Which step in this process allows me to erroneously conclude that $i = 1$












11












$begingroup$


I was playing around with imaginary numbers and exponents and came up with this:



$$ i = sqrt{-1} $$



$$ sqrt{-1} = (-1)^{1/2} $$
$$ (-1)^{1/2} = (-1)^{2/4} $$
$$ (-1)^{2/4} = ((-1)^{2})^{1/4} $$
$$ ((-1)^{2})^{1/4} = (1)^{1/4} $$
$$ (1)^{1/4} = 1 $$



Since we know that $i ne 1$, one of these steps must be incorrect, but I can't figure out which one it could be. Where have I gone wrong?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
    $endgroup$
    – gary
    Nov 22 '11 at 3:57






  • 2




    $begingroup$
    You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
    $endgroup$
    – GEdgar
    Nov 22 '11 at 4:13






  • 6




    $begingroup$
    @GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
    $endgroup$
    – Reid
    Nov 22 '11 at 5:13


















11












$begingroup$


I was playing around with imaginary numbers and exponents and came up with this:



$$ i = sqrt{-1} $$



$$ sqrt{-1} = (-1)^{1/2} $$
$$ (-1)^{1/2} = (-1)^{2/4} $$
$$ (-1)^{2/4} = ((-1)^{2})^{1/4} $$
$$ ((-1)^{2})^{1/4} = (1)^{1/4} $$
$$ (1)^{1/4} = 1 $$



Since we know that $i ne 1$, one of these steps must be incorrect, but I can't figure out which one it could be. Where have I gone wrong?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
    $endgroup$
    – gary
    Nov 22 '11 at 3:57






  • 2




    $begingroup$
    You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
    $endgroup$
    – GEdgar
    Nov 22 '11 at 4:13






  • 6




    $begingroup$
    @GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
    $endgroup$
    – Reid
    Nov 22 '11 at 5:13
















11












11








11





$begingroup$


I was playing around with imaginary numbers and exponents and came up with this:



$$ i = sqrt{-1} $$



$$ sqrt{-1} = (-1)^{1/2} $$
$$ (-1)^{1/2} = (-1)^{2/4} $$
$$ (-1)^{2/4} = ((-1)^{2})^{1/4} $$
$$ ((-1)^{2})^{1/4} = (1)^{1/4} $$
$$ (1)^{1/4} = 1 $$



Since we know that $i ne 1$, one of these steps must be incorrect, but I can't figure out which one it could be. Where have I gone wrong?










share|cite|improve this question











$endgroup$




I was playing around with imaginary numbers and exponents and came up with this:



$$ i = sqrt{-1} $$



$$ sqrt{-1} = (-1)^{1/2} $$
$$ (-1)^{1/2} = (-1)^{2/4} $$
$$ (-1)^{2/4} = ((-1)^{2})^{1/4} $$
$$ ((-1)^{2})^{1/4} = (1)^{1/4} $$
$$ (1)^{1/4} = 1 $$



Since we know that $i ne 1$, one of these steps must be incorrect, but I can't figure out which one it could be. Where have I gone wrong?







complex-numbers exponentiation fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '11 at 8:02









Srivatsan

20.9k371125




20.9k371125










asked Nov 22 '11 at 3:14









Peter OlsonPeter Olson

1,2721533




1,2721533








  • 1




    $begingroup$
    Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
    $endgroup$
    – gary
    Nov 22 '11 at 3:57






  • 2




    $begingroup$
    You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
    $endgroup$
    – GEdgar
    Nov 22 '11 at 4:13






  • 6




    $begingroup$
    @GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
    $endgroup$
    – Reid
    Nov 22 '11 at 5:13
















  • 1




    $begingroup$
    Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
    $endgroup$
    – gary
    Nov 22 '11 at 3:57






  • 2




    $begingroup$
    You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
    $endgroup$
    – GEdgar
    Nov 22 '11 at 4:13






  • 6




    $begingroup$
    @GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
    $endgroup$
    – Reid
    Nov 22 '11 at 5:13










1




1




$begingroup$
Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
$endgroup$
– gary
Nov 22 '11 at 3:57




$begingroup$
Once you start talking about complex n-th roots, you need to be careful; start defining branches, etc. e.g., when you say $(-1)^{1/2}$ , this is not a well-defined, or uniquely-defined object, since you have two candidates for the value of $(-1)^{1/2}$. You then create even more ambiguity by referring to $(-1)^{2/4}$, which has 4 possible candidates for the value. Basically you are implicitly using/assuming set equalities (a set for each possible solution of an n-th root) that do not really hold, and then treating sets as numbers.
$endgroup$
– gary
Nov 22 '11 at 3:57




2




2




$begingroup$
You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
$endgroup$
– GEdgar
Nov 22 '11 at 4:13




$begingroup$
You should cite a reputable reference for the properties of exponents you are using. They certainly are NOT taught in high-school algebra (where the base is required to be a positive real number)! As Jonas indicates, you will be unable to find such a reference! In fact, why not just go: $1^2 = (-1)^2$, take power $1/2$ on both sides, conclude $1=-1$.
$endgroup$
– GEdgar
Nov 22 '11 at 4:13




6




6




$begingroup$
@GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
$endgroup$
– Reid
Nov 22 '11 at 5:13






$begingroup$
@GEdgar: In my experience with high school (just a year ago for me), almost no properties are given restrictions except when their formal definitions are introduced at the very beginning, then they're never mentioned again. So, while the restrictions placed on a property may be "taught" in high school, that's not a guarantee: $sqrt{a} cdot sqrt{b} = sqrt{ab}$ is one that comes to mind almost immediately as having a restriction that's almost never mentioned in a high school class.
$endgroup$
– Reid
Nov 22 '11 at 5:13












4 Answers
4






active

oldest

votes


















18












$begingroup$

Exposing the scam becomes easier if we run an analogous one in the reals:



$-1 = (-1)^1$

$phantom{-1} = (-1)^{2/2}$

$phantom{-1} = ((-1)^2)^{1/2}$

$phantom{-1} = (1)^{1/2}$

$phantom{-1} = 1$



One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.






share|cite|improve this answer









$endgroup$





















    10












    $begingroup$

    It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.



    Already with $i=sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      The general rule $left(x^aright)^b = x^{acdot b}$ stands only for $left(a,bright) in mathbb{Z}^2$.



      Also the notation $sqrt{ times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$.
        The argument now reads:



        $ X = sqrt{X^2} $


        $ sqrt{X^2} = (X^2)^{1/2} $


        $ (X^2)^{1/2} = (X^2)^{2/4} $


        $ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $


        $ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $


        $ (X^4)^{1/4} = 1 $



        Some vertical equality signs are implied but not written.



        All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.

        In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.



        The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          The last step relies also on $X^4=1$.
          $endgroup$
          – Jonas Meyer
          Nov 22 '11 at 6:57











        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        18












        $begingroup$

        Exposing the scam becomes easier if we run an analogous one in the reals:



        $-1 = (-1)^1$

        $phantom{-1} = (-1)^{2/2}$

        $phantom{-1} = ((-1)^2)^{1/2}$

        $phantom{-1} = (1)^{1/2}$

        $phantom{-1} = 1$



        One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.






        share|cite|improve this answer









        $endgroup$


















          18












          $begingroup$

          Exposing the scam becomes easier if we run an analogous one in the reals:



          $-1 = (-1)^1$

          $phantom{-1} = (-1)^{2/2}$

          $phantom{-1} = ((-1)^2)^{1/2}$

          $phantom{-1} = (1)^{1/2}$

          $phantom{-1} = 1$



          One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.






          share|cite|improve this answer









          $endgroup$
















            18












            18








            18





            $begingroup$

            Exposing the scam becomes easier if we run an analogous one in the reals:



            $-1 = (-1)^1$

            $phantom{-1} = (-1)^{2/2}$

            $phantom{-1} = ((-1)^2)^{1/2}$

            $phantom{-1} = (1)^{1/2}$

            $phantom{-1} = 1$



            One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.






            share|cite|improve this answer









            $endgroup$



            Exposing the scam becomes easier if we run an analogous one in the reals:



            $-1 = (-1)^1$

            $phantom{-1} = (-1)^{2/2}$

            $phantom{-1} = ((-1)^2)^{1/2}$

            $phantom{-1} = (1)^{1/2}$

            $phantom{-1} = 1$



            One can now see the deception, going from the second line to the third: if we've decided to interpret $x^{1/2}$ always as positive square root (or indeed if we've made any choice of square roots in order to regard $x mapsto x^{1/2}$ as a function), then $x^{2/2} = (x^2)^{1/2}$ will not necessarily hold.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 '11 at 4:10









            Peter LeFanu LumsdainePeter LeFanu Lumsdaine

            5,76341840




            5,76341840























                10












                $begingroup$

                It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.



                Already with $i=sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.






                share|cite|improve this answer











                $endgroup$


















                  10












                  $begingroup$

                  It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.



                  Already with $i=sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.






                  share|cite|improve this answer











                  $endgroup$
















                    10












                    10








                    10





                    $begingroup$

                    It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.



                    Already with $i=sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.






                    share|cite|improve this answer











                    $endgroup$



                    It is true that $i$ and $1$ are both fourth roots of $1$, just not the same fourth root. You have chosen a branch of the fourth root function, and you have assumed algebraic properties of it that don't exist. You need to define $x^{1/4}$, $sqrt x$, etc.; there is no way to do so that is consistent with your steps. In general, don't assume $x^{ab}=(x^a)^b$ without reason.



                    Already with $i=sqrt{-1}$ I am suspicious. There is more than one squareroot of $-1$, and there can be trouble with that notation if you aren't careful. Rewriting that as $(-1)^{1/2}$ doesn't really change anything, but changing the exponent to $2/4$ indicates more trouble as you get ready to take fourth roots. There are 4 complex fourth roots of a nonzero complex number, and you are later going to make a choice about which to take. The step where $i$ becomes $((-1)^2)^{1/4}$ is definitely trouble; $(x^2)^{1/4}=x^{1/2}$ is not an identity for complex numbers. It will work sometimes, depending on what branches of the square root and fourth root functions you take and what $x$ is, and other times it will not work. You can think of your work as a proof that it does not hold with your choice of $(-1)^{1/2}=i$ and $1^{1/4}=1$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 22 '11 at 3:38

























                    answered Nov 22 '11 at 3:33









                    Jonas MeyerJonas Meyer

                    40.3k6146254




                    40.3k6146254























                        2












                        $begingroup$

                        The general rule $left(x^aright)^b = x^{acdot b}$ stands only for $left(a,bright) in mathbb{Z}^2$.



                        Also the notation $sqrt{ times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          The general rule $left(x^aright)^b = x^{acdot b}$ stands only for $left(a,bright) in mathbb{Z}^2$.



                          Also the notation $sqrt{ times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            The general rule $left(x^aright)^b = x^{acdot b}$ stands only for $left(a,bright) in mathbb{Z}^2$.



                            Also the notation $sqrt{ times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.






                            share|cite|improve this answer









                            $endgroup$



                            The general rule $left(x^aright)^b = x^{acdot b}$ stands only for $left(a,bright) in mathbb{Z}^2$.



                            Also the notation $sqrt{ times }$ is abusive and should not be used unless speaking of positive real numbers, because each complex number has two square roots (0 has only one) but for positive reals we can arbitrarily decide that this notation maps to the real positive one.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 '11 at 9:20









                            BenoitBenoit

                            31318




                            31318























                                1












                                $begingroup$

                                Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$.
                                The argument now reads:



                                $ X = sqrt{X^2} $


                                $ sqrt{X^2} = (X^2)^{1/2} $


                                $ (X^2)^{1/2} = (X^2)^{2/4} $


                                $ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $


                                $ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $


                                $ (X^4)^{1/4} = 1 $



                                Some vertical equality signs are implied but not written.



                                All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.

                                In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.



                                The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  The last step relies also on $X^4=1$.
                                  $endgroup$
                                  – Jonas Meyer
                                  Nov 22 '11 at 6:57
















                                1












                                $begingroup$

                                Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$.
                                The argument now reads:



                                $ X = sqrt{X^2} $


                                $ sqrt{X^2} = (X^2)^{1/2} $


                                $ (X^2)^{1/2} = (X^2)^{2/4} $


                                $ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $


                                $ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $


                                $ (X^4)^{1/4} = 1 $



                                Some vertical equality signs are implied but not written.



                                All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.

                                In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.



                                The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  The last step relies also on $X^4=1$.
                                  $endgroup$
                                  – Jonas Meyer
                                  Nov 22 '11 at 6:57














                                1












                                1








                                1





                                $begingroup$

                                Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$.
                                The argument now reads:



                                $ X = sqrt{X^2} $


                                $ sqrt{X^2} = (X^2)^{1/2} $


                                $ (X^2)^{1/2} = (X^2)^{2/4} $


                                $ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $


                                $ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $


                                $ (X^4)^{1/4} = 1 $



                                Some vertical equality signs are implied but not written.



                                All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.

                                In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.



                                The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.






                                share|cite|improve this answer









                                $endgroup$



                                Replace every mention (outside the exponent) of $i$, $-1$ and $1$ respectively by $X$, $X^2$ and $X^4$. This does not involve any ambiguity about signs or square roots or branch cuts, and uses only the always valid identities $i^2 = -1$, $(-1)^2 = 1$, and $(a^2)^2 = a^4$.
                                The argument now reads:



                                $ X = sqrt{X^2} $


                                $ sqrt{X^2} = (X^2)^{1/2} $


                                $ (X^2)^{1/2} = (X^2)^{2/4} $


                                $ (X^2)^{2/4} = ((X^2)^{2})^{1/4} $


                                $ ((X^2)^{2})^{1/4} = (X^4)^{1/4} $


                                $ (X^4)^{1/4} = 1 $



                                Some vertical equality signs are implied but not written.



                                All the equalities are either correct, or deduced from an axiom that $(T^{n})^{1/n} = T = (T^{1/n})^{n}$.

                                In each use of the axiom, $T$ is $X$, $X^2$ or $X^4$ and $n$ is $1,2$ or $4$.



                                The second equation in the axiom is correct. The first is not. There are difficulties in consistently choosing the sign of the square root, when working with complex numbers, and the first equation (in effect) pretends that a consistent choice of $n$-th root can always be made.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 22 '11 at 6:51









                                zyxzyx

                                31.5k33698




                                31.5k33698












                                • $begingroup$
                                  The last step relies also on $X^4=1$.
                                  $endgroup$
                                  – Jonas Meyer
                                  Nov 22 '11 at 6:57


















                                • $begingroup$
                                  The last step relies also on $X^4=1$.
                                  $endgroup$
                                  – Jonas Meyer
                                  Nov 22 '11 at 6:57
















                                $begingroup$
                                The last step relies also on $X^4=1$.
                                $endgroup$
                                – Jonas Meyer
                                Nov 22 '11 at 6:57




                                $begingroup$
                                The last step relies also on $X^4=1$.
                                $endgroup$
                                – Jonas Meyer
                                Nov 22 '11 at 6:57


















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