properties that real numbers hold but complex numbers does notWild automorphisms of the complex numbersTotal ordering on complex numbersWhat is a definable set?The Fundamental Theorem of Algebra and Complex NumbersHow are complex numbers useful to real number mathematics?Using the properties of real numbers, verify that complex numbers are associative and there exists an additive inverseComplex Numbers $stackrel?= mathbbR^ 2$If $z$ and $w$ are complex numbers can we use the proof in $mathbbR$ to demonstrate that $|z w|=|z||w|$?Complex Numbers Questions…Limit defintion of a function that is $mathbbR$-differentiable but not $mathbbC$ differentiable.Is there a natural way to prove trig identities also hold for complex numbers?

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properties that real numbers hold but complex numbers does not


Wild automorphisms of the complex numbersTotal ordering on complex numbersWhat is a definable set?The Fundamental Theorem of Algebra and Complex NumbersHow are complex numbers useful to real number mathematics?Using the properties of real numbers, verify that complex numbers are associative and there exists an additive inverseComplex Numbers $stackrel?= mathbbR^ 2$If $z$ and $w$ are complex numbers can we use the proof in $mathbbR$ to demonstrate that $|z w|=|z||w|$?Complex Numbers Questions…Limit defintion of a function that is $mathbbR$-differentiable but not $mathbbC$ differentiable.Is there a natural way to prove trig identities also hold for complex numbers?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


I need to find few examples about differences between real numbers and complex numbers like



1) if $x in mathbb R $ then $x^2 geq0$ is true



if $z in mathbb C $ then $z^2 geq0$ is false



2) let $a in mathbb R/0, 1 $ if $a^x =a^y$ then $x=y$ is true



let $ain mathbb z/0, 1 in mathbb C $ if $a^x =a^y$ then $x=y$ is false



but these examples "not cool enough feels like so trivial" can you recommend some properties like these



Thanks.










share|cite|improve this question









$endgroup$









  • 9




    $begingroup$
    If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
    $endgroup$
    – Arthur
    12 hours ago










  • $begingroup$
    true, but we can say that the statement is false right
    $endgroup$
    – Bad English
    12 hours ago







  • 4




    $begingroup$
    No. It is nonsensical. It's neither true nor false.
    $endgroup$
    – Arthur
    12 hours ago






  • 3




    $begingroup$
    This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
    $endgroup$
    – Mark Kamsma
    11 hours ago










  • $begingroup$
    @Arthur wow it's good to learn new things, thanks
    $endgroup$
    – Bad English
    11 hours ago

















2












$begingroup$


I need to find few examples about differences between real numbers and complex numbers like



1) if $x in mathbb R $ then $x^2 geq0$ is true



if $z in mathbb C $ then $z^2 geq0$ is false



2) let $a in mathbb R/0, 1 $ if $a^x =a^y$ then $x=y$ is true



let $ain mathbb z/0, 1 in mathbb C $ if $a^x =a^y$ then $x=y$ is false



but these examples "not cool enough feels like so trivial" can you recommend some properties like these



Thanks.










share|cite|improve this question









$endgroup$









  • 9




    $begingroup$
    If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
    $endgroup$
    – Arthur
    12 hours ago










  • $begingroup$
    true, but we can say that the statement is false right
    $endgroup$
    – Bad English
    12 hours ago







  • 4




    $begingroup$
    No. It is nonsensical. It's neither true nor false.
    $endgroup$
    – Arthur
    12 hours ago






  • 3




    $begingroup$
    This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
    $endgroup$
    – Mark Kamsma
    11 hours ago










  • $begingroup$
    @Arthur wow it's good to learn new things, thanks
    $endgroup$
    – Bad English
    11 hours ago













2












2








2





$begingroup$


I need to find few examples about differences between real numbers and complex numbers like



1) if $x in mathbb R $ then $x^2 geq0$ is true



if $z in mathbb C $ then $z^2 geq0$ is false



2) let $a in mathbb R/0, 1 $ if $a^x =a^y$ then $x=y$ is true



let $ain mathbb z/0, 1 in mathbb C $ if $a^x =a^y$ then $x=y$ is false



but these examples "not cool enough feels like so trivial" can you recommend some properties like these



Thanks.










share|cite|improve this question









$endgroup$




I need to find few examples about differences between real numbers and complex numbers like



1) if $x in mathbb R $ then $x^2 geq0$ is true



if $z in mathbb C $ then $z^2 geq0$ is false



2) let $a in mathbb R/0, 1 $ if $a^x =a^y$ then $x=y$ is true



let $ain mathbb z/0, 1 in mathbb C $ if $a^x =a^y$ then $x=y$ is false



but these examples "not cool enough feels like so trivial" can you recommend some properties like these



Thanks.







complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 12 hours ago









Bad EnglishBad English

3563 silver badges14 bronze badges




3563 silver badges14 bronze badges










  • 9




    $begingroup$
    If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
    $endgroup$
    – Arthur
    12 hours ago










  • $begingroup$
    true, but we can say that the statement is false right
    $endgroup$
    – Bad English
    12 hours ago







  • 4




    $begingroup$
    No. It is nonsensical. It's neither true nor false.
    $endgroup$
    – Arthur
    12 hours ago






  • 3




    $begingroup$
    This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
    $endgroup$
    – Mark Kamsma
    11 hours ago










  • $begingroup$
    @Arthur wow it's good to learn new things, thanks
    $endgroup$
    – Bad English
    11 hours ago












  • 9




    $begingroup$
    If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
    $endgroup$
    – Arthur
    12 hours ago










  • $begingroup$
    true, but we can say that the statement is false right
    $endgroup$
    – Bad English
    12 hours ago







  • 4




    $begingroup$
    No. It is nonsensical. It's neither true nor false.
    $endgroup$
    – Arthur
    12 hours ago






  • 3




    $begingroup$
    This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
    $endgroup$
    – Mark Kamsma
    11 hours ago










  • $begingroup$
    @Arthur wow it's good to learn new things, thanks
    $endgroup$
    – Bad English
    11 hours ago







9




9




$begingroup$
If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
$endgroup$
– Arthur
12 hours ago




$begingroup$
If $z in mathbb C $ then $z^2 geq0$ isn't just false. It makes no sense to ask about.
$endgroup$
– Arthur
12 hours ago












$begingroup$
true, but we can say that the statement is false right
$endgroup$
– Bad English
12 hours ago





$begingroup$
true, but we can say that the statement is false right
$endgroup$
– Bad English
12 hours ago





4




4




$begingroup$
No. It is nonsensical. It's neither true nor false.
$endgroup$
– Arthur
12 hours ago




$begingroup$
No. It is nonsensical. It's neither true nor false.
$endgroup$
– Arthur
12 hours ago




3




3




$begingroup$
This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
$endgroup$
– Mark Kamsma
11 hours ago




$begingroup$
This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not.
$endgroup$
– Mark Kamsma
11 hours ago












$begingroup$
@Arthur wow it's good to learn new things, thanks
$endgroup$
– Bad English
11 hours ago




$begingroup$
@Arthur wow it's good to learn new things, thanks
$endgroup$
– Bad English
11 hours ago










4 Answers
4






active

oldest

votes


















7













$begingroup$

The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.






share|cite|improve this answer











$endgroup$














  • $begingroup$
    It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
    $endgroup$
    – Calum Gilhooley
    5 hours ago










  • $begingroup$
    @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
    $endgroup$
    – Matthew Daly
    4 hours ago


















4













$begingroup$

The field of complex numbers is algebraically closed, but the field of real numbers is not.






share|cite|improve this answer









$endgroup$






















    3













    $begingroup$

    Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:



    Let's look at just the additive and multiplicative behavior - that is, we're considering $mathbbR$ and $mathbbC$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $alpha(x+y)=alpha(x)+alpha(y)$, and so forth.



    Every field has at least one automorphism, namely the identity (the trivial automorphism). $mathbbC$ has one obvious nontrivial automorphism, namely conjugation $$a+bimapsto a-biquad (a,binmathbbR),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $mathbbR$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $mathbbR$ is definable (which has a precise general meaning, incidentally) just from the field structure: $xge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $mathbbQ$ in $mathbbR$, we rule out any nontrivial automorphisms.



    The two points where this breaks down for $mathbbC$ are:



    • The relation $exists z(y+z^2=x)$ does not define an ordering on $mathbbC$ (indeed, $mathbbC$ as a field cannot be definably ordered at all).


    • $mathbbQ$ is not in fact dense in $mathbbC$. Even without a definable ordering, if $mathbbQ$ were dense in $mathbbC$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).






    share|cite|improve this answer









    $endgroup$






















      0













      $begingroup$

      Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).






      share|cite|improve this answer









      $endgroup$

















        Your Answer








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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7













        $begingroup$

        The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.






        share|cite|improve this answer











        $endgroup$














        • $begingroup$
          It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
          $endgroup$
          – Calum Gilhooley
          5 hours ago










        • $begingroup$
          @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
          $endgroup$
          – Matthew Daly
          4 hours ago















        7













        $begingroup$

        The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.






        share|cite|improve this answer











        $endgroup$














        • $begingroup$
          It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
          $endgroup$
          – Calum Gilhooley
          5 hours ago










        • $begingroup$
          @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
          $endgroup$
          – Matthew Daly
          4 hours ago













        7














        7










        7







        $begingroup$

        The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.






        share|cite|improve this answer











        $endgroup$



        The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 10 hours ago









        Matthew DalyMatthew Daly

        4,1794 silver badges26 bronze badges




        4,1794 silver badges26 bronze badges














        • $begingroup$
          It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
          $endgroup$
          – Calum Gilhooley
          5 hours ago










        • $begingroup$
          @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
          $endgroup$
          – Matthew Daly
          4 hours ago
















        • $begingroup$
          It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
          $endgroup$
          – Calum Gilhooley
          5 hours ago










        • $begingroup$
          @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
          $endgroup$
          – Matthew Daly
          4 hours ago















        $begingroup$
        It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
        $endgroup$
        – Calum Gilhooley
        5 hours ago




        $begingroup$
        It is worth noting that the lexicographical order on $mathbbR^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $mathbbC$. See Ordered vector space - Wikipedia.
        $endgroup$
        – Calum Gilhooley
        5 hours ago












        $begingroup$
        @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
        $endgroup$
        – Matthew Daly
        4 hours ago




        $begingroup$
        @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual.
        $endgroup$
        – Matthew Daly
        4 hours ago













        4













        $begingroup$

        The field of complex numbers is algebraically closed, but the field of real numbers is not.






        share|cite|improve this answer









        $endgroup$



















          4













          $begingroup$

          The field of complex numbers is algebraically closed, but the field of real numbers is not.






          share|cite|improve this answer









          $endgroup$

















            4














            4










            4







            $begingroup$

            The field of complex numbers is algebraically closed, but the field of real numbers is not.






            share|cite|improve this answer









            $endgroup$



            The field of complex numbers is algebraically closed, but the field of real numbers is not.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 12 hours ago









            WuestenfuxWuestenfux

            9,9622 gold badges6 silver badges16 bronze badges




            9,9622 gold badges6 silver badges16 bronze badges
























                3













                $begingroup$

                Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:



                Let's look at just the additive and multiplicative behavior - that is, we're considering $mathbbR$ and $mathbbC$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $alpha(x+y)=alpha(x)+alpha(y)$, and so forth.



                Every field has at least one automorphism, namely the identity (the trivial automorphism). $mathbbC$ has one obvious nontrivial automorphism, namely conjugation $$a+bimapsto a-biquad (a,binmathbbR),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $mathbbR$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $mathbbR$ is definable (which has a precise general meaning, incidentally) just from the field structure: $xge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $mathbbQ$ in $mathbbR$, we rule out any nontrivial automorphisms.



                The two points where this breaks down for $mathbbC$ are:



                • The relation $exists z(y+z^2=x)$ does not define an ordering on $mathbbC$ (indeed, $mathbbC$ as a field cannot be definably ordered at all).


                • $mathbbQ$ is not in fact dense in $mathbbC$. Even without a definable ordering, if $mathbbQ$ were dense in $mathbbC$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).






                share|cite|improve this answer









                $endgroup$



















                  3













                  $begingroup$

                  Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:



                  Let's look at just the additive and multiplicative behavior - that is, we're considering $mathbbR$ and $mathbbC$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $alpha(x+y)=alpha(x)+alpha(y)$, and so forth.



                  Every field has at least one automorphism, namely the identity (the trivial automorphism). $mathbbC$ has one obvious nontrivial automorphism, namely conjugation $$a+bimapsto a-biquad (a,binmathbbR),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $mathbbR$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $mathbbR$ is definable (which has a precise general meaning, incidentally) just from the field structure: $xge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $mathbbQ$ in $mathbbR$, we rule out any nontrivial automorphisms.



                  The two points where this breaks down for $mathbbC$ are:



                  • The relation $exists z(y+z^2=x)$ does not define an ordering on $mathbbC$ (indeed, $mathbbC$ as a field cannot be definably ordered at all).


                  • $mathbbQ$ is not in fact dense in $mathbbC$. Even without a definable ordering, if $mathbbQ$ were dense in $mathbbC$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).






                  share|cite|improve this answer









                  $endgroup$

















                    3














                    3










                    3







                    $begingroup$

                    Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:



                    Let's look at just the additive and multiplicative behavior - that is, we're considering $mathbbR$ and $mathbbC$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $alpha(x+y)=alpha(x)+alpha(y)$, and so forth.



                    Every field has at least one automorphism, namely the identity (the trivial automorphism). $mathbbC$ has one obvious nontrivial automorphism, namely conjugation $$a+bimapsto a-biquad (a,binmathbbR),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $mathbbR$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $mathbbR$ is definable (which has a precise general meaning, incidentally) just from the field structure: $xge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $mathbbQ$ in $mathbbR$, we rule out any nontrivial automorphisms.



                    The two points where this breaks down for $mathbbC$ are:



                    • The relation $exists z(y+z^2=x)$ does not define an ordering on $mathbbC$ (indeed, $mathbbC$ as a field cannot be definably ordered at all).


                    • $mathbbQ$ is not in fact dense in $mathbbC$. Even without a definable ordering, if $mathbbQ$ were dense in $mathbbC$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).






                    share|cite|improve this answer









                    $endgroup$



                    Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:



                    Let's look at just the additive and multiplicative behavior - that is, we're considering $mathbbR$ and $mathbbC$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $alpha(x+y)=alpha(x)+alpha(y)$, and so forth.



                    Every field has at least one automorphism, namely the identity (the trivial automorphism). $mathbbC$ has one obvious nontrivial automorphism, namely conjugation $$a+bimapsto a-biquad (a,binmathbbR),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $mathbbR$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $mathbbR$ is definable (which has a precise general meaning, incidentally) just from the field structure: $xge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $mathbbQ$ in $mathbbR$, we rule out any nontrivial automorphisms.



                    The two points where this breaks down for $mathbbC$ are:



                    • The relation $exists z(y+z^2=x)$ does not define an ordering on $mathbbC$ (indeed, $mathbbC$ as a field cannot be definably ordered at all).


                    • $mathbbQ$ is not in fact dense in $mathbbC$. Even without a definable ordering, if $mathbbQ$ were dense in $mathbbC$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    Noah SchweberNoah Schweber

                    139k10 gold badges166 silver badges315 bronze badges




                    139k10 gold badges166 silver badges315 bronze badges
























                        0













                        $begingroup$

                        Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).






                        share|cite|improve this answer









                        $endgroup$



















                          0













                          $begingroup$

                          Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).






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                            $begingroup$

                            Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).






                            share|cite|improve this answer









                            $endgroup$



                            Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









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