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.
complex-numbers definition proof-theory
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.
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up vote
9
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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
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
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
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
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
complex-numbers definition proof-theory
edited 2 days ago
Xander Henderson
13.7k93552
13.7k93552
asked Nov 17 at 0:53
ashold7
697
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
add a comment |
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
add a comment |
6 Answers
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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.
add a comment |
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).
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
add a comment |
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 $mathbb R$.
add a comment |
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.
add a comment |
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.
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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?
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
edited 2 days ago
Akiva Weinberger
13.6k12164
13.6k12164
answered Nov 17 at 1:02
Joey Kilpatrick
1,014121
1,014121
add a comment |
add a comment |
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).
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
add a comment |
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).
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
add a comment |
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).
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).
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
add a comment |
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
add a comment |
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 $mathbb R$.
add a comment |
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 $mathbb R$.
add a comment |
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 $mathbb R$.
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 $mathbb R$.
edited 2 days ago
Hanno
1,819424
1,819424
answered 2 days ago
Sujit Bhattacharyya
784216
784216
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 17 at 1:36
JoshuaZ
1,1401010
1,1401010
add a comment |
add a comment |
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.
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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.
add a comment |
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.
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.
answered 2 days ago
YiFan
1,5041311
1,5041311
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2
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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?
add a comment |
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?
add a comment |
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?
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?
answered 2 days ago
tonychow0929
15612
15612
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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