi
    I was wondering why is negative one in brackets to the power of zero equal to positive one( (-1)^0 = 1 ) while negative one without brackets to the power of zero is equal to negative one( -1^0 = -1 ).

    2. Relevant equations
    none

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 4, 2017 #2

    PeroK

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    ##x^0 = 1## for all ##x \ne 0##

    The ##-## outisde the bracket is evaluated after the power is taken.
     
  4. Feb 4, 2017 #3

    berkeman

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    You can read about the Order of Operations here:

    http://www.studygs.net/pemdas/

    :smile:
     
  5. Feb 4, 2017 #4

    Mark44

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    This is so for the same reason that ##-2^2= -4## while ##(-2)^2 = 4##. As already mentioned, the reason is due to the order in which operations are applied. Without parentheses, the exponent operation takes precedence over the sign change operation. To force the sign change to happen before exponentiation, you need to use parentheses.

    (I call these -- [ ] -- brackets, and these -- ( ) -- parentheses. )
     
  6. Feb 5, 2017 #5

    rcgldr

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    This is an issue with syntax and order of operations, not logic. The statement negative one raised to any even power, including zero, equals positive one is true. In the case of the language APL, this isn't an issue since different symbols are used for negative sign and negation or subtract: ‾1 means negative 1, while -a means to negate a and a-b means a subtract b. in APL, a * b means raise a to power b (aⅹb is used for multiplication), and ‾1 * 0 returns a 1, no parenthesis needed.

    On a side note, APL evaluates expressions right to left without precedence, except for parenthesis.
     
  7. Feb 6, 2017 #6
    Thanks everyone
    I still don' t get it, could someone break it down for me I read about the order of operations but this still does not make any sense to me.

    " Without parentheses, the exponent operation takes precedence over the sign change operation"
    What sign change operation?
     
  8. Feb 6, 2017 #7

    rcgldr

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    The precedence of exponent operation before negation (sign change) is standard for mathematical formulas and most (but not all) programming languages. Consider the expression of a polynomial term such as ax^2, it means a(x^2), not (ax)^2, and if used in the form 0 - ax^2, it means 0 - (a(x^2)). Getting back to the example, this usage means that the order of operations for the expression -1^0 is to first evaluate 1^0 == 1, followed by evaluating -(1^0) == -(1) = -1.
     
  9. Feb 6, 2017 #8
    -1^0= -(1^0)
    Now a^0=1
    Therefore,
    (-1^0)=1 {Note that here a= (-1) }
    Mod edit: As noted by DrClaude, (-1^0) = -1, not 1. In this case (-1^0) is exactly the same as -1^0. That is, the parentheses don't have any effect.
    But there is a minus sign in the first one.
    So,
    -1^0= -(1^0)= -(1)= -1 {Here a=1 the minus is not included with a}
    Hope this solves your doubt.
     
    Last edited by a moderator: Feb 6, 2017
  10. Feb 6, 2017 #9

    DrClaude

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    You mean: (-1)^0 = 1
     
  11. Feb 6, 2017 #10
    Yes that's what I am trying to say
    Thanks
     
  12. Feb 6, 2017 #11
    Thanks for everyone

    @rcqldr and shihab-kol
    "-1^0= -(1^0)= -(1)= -1"
    Thanks for the breakdown helped at lot.
    But I still don't quite get how (-1)^0 equals 1 how exactly do the braces change the sign?
    if the braces take precedence then evaluating (-1) first still leaves you with -1^0,
    Now if the exponent takes precedence how does it change the sign when working with braces?
     
  13. Feb 6, 2017 #12

    Mark44

    Staff: Mentor

    Any finite, nonzero number raised to the power 0 is 1. In symbols, ##a^0 = 1##, for a finite and not equal to zero.
    Your question isn't very clear.
    ##-1^0## is the same as ##-(1^0) = -(1) = -1##.

    ##-1^0## means the negative of ##1^0##, or -1
    ##(-1)^0## means the quantity negative 1 is being raised to the power zero. As I said above, any nonzero, finite number raised to power zero is 1.
     
  14. Feb 7, 2017 #13
    Let us take a real number a {A real number includes all numbers, whether integers or rationals, and so(-1) is also included}
    a^0= a^(1-1) ..........{As 0= (1-1)}
    Now,
    a^(1-1)= a^1×a^(-1) ............{As a^x × a^x= a^(x+x)= a^2x Here I am only reversing the thing i.e. I am taking a^2x first and breaking it}
    a^1 × 1/a^1 ..............{As a^(-x)= 1/a^x}
    Now, they divide each other and so we are left with only 1.
    Now, in your case this a = (-1)
    -1 is a real number like a and so though it is a negative integer it still follows the above rule. You can check this by replacing a with -1 in the above process.
    It will still come out as 1 as -1/-1 = +1 (the minuses cancel out)
    I hope this clears your doubts.
    Cheers.
     
  15. Feb 7, 2017 #14

    PeroK

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    It no different from:

    ##-(2^2) = -(4) = -4##

    But

    ##(-2)^2 = (-2) \times (-2) = 4##

    PS There is also a thing called "order of operations", which tells you that:

    ##-2^2##

    Should be interpreted as ##-(2^2)##. So, in these cases you have to know this - the rule being that you do the power first and then apply the negative.
     
    Last edited: Feb 7, 2017
  16. Feb 7, 2017 #15
    Thanks everyone, now it all makes sense.
     
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