Does a mathematical proof in $mathbb{C}$ imply a proof in the $mathbb{R}$? [on hold]











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Does every proof in the complex numbers also prove the statement in the real numbers? I thought it might be true, because the real numbers are part of the complex numbers.










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put on hold as unclear what you're asking by Matthew Towers, quid 2 days ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 15




    In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
    – MJD
    Nov 17 at 0:55






  • 3




    If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
    – fleablood
    Nov 17 at 1:57






  • 4




    " I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
    – fleablood
    Nov 17 at 2:03






  • 2




    As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
    – spaceisdarkgreen
    Nov 17 at 2:58






  • 1




    If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
    – fleablood
    2 days ago















up vote
9
down vote

favorite
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Does every proof in the complex numbers also prove the statement in the real numbers? I thought it might be true, because the real numbers are part of the complex numbers.










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put on hold as unclear what you're asking by Matthew Towers, quid 2 days ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 15




    In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
    – MJD
    Nov 17 at 0:55






  • 3




    If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
    – fleablood
    Nov 17 at 1:57






  • 4




    " I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
    – fleablood
    Nov 17 at 2:03






  • 2




    As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
    – spaceisdarkgreen
    Nov 17 at 2:58






  • 1




    If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
    – fleablood
    2 days ago













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Does every proof in the complex numbers also prove the statement in the real numbers? I thought it might be true, because the real numbers are part of the complex numbers.










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Does every proof in the complex numbers also prove the statement in the real numbers? I thought it might be true, because the real numbers are part of the complex numbers.







complex-numbers definition proof-theory






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edited 2 days ago









Xander Henderson

13.7k93552




13.7k93552










asked Nov 17 at 0:53









ashold7

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697




put on hold as unclear what you're asking by Matthew Towers, quid 2 days ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Matthew Towers, quid 2 days ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 15




    In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
    – MJD
    Nov 17 at 0:55






  • 3




    If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
    – fleablood
    Nov 17 at 1:57






  • 4




    " I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
    – fleablood
    Nov 17 at 2:03






  • 2




    As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
    – spaceisdarkgreen
    Nov 17 at 2:58






  • 1




    If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
    – fleablood
    2 days ago














  • 15




    In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
    – MJD
    Nov 17 at 0:55






  • 3




    If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
    – fleablood
    Nov 17 at 1:57






  • 4




    " I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
    – fleablood
    Nov 17 at 2:03






  • 2




    As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
    – spaceisdarkgreen
    Nov 17 at 2:58






  • 1




    If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
    – fleablood
    2 days ago








15




15




In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
– MJD
Nov 17 at 0:55




In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true.
– MJD
Nov 17 at 0:55




3




3




If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
– fleablood
Nov 17 at 1:57




If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero.
– fleablood
Nov 17 at 1:57




4




4




" I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
– fleablood
Nov 17 at 2:03




" I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense.
– fleablood
Nov 17 at 2:03




2




2




As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
– spaceisdarkgreen
Nov 17 at 2:58




As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$"
– spaceisdarkgreen
Nov 17 at 2:58




1




1




If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
– fleablood
2 days ago




If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way.
– fleablood
2 days ago










6 Answers
6






active

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up vote
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If we show something is true for every complex number $zin mathbb{C}$, then we have shown that it is true for every $xin mathbb{R}$ since $mathbb{R}subsetmathbb{C}$.



However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $zin mathbb{C}$ such that $z^2=-1$, but there is no real number that has this property.






share|cite|improve this answer






























    up vote
    16
    down vote













    Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.



    This is not a complete answer but you need to think about "some" and "all".



    If X is true for all complex, it is true for all reals.



    If Y is true for some complex, it may or may not be true for some, all, or no reals.



    If W is true for no complex, it is not true for any reals.



    If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.



    If B is true for some reals then it is true for some complex.



    And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).






    share|cite|improve this answer

















    • 5




      I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
      – spaceisdarkgreen
      2 days ago








    • 1




      That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
      – fleablood
      2 days ago


















    up vote
    6
    down vote













    There is a problem in Kreyszig's Functional Analysis:



    Let $X$ be an inner product space over $mathbb{C}$, and $T:Xto X$ is a linear map. If $langle x,Txrangle =0:forall xin X$, then $T$ is the null transformation.



    This is not true for inner product spaces over $mathbb R$.






    share|cite|improve this answer






























      up vote
      4
      down vote













      This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.






      share|cite|improve this answer




























        up vote
        2
        down vote













        Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $mathbb C$ but not in $mathbb R$: the fundamental theorem of algebra. It states that a polynomial of degree $n$ always has exactly $n$ roots in $mathbb C$, which is however clearly not true in $mathbb R$ except degenerate cases. In fact, if FTA were to be true in $mathbb R$, it would be a strictly stronger statement and would imply FTA in $mathbb C$, not the other way round.






        share|cite|improve this answer




























          up vote
          2
          down vote













          It is more or less dangerous to think of proofs in complex analysis in real sense, because things in complex analysis are vastly different from those in real analysis.



          A function mapping from $mathbb{C}$ to $mathbb{C}$ that we concern in complex analysis is usually in terms of $z in mathbb{C}$ instead of individual $x, y in mathbb{R}$ even though we have $z = x+yi$ as usual. A differentiable (holomorphic) function in complex sense is a much stronger. Some results could be surprising for those who just started learning complex analysis.



          For example, a real differentiable function may not be twice differentiable, and its derivative may not even be continuous (that's why we have different regularity conditions like $C^1$, $C^2$ up to $C^infty$ and $C^omega$). However, holomorphic functions mapping from $mathbb{C}$ to $mathbb{C}$ are automatically differentiable infinitely many times. Liouville's Theorem states that any bounded entire (holomorphic in $mathbb{C}$) function is constant. This is certainly not true in real analysis: how would real analysis become if all bounded functions mapping from $mathbb{R}$ to $mathbb{R}$ that are differentiable on $mathbb{R}$ (e.g. $sin$, $cos$, $arctan$) are constant?






          share|cite|improve this answer




























            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            30
            down vote













            If we show something is true for every complex number $zin mathbb{C}$, then we have shown that it is true for every $xin mathbb{R}$ since $mathbb{R}subsetmathbb{C}$.



            However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $zin mathbb{C}$ such that $z^2=-1$, but there is no real number that has this property.






            share|cite|improve this answer



























              up vote
              30
              down vote













              If we show something is true for every complex number $zin mathbb{C}$, then we have shown that it is true for every $xin mathbb{R}$ since $mathbb{R}subsetmathbb{C}$.



              However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $zin mathbb{C}$ such that $z^2=-1$, but there is no real number that has this property.






              share|cite|improve this answer

























                up vote
                30
                down vote










                up vote
                30
                down vote









                If we show something is true for every complex number $zin mathbb{C}$, then we have shown that it is true for every $xin mathbb{R}$ since $mathbb{R}subsetmathbb{C}$.



                However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $zin mathbb{C}$ such that $z^2=-1$, but there is no real number that has this property.






                share|cite|improve this answer














                If we show something is true for every complex number $zin mathbb{C}$, then we have shown that it is true for every $xin mathbb{R}$ since $mathbb{R}subsetmathbb{C}$.



                However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $zin mathbb{C}$ such that $z^2=-1$, but there is no real number that has this property.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago









                Akiva Weinberger

                13.6k12164




                13.6k12164










                answered Nov 17 at 1:02









                Joey Kilpatrick

                1,014121




                1,014121






















                    up vote
                    16
                    down vote













                    Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.



                    This is not a complete answer but you need to think about "some" and "all".



                    If X is true for all complex, it is true for all reals.



                    If Y is true for some complex, it may or may not be true for some, all, or no reals.



                    If W is true for no complex, it is not true for any reals.



                    If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.



                    If B is true for some reals then it is true for some complex.



                    And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).






                    share|cite|improve this answer

















                    • 5




                      I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                      – spaceisdarkgreen
                      2 days ago








                    • 1




                      That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                      – fleablood
                      2 days ago















                    up vote
                    16
                    down vote













                    Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.



                    This is not a complete answer but you need to think about "some" and "all".



                    If X is true for all complex, it is true for all reals.



                    If Y is true for some complex, it may or may not be true for some, all, or no reals.



                    If W is true for no complex, it is not true for any reals.



                    If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.



                    If B is true for some reals then it is true for some complex.



                    And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).






                    share|cite|improve this answer

















                    • 5




                      I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                      – spaceisdarkgreen
                      2 days ago








                    • 1




                      That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                      – fleablood
                      2 days ago













                    up vote
                    16
                    down vote










                    up vote
                    16
                    down vote









                    Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.



                    This is not a complete answer but you need to think about "some" and "all".



                    If X is true for all complex, it is true for all reals.



                    If Y is true for some complex, it may or may not be true for some, all, or no reals.



                    If W is true for no complex, it is not true for any reals.



                    If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.



                    If B is true for some reals then it is true for some complex.



                    And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).






                    share|cite|improve this answer












                    Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.



                    This is not a complete answer but you need to think about "some" and "all".



                    If X is true for all complex, it is true for all reals.



                    If Y is true for some complex, it may or may not be true for some, all, or no reals.



                    If W is true for no complex, it is not true for any reals.



                    If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.



                    If B is true for some reals then it is true for some complex.



                    And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    fleablood

                    65.4k22682




                    65.4k22682








                    • 5




                      I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                      – spaceisdarkgreen
                      2 days ago








                    • 1




                      That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                      – fleablood
                      2 days ago














                    • 5




                      I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                      – spaceisdarkgreen
                      2 days ago








                    • 1




                      That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                      – fleablood
                      2 days ago








                    5




                    5




                    I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                    – spaceisdarkgreen
                    2 days ago






                    I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'.
                    – spaceisdarkgreen
                    2 days ago






                    1




                    1




                    That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                    – fleablood
                    2 days ago




                    That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory.
                    – fleablood
                    2 days ago










                    up vote
                    6
                    down vote













                    There is a problem in Kreyszig's Functional Analysis:



                    Let $X$ be an inner product space over $mathbb{C}$, and $T:Xto X$ is a linear map. If $langle x,Txrangle =0:forall xin X$, then $T$ is the null transformation.



                    This is not true for inner product spaces over $mathbb R$.






                    share|cite|improve this answer



























                      up vote
                      6
                      down vote













                      There is a problem in Kreyszig's Functional Analysis:



                      Let $X$ be an inner product space over $mathbb{C}$, and $T:Xto X$ is a linear map. If $langle x,Txrangle =0:forall xin X$, then $T$ is the null transformation.



                      This is not true for inner product spaces over $mathbb R$.






                      share|cite|improve this answer

























                        up vote
                        6
                        down vote










                        up vote
                        6
                        down vote









                        There is a problem in Kreyszig's Functional Analysis:



                        Let $X$ be an inner product space over $mathbb{C}$, and $T:Xto X$ is a linear map. If $langle x,Txrangle =0:forall xin X$, then $T$ is the null transformation.



                        This is not true for inner product spaces over $mathbb R$.






                        share|cite|improve this answer














                        There is a problem in Kreyszig's Functional Analysis:



                        Let $X$ be an inner product space over $mathbb{C}$, and $T:Xto X$ is a linear map. If $langle x,Txrangle =0:forall xin X$, then $T$ is the null transformation.



                        This is not true for inner product spaces over $mathbb R$.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 2 days ago









                        Hanno

                        1,819424




                        1,819424










                        answered 2 days ago









                        Sujit Bhattacharyya

                        784216




                        784216






















                            up vote
                            4
                            down vote













                            This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.






                            share|cite|improve this answer

























                              up vote
                              4
                              down vote













                              This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.






                              share|cite|improve this answer























                                up vote
                                4
                                down vote










                                up vote
                                4
                                down vote









                                This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.






                                share|cite|improve this answer












                                This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 17 at 1:36









                                JoshuaZ

                                1,1401010




                                1,1401010






















                                    up vote
                                    2
                                    down vote













                                    Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $mathbb C$ but not in $mathbb R$: the fundamental theorem of algebra. It states that a polynomial of degree $n$ always has exactly $n$ roots in $mathbb C$, which is however clearly not true in $mathbb R$ except degenerate cases. In fact, if FTA were to be true in $mathbb R$, it would be a strictly stronger statement and would imply FTA in $mathbb C$, not the other way round.






                                    share|cite|improve this answer

























                                      up vote
                                      2
                                      down vote













                                      Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $mathbb C$ but not in $mathbb R$: the fundamental theorem of algebra. It states that a polynomial of degree $n$ always has exactly $n$ roots in $mathbb C$, which is however clearly not true in $mathbb R$ except degenerate cases. In fact, if FTA were to be true in $mathbb R$, it would be a strictly stronger statement and would imply FTA in $mathbb C$, not the other way round.






                                      share|cite|improve this answer























                                        up vote
                                        2
                                        down vote










                                        up vote
                                        2
                                        down vote









                                        Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $mathbb C$ but not in $mathbb R$: the fundamental theorem of algebra. It states that a polynomial of degree $n$ always has exactly $n$ roots in $mathbb C$, which is however clearly not true in $mathbb R$ except degenerate cases. In fact, if FTA were to be true in $mathbb R$, it would be a strictly stronger statement and would imply FTA in $mathbb C$, not the other way round.






                                        share|cite|improve this answer












                                        Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $mathbb C$ but not in $mathbb R$: the fundamental theorem of algebra. It states that a polynomial of degree $n$ always has exactly $n$ roots in $mathbb C$, which is however clearly not true in $mathbb R$ except degenerate cases. In fact, if FTA were to be true in $mathbb R$, it would be a strictly stronger statement and would imply FTA in $mathbb C$, not the other way round.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 2 days ago









                                        YiFan

                                        1,5041311




                                        1,5041311






















                                            up vote
                                            2
                                            down vote













                                            It is more or less dangerous to think of proofs in complex analysis in real sense, because things in complex analysis are vastly different from those in real analysis.



                                            A function mapping from $mathbb{C}$ to $mathbb{C}$ that we concern in complex analysis is usually in terms of $z in mathbb{C}$ instead of individual $x, y in mathbb{R}$ even though we have $z = x+yi$ as usual. A differentiable (holomorphic) function in complex sense is a much stronger. Some results could be surprising for those who just started learning complex analysis.



                                            For example, a real differentiable function may not be twice differentiable, and its derivative may not even be continuous (that's why we have different regularity conditions like $C^1$, $C^2$ up to $C^infty$ and $C^omega$). However, holomorphic functions mapping from $mathbb{C}$ to $mathbb{C}$ are automatically differentiable infinitely many times. Liouville's Theorem states that any bounded entire (holomorphic in $mathbb{C}$) function is constant. This is certainly not true in real analysis: how would real analysis become if all bounded functions mapping from $mathbb{R}$ to $mathbb{R}$ that are differentiable on $mathbb{R}$ (e.g. $sin$, $cos$, $arctan$) are constant?






                                            share|cite|improve this answer

























                                              up vote
                                              2
                                              down vote













                                              It is more or less dangerous to think of proofs in complex analysis in real sense, because things in complex analysis are vastly different from those in real analysis.



                                              A function mapping from $mathbb{C}$ to $mathbb{C}$ that we concern in complex analysis is usually in terms of $z in mathbb{C}$ instead of individual $x, y in mathbb{R}$ even though we have $z = x+yi$ as usual. A differentiable (holomorphic) function in complex sense is a much stronger. Some results could be surprising for those who just started learning complex analysis.



                                              For example, a real differentiable function may not be twice differentiable, and its derivative may not even be continuous (that's why we have different regularity conditions like $C^1$, $C^2$ up to $C^infty$ and $C^omega$). However, holomorphic functions mapping from $mathbb{C}$ to $mathbb{C}$ are automatically differentiable infinitely many times. Liouville's Theorem states that any bounded entire (holomorphic in $mathbb{C}$) function is constant. This is certainly not true in real analysis: how would real analysis become if all bounded functions mapping from $mathbb{R}$ to $mathbb{R}$ that are differentiable on $mathbb{R}$ (e.g. $sin$, $cos$, $arctan$) are constant?






                                              share|cite|improve this answer























                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                It is more or less dangerous to think of proofs in complex analysis in real sense, because things in complex analysis are vastly different from those in real analysis.



                                                A function mapping from $mathbb{C}$ to $mathbb{C}$ that we concern in complex analysis is usually in terms of $z in mathbb{C}$ instead of individual $x, y in mathbb{R}$ even though we have $z = x+yi$ as usual. A differentiable (holomorphic) function in complex sense is a much stronger. Some results could be surprising for those who just started learning complex analysis.



                                                For example, a real differentiable function may not be twice differentiable, and its derivative may not even be continuous (that's why we have different regularity conditions like $C^1$, $C^2$ up to $C^infty$ and $C^omega$). However, holomorphic functions mapping from $mathbb{C}$ to $mathbb{C}$ are automatically differentiable infinitely many times. Liouville's Theorem states that any bounded entire (holomorphic in $mathbb{C}$) function is constant. This is certainly not true in real analysis: how would real analysis become if all bounded functions mapping from $mathbb{R}$ to $mathbb{R}$ that are differentiable on $mathbb{R}$ (e.g. $sin$, $cos$, $arctan$) are constant?






                                                share|cite|improve this answer












                                                It is more or less dangerous to think of proofs in complex analysis in real sense, because things in complex analysis are vastly different from those in real analysis.



                                                A function mapping from $mathbb{C}$ to $mathbb{C}$ that we concern in complex analysis is usually in terms of $z in mathbb{C}$ instead of individual $x, y in mathbb{R}$ even though we have $z = x+yi$ as usual. A differentiable (holomorphic) function in complex sense is a much stronger. Some results could be surprising for those who just started learning complex analysis.



                                                For example, a real differentiable function may not be twice differentiable, and its derivative may not even be continuous (that's why we have different regularity conditions like $C^1$, $C^2$ up to $C^infty$ and $C^omega$). However, holomorphic functions mapping from $mathbb{C}$ to $mathbb{C}$ are automatically differentiable infinitely many times. Liouville's Theorem states that any bounded entire (holomorphic in $mathbb{C}$) function is constant. This is certainly not true in real analysis: how would real analysis become if all bounded functions mapping from $mathbb{R}$ to $mathbb{R}$ that are differentiable on $mathbb{R}$ (e.g. $sin$, $cos$, $arctan$) are constant?







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 2 days ago









                                                tonychow0929

                                                15612




                                                15612















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