Proof that the square root of a negative number is real.











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So I stumbled upon this weird result when experimenting with fractional exponents.
Suppose you have some negative, real number, for example -8. We know $sqrt{-8}$ is not a real number.But
$$sqrt{-8} = {(-8)}^{frac1{2}}$$ and $$frac1{2} = frac2{4}$$ so
$$sqrt{-8} = {(-8)}^frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!

I can't seem to find out what the error of the proof is, any thoughts are appreciated!






roots exponentiation fake-proofs







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  • 1




    it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
    – mathworker21
    Nov 23 at 12:43

















up vote
2
down vote

favorite












So I stumbled upon this weird result when experimenting with fractional exponents.
Suppose you have some negative, real number, for example -8. We know $sqrt{-8}$ is not a real number.But
$$sqrt{-8} = {(-8)}^{frac1{2}}$$ and $$frac1{2} = frac2{4}$$ so
$$sqrt{-8} = {(-8)}^frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!

I can't seem to find out what the error of the proof is, any thoughts are appreciated!






roots exponentiation fake-proofs







share|cite|improve this question




















  • 1




    it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
    – mathworker21
    Nov 23 at 12:43















up vote
2
down vote

favorite









up vote
2
down vote

favorite











So I stumbled upon this weird result when experimenting with fractional exponents.
Suppose you have some negative, real number, for example -8. We know $sqrt{-8}$ is not a real number.But
$$sqrt{-8} = {(-8)}^{frac1{2}}$$ and $$frac1{2} = frac2{4}$$ so
$$sqrt{-8} = {(-8)}^frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!

I can't seem to find out what the error of the proof is, any thoughts are appreciated!






roots exponentiation fake-proofs







share|cite|improve this question















So I stumbled upon this weird result when experimenting with fractional exponents.
Suppose you have some negative, real number, for example -8. We know $sqrt{-8}$ is not a real number.But
$$sqrt{-8} = {(-8)}^{frac1{2}}$$ and $$frac1{2} = frac2{4}$$ so
$$sqrt{-8} = {(-8)}^frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!

I can't seem to find out what the error of the proof is, any thoughts are appreciated!






roots exponentiation fake-proofs




roots exponentiation fake-proofs






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share|cite|improve this question













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edited Nov 23 at 13:02









Henning Makholm

236k16300534




236k16300534










asked Nov 23 at 12:38









Magnus E-F

253




253








  • 1




    it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
    – mathworker21
    Nov 23 at 12:43
















  • 1




    it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
    – mathworker21
    Nov 23 at 12:43










1




1




it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
– mathworker21
Nov 23 at 12:43






it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$.
– mathworker21
Nov 23 at 12:43












3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.



The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $sqrt8$ and $isqrt 8$, but that does not make those two outputs equal.



If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.






share|cite|improve this answer






























    up vote
    1
    down vote













    A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.






    share|cite|improve this answer





















    • I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
      – Magnus E-F
      Nov 23 at 12:58


















    up vote
    0
    down vote













    You tell it yourself, $sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of



    $$left(sqrt{-8}right)^2 ?$$



    Mathematicians have found answers to this question, but it turns out that



    $$left(sqrt{-8}right)^{2}=sqrt{(-8)^2}$$ cannot hold !






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.



      The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $sqrt8$ and $isqrt 8$, but that does not make those two outputs equal.



      If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.



        The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $sqrt8$ and $isqrt 8$, but that does not make those two outputs equal.



        If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.



          The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $sqrt8$ and $isqrt 8$, but that does not make those two outputs equal.



          If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.






          share|cite|improve this answer














          You're using the power rule $a^{bc} = (a^b)^c$, but this rule does not hold in general unless $a$ is a positive real (and $b, c$ are both real too), or $b, c$ are both integers.



          The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $(a^b)^c$ come out the same as $a^{bc}$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $sqrt8$ and $isqrt 8$, but that does not make those two outputs equal.



          If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $a^{bc}= (a^b)^c$ fail for some values.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 13:00

























          answered Nov 23 at 12:54









          Henning Makholm

          236k16300534




          236k16300534






















              up vote
              1
              down vote













              A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.






              share|cite|improve this answer





















              • I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
                – Magnus E-F
                Nov 23 at 12:58















              up vote
              1
              down vote













              A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.






              share|cite|improve this answer





















              • I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
                – Magnus E-F
                Nov 23 at 12:58













              up vote
              1
              down vote










              up vote
              1
              down vote









              A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.






              share|cite|improve this answer












              A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $x^2=-8$ implies $x^4=+64$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $-8$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 23 at 12:46









              Oscar Lanzi

              12k12036




              12k12036












              • I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
                – Magnus E-F
                Nov 23 at 12:58


















              • I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
                – Magnus E-F
                Nov 23 at 12:58
















              I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
              – Magnus E-F
              Nov 23 at 12:58




              I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations.
              – Magnus E-F
              Nov 23 at 12:58










              up vote
              0
              down vote













              You tell it yourself, $sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of



              $$left(sqrt{-8}right)^2 ?$$



              Mathematicians have found answers to this question, but it turns out that



              $$left(sqrt{-8}right)^{2}=sqrt{(-8)^2}$$ cannot hold !






              share|cite|improve this answer

























                up vote
                0
                down vote













                You tell it yourself, $sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of



                $$left(sqrt{-8}right)^2 ?$$



                Mathematicians have found answers to this question, but it turns out that



                $$left(sqrt{-8}right)^{2}=sqrt{(-8)^2}$$ cannot hold !






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You tell it yourself, $sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of



                  $$left(sqrt{-8}right)^2 ?$$



                  Mathematicians have found answers to this question, but it turns out that



                  $$left(sqrt{-8}right)^{2}=sqrt{(-8)^2}$$ cannot hold !






                  share|cite|improve this answer












                  You tell it yourself, $sqrt{-8}$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of



                  $$left(sqrt{-8}right)^2 ?$$



                  Mathematicians have found answers to this question, but it turns out that



                  $$left(sqrt{-8}right)^{2}=sqrt{(-8)^2}$$ cannot hold !







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 13:09









                  Yves Daoust

                  123k668219




                  123k668219






























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