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;
$begingroup$
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.
complex-numbers
$endgroup$
add a comment |
$begingroup$
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.
complex-numbers
$endgroup$
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
add a comment |
$begingroup$
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.
complex-numbers
$endgroup$
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
complex-numbers
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The field of complex numbers is algebraically closed, but the field of real numbers is not.
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
The field of complex numbers is algebraically closed, but the field of real numbers is not.
$endgroup$
add a comment |
$begingroup$
The field of complex numbers is algebraically closed, but the field of real numbers is not.
$endgroup$
add a comment |
$begingroup$
The field of complex numbers is algebraically closed, but the field of real numbers is not.
$endgroup$
The field of complex numbers is algebraically closed, but the field of real numbers is not.
answered 12 hours ago
WuestenfuxWuestenfux
9,9622 gold badges6 silver badges16 bronze badges
9,9622 gold badges6 silver badges16 bronze badges
add a comment |
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
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).
answered 3 hours ago
Noah SchweberNoah Schweber
139k10 gold badges166 silver badges315 bronze badges
139k10 gold badges166 silver badges315 bronze badges
add a comment |
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
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).
answered 3 hours ago
G CabG Cab
23k3 gold badges13 silver badges45 bronze badges
23k3 gold badges13 silver badges45 bronze badges
add a comment |
add a comment |
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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
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@Arthur wow it's good to learn new things, thanks
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– Bad English
11 hours ago