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Why 1=-1

  1. Jun 17, 2008 #1
    Just kidding, I know there's a flaw in my calculation (as a matter of fact, I'm quite sure what).
    Now, with that said;

    1^0.25=i, because i^4=1
    Based on my formula: a^x=b & b^y=a ***
    Where: x=1/y & y=1/x

    Further on, if:
    1^0.25=i
    then:
    1^4=i=1
    So:
    i^2 & 1^3=-1
    But 1^2 & 1^3=1
    So 1=-1

    ***Possibly only works for real numbers, not Complex nor Imaginary numbers, who probably has different rules set, which I'm not taught nor read about

    And I've got the impression of this forum lacking flamers; thank god for that; meaning I don't need to explain myself further more :D
    Thanks <3
    Just in case: this is really just a question: where's my flaw.
    Is it my formula who is wrong, or does it just only work for real numbers?
     
  2. jcsd
  3. Jun 17, 2008 #2

    D H

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    HallsofIvy already addressed this in your other thread on basically the same issue:
     
  4. Jun 17, 2008 #3
    Oh, I'm sorry, didn't pay attention.
    I'm "polen" allergic (idk the English word.. like grass and stuff), and I've been working for 9h just now, and I've been up for a while, from like 6 am, so I'm... totally... gone? Wasted? xD
    Thanks for your reply anyways.
    So my formula is correct for all real numbers then?

    Btw: how come it does not apply to complex and imaginary numbers?
    Oh.. and I don't get the similarity of a^x*b^x=(ab)^x and my formula..
    3^2*3^2=(3*3)^2 ?

    Edit: hahahahaha, I've made a thread of almost same content xD
    Sorry, sorry, sorry, delete thread plx.
     
    Last edited: Jun 17, 2008
  5. Jun 17, 2008 #4
    The word is "pollen", in case you're still wondering.
     
  6. Jun 17, 2008 #5
    Isn't the error rather assuming the solutions of x^4=1 are all equal?
     
  7. Jun 18, 2008 #6
    *-<|
    That made absolutely no sense.
     
  8. Jun 18, 2008 #7

    tiny-tim

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    "I have hayfever" …

    Bless you! :smile:

    (try loratadine … should be less than 20¢ a day … non-drowsy :wink: )
     
  9. Jun 18, 2008 #8
    I have that effect on many people, but still 1 has 4 quartic roots (1,-1,i and -i), none of which are equal to eachother. To demonstrate, we have i^4 = 1^4, take quartic root on each side, you get 1=i, which is false.
     
  10. Jun 18, 2008 #9
    You're a real quicky, aren't you..
    My question isn't whether or not they're equal, my question is why!

    @Topic
    I wrote this today, to figure out if a^x*b^x=(ab)^x only works for real numbers.
    The conclusion was: no, they work for complex numbers too:
    (i+1)^2(i/2)^2=((i+1)(i/2))^2
    -0,5i=-0,5i

    Though, I still don't get why that formula has anything to do with my:
    a^x=b
    a=b^y
    x=1/y
    y=1/x

    Which at least works for real numbers:
    a=7
    x=3
    b=7^3=343
    343^y=343^(1/x)=343^(1/3)=7=a

    Give me more than ignorant words to describe why it does not apply to complex values.
     
  11. Jun 18, 2008 #10

    D H

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    It doesn't even necessarily apply to real numbers. Just because (-1)2=(1)2 does not mean that -1=1.
     
  12. Jun 18, 2008 #11

    matt grime

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    Dark Fire: just because you verified something for one case does not mean it is true.

    Look, raising to the n'th power viewed as a function from C to C, generically, is an n to 1 map. It doesn't have an inverse. One can take a choice of root, a so-called branch, but it won't behave exactly as you seem to wish it to (this choice is what you did when you said the 1^0.25=i). It is up to you to justify why something should be true, since a priori there is no justifiable reason why it should.


    If you want to "see" what is going on, then you need to visualize the corrresponding Riemann surface. Imagine taking the complex plane and cutting along the positive real axis, then 'stretching and rotating' so you get something that looks a little like a helical screw, with four turns. This is the way to make sense of the 1 to n "function" that is raising to the power 0.25.
     
    Last edited: Jun 18, 2008
  13. Jun 27, 2008 #12
    First of all, x^(1/n) has a name. It is called the "nth root of x". For example, x^(1/7) is the 7th root of x. The thing you apparently don't completely understand is that there is more than one nth root of a number. In fact, there are exactly n (if you don't mind complex numbers). 1^(1/2) is either 1 or -1 (your choice), and 1^(1/4) is 1, -1, i, or -i. This doesn't mean they are equal.

    Another example would be if someone asked you where you live, but you have two houses. Either house would be a correct answer, but that doesn't make them the same house.

    The fundamental problem is that you seem to think that all answers to a question must be the same. Actually, the rule is: all answers to a question must be the same if the question is known to have only one answer. You are asking a question with more than one answer, so setting all of its answers equal to one another is pointless.
     
  14. Jun 27, 2008 #13

    D H

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    Xezlec, nice answer --- but, if you look at any post by the OP, you will see that the OP's name is stricken (has a horizontal line through the middle). This means the OP is taking a respite from PF for a while, and answering the thread is a bit moot.
     
  15. Jun 27, 2008 #14

    Garth

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    Well then, as this thread exists I cannot resist starting another discussion about -1 = 1 !

    How about:

    1 = 1
    1 = [itex]+\sqrt{(+1)}[/itex]

    But +1 = (-1).(-1), therefore

    1 = [itex]+\sqrt{(-1).(-1)}[/itex]
    1 = [itex]+\sqrt{(-1)}.\sqrt{(-1)}[/itex]

    put i = [itex]+\sqrt{-1}[/itex]

    Therefore from above

    1 = i.i
    1 = i2
    1 =-1.

    Just food for thought - something we resolved in the Sixth Form (High School). :wink:

    Garth
     
    Last edited: Jun 27, 2008
  16. Jun 27, 2008 #15

    D H

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    The function
    [tex]f(x) = \sqrt x[/tex]
    is defined for real, non-negative x only. The last step is not valid.
     
  17. Jun 27, 2008 #16
    sqrt(a*b)=sqrt(a)*sqrt(b) only for non-negative reals a,b
     
  18. Jun 27, 2008 #17

    Garth

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    Are you saying
    [tex]i = \sqrt{-1}[/tex]
    is not valid, i..e not allowing the whole of complex analysis?

    Garth
     
  19. Jun 27, 2008 #18

    Garth

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    That is correct, in fact if a < 0 and b < 0 then a.b > 0 but

    [tex]\sqrt{a.b} = [+\sqrt{a}].[-\sqrt{b}][/tex]

    but can you say why?

    It is instructive to do it on the Argand diagram.

    It is a useful insight into a potential 'sign' trap in algebra.

    Garth
     
    Last edited: Jun 27, 2008
  20. Jun 27, 2008 #19

    D H

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    Of course not. It is the square root sign that is the problem, as this is short for the square root function. Saying [itex]i = \sqrt{-1}[/itex] is abuse of notation. Abuse of notation is fine so long as you acknowledge it as such.

    A simple way around this "paradox" is to note that [itex]a^rb^r = (ab)^r[/itex] is valid only for two sets of cases: (1) r is a positive integer, or (2) a and b are positive reals and r is real.

    This answer begs the question: Why is that relation only valid for those limited cases? A better answer in my opinion is that you are asking for trouble any time you use a multivalued function in an equality and don't pay attention to the fact that the "function" is multivalued. For example, using the identity [itex]\sin^{-1}(\sin x) = x[/itex], [itex]\sin 2\pi = 0[/itex], and [itex]\asin 0 = 0[/itex] lets me conclude [itex]0=2\pi[/itex]. Saying
    [tex]1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot\sqrt{-1} = i^2 = -1[/tex]
    leads to a contradiction because you are using a multivalued function and pretending it is a true function.
     
  21. Jun 27, 2008 #20
    Oh.

    *shrug*
     
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