Why does every subfield of Complex number have a copy of Q?

In summary, the conversation discusses the proof that every subfield of complex numbers contains a copy of rational numbers. The proof begins by assuming that 0 and 1 are elements of the subfield and then shows that for any number p in the subfield, its inverse and negative must also be in the subfield. This means that every number of the form p/q (where p and q are non-zero) is also in the subfield. However, there is still a need to show that there are no cycles in the subfield that would prevent all rational numbers from being included. The conversation concludes with the suggestion to start with 1 and construct all natural numbers, integers, and quotients to show why the subfield must contain more
  • #1
Buffu
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Why does every subfield of Complex number have a copy of rational numbers ?

Here's my proof,

Let ##F## is a subield of ##\Bbb C##. I can assume that ##0, 1 \in F##.
Lets say a number ##p \in F##, then ##1/p \ p \ne 0## and ##-p## must be in ##F##.

Now since ##F## is subfield of ##\Bbb C## it has the operation of mutiplication and addition.

So for some ##q \in F## there exist ##q/p## and ##p/q##( ##p,q \ne 0## ) in ##F##. Therefore every number of form ##p/q, q \ne 0## exist in ##F##.

Is this enough ?
 
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  • #2
If your goal is to learn how to do formal math proofs, here are a couple of comments:
Is there something in the problem statement that rules out the trivial subfield {0}?
Buffu said:
Let ##F## is a subield of ##\Bbb C##. I can assume that ##0, 1 \in F##.
"assume" should raise a red flag. You should prove this.
Lets say a number ##p \in F##, then ##1/p \ p \ne 0## and ##-p## must be in ##F##.

Now since ##F## is subfield of ##\Bbb C## it has the operation of mutiplication and addition.

So for some ##q \in F## there exist ##q/p## and ##p/q##( ##p,q \ne 0## ) in ##F##. Therefore every number of form ##p/q, q \ne 0## exist in ##F##.

Is this enough ?
You should start with an arbitrary element of Q and show that it is in F.
 
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  • #3
Buffu said:
Why does every subfield of Complex number have a copy of rational numbers ?

Here's my proof,

Let ##F## is a subield of ##\Bbb C##. I can assume that ##0, 1 \in F##.
Lets say a number ##p \in F##, then ##1/p \ p \ne 0## and ##-p## must be in ##F##.

Now since ##F## is subfield of ##\Bbb C## it has the operation of mutiplication and addition.

So for some ##q \in F## there exist ##q/p## and ##p/q##( ##p,q \ne 0## ) in ##F##. Therefore every number of form ##p/q, q \ne 0## exist in ##F##.

Is this enough ?
As already mentioned, I would replace "assume" by an argument why this must be true, too. Besides this, the only question open is, why can't it happen that you get cycles in ##F##, i.e. do you obtain all quotients or might it happen that for ##p \in F## you get ##p \cdot n \in \{0,1\}## which would prevent you from getting all numbers you need. Why is this impossible? And why is there a ##p \in F## at all? ##0## and ##1## are basically clear, but why are there other elements? ##\mathbb{Z}_2## is also a field. Can it be a subfield of ##\mathbb{C}## ?
 
  • #4
FactChecker said:
If your goal is to learn how to do formal math proofs, here are a couple of comments:
Is there something in the problem statement that rules out the trivial subfield {0}?
"assume" should raise a red flag. You should prove this.You should start with an arbitrary element of Q and show that it is in F.

Original question :
upload_2017-6-11_21-3-43.png


I think the author assumes that we are talking about non trivial fields.

fresh_42 said:
As already mentioned, I would replace "assume" by an argument why this must be true, too.
I assumed ##0,1## because the definition of field given in book says that a field must contain additive and multiplicative inverse.

fresh_42 said:
Besides this, the only question open is, why can't it happen that you get cycles in ##F##, i.e. do you obtain all quotients or might it happen that for ##p \in F## you get ##p \cdot n \in \{0,1\}## which would prevent you from getting all numbers you need. Why is this impossible? And why is there a ##p \in F## at all? ##0## and ##1## are basically clear, but why are there other elements? ##\mathbb{Z}_2## is also a field. Can it be a subfield of ##\mathbb{C}## ?

What does ##p \cdot n ## mean ?
Is ##\Bbb Z_2## same as ##\Bbb Z^2## ?
 
  • #5
Buffu said:
Original question : View attachment 205245

I think the author assumes that we are talking about non trivial fields.
There is no such thing as a trivial field. Every field of characteristic zero contains the rational numbers. It is called the prime field of characteristic zero.
I assumed ##0,1## because the definition of field given in book says that a field must contain additive and multiplicative inverse.
Yes. Only the wording has been strange. You do not assume, you know that ##0,1## are in any field.
What does ##p \cdot n ## mean ?
It means ##p+p+ \ldots + p## , ##n-## times.
Is ##\Bbb Z_2## same as ##\Bbb Z^2## ?
No. ##\mathbb{Z}_2 = \{0,1\}##, the smallest possible field. ##\mathbb{Z}^2 = \mathbb{Z} \times \mathbb{Z}## is a ring and not a field.

You have to find an argument to get all rational numbers. But addition only gets you all, if there are no cycles, no numbers that turn to zero by addition like in ##\mathbb{Z}_2## in which ##1+1=0##. So why are all elements ##\frac{1}{p},\frac{2}{p}, \frac{3}{p}, \ldots ## different? It's not really hard to say why, it simply has to be mentioned. ##\mathbb{Z}_2 = \{0,1\} \subseteq \mathbb{C}## as a set, but not as a field. Where's the difference? And there is still an argument missing, why e.g. ##p \in F - \{0,1\}##. Again nothing complicated, just a short reason for it.

You can start with ##0## and ##1## which you have and then construct first all naturals (half group), then all integers (group, integral domain) and at last all quotients (field). There must be a reason why ##\mathbb{Z}_2## isn't the smallest subfield, which?
 
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  • #6
Start with ## 1##. Then ##1+1 ## is in ##F##, and so is ## \frac {1} {1+1} ## ...
 
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  • #7
Buffu said:
Original question : View attachment 205245
I think the author assumes that we are talking about non trivial fields.
Sorry. I stand corrected. The inclusion of different additive and multiplicative identity elements (0 and 1) is one of the field axioms. So @fresh_42 is correct -- there is no "trivial" field {0}.
 
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  • #8
fresh_42 said:
As already mentioned, I would replace "assume" by an argument why this must be true, too. Besides this, the only question open is, why can't it happen that you get cycles in ##F##, i.e. do you obtain all quotients or might it happen that for ##p \in F## you get ##p \cdot n \in \{0,1\}## which would prevent you from getting all numbers you need. Why is this impossible? And why is there a ##p \in F## at all? ##0## and ##1## are basically clear, but why are there other elements? ##\mathbb{Z}_2## is also a field. Can it be a subfield of ##\mathbb{C}## ?

Ok, then ##\Bbb Z_2## cannot be a subfield because a subfield of complex number has characteristics zero and ##\Bbb Z_2## has a characteristics of 2.
fresh_42 said:
There is no such thing as a trivial field. Every field of characteristic zero contains the rational numbers. It is called the prime field of characteristic zero.

Yes. Only the wording has been strange. You do not assume, you know that ##0,1## are in any field.

It means ##p+p+ \ldots + p## , ##n-## times.

No. ##\mathbb{Z}_2 = \{0,1\}##, the smallest possible field. ##\mathbb{Z}^2 = \mathbb{Z} \times \mathbb{Z}## is a ring and not a field.

You have to find an argument to get all rational numbers. But addition only gets you all, if there are no cycles, no numbers that turn to zero by addition like in ##\mathbb{Z}_2## in which ##1+1=0##. So why are all elements ##\frac{1}{p},\frac{2}{p}, \frac{3}{p}, \ldots ## different? It's not really hard to say why, it simply has to be mentioned. ##\mathbb{Z}_2 = \{0,1\} \subseteq \mathbb{C}## as a set, but not as a field. Where's the difference? And there is still an argument missing, why e.g. ##p \in F - \{0,1\}##. Again nothing complicated, just a short reason for it.

You can start with ##0## and ##1## which you have and then construct first all naturals (half group), then all integers (group, integral domain) and at last all quotients (field). There must be a reason why ##\mathbb{Z}_2## isn't the smallest subfield, which?
Ok here is my proof again.

Lets start with ##1##. Since ##1 + 1## cannot be zero, so we introduced a new element ##2##.
Next we to ##2 + 1## this also should not be zero because so we introduced element ##3##, and so on till be get all natural number.

There also exist multiplicative and additive inverse of each element in ##F##. Since we already know that any natural number added to any other won't give ##0##, so we introduced new elements ##-1, -2, -3, \cdots ## corresponding to ##1, 2, 3\cdots ##.
Similarly we get multiplicative inverses ##... -1/2, -1/1 , 1/1, 1/2 ...## for each element in ##F##.

Therefore we have got ##1/q## ( ##q \ne 0## ) in ##F## where ##q \in \Bbb Z##.
To get ##p/q## where ##p \in \Bbb Z## we just multiply ##p## and ##1/q##.

Thus we got every element of form ##p/q## where ##p,q \in \Bbb Z## and ##q \ne 0##.

Puff, is this correct now ?
 
  • #9
Yes. Only a minor remark: "because ##\operatorname{char} \mathbb{C} = 0## and a subfield cannot have another" provokes the question why? The reason is, that an equation ##1+ \ldots +1 = 0## in the subfield ##F## is also an equation in ##\mathbb{C}## and therefore it is impossible. I know it is kind of hair splitting, but it is not trivial, because it uses the fact, that the embedding ##F \subseteq \mathbb{C}## is injective. If you consider it trivial, then the entire exercise is.
 
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  • #10
fresh_42 said:
Yes. Only a minor remark: "because ##\operatorname{char} \mathbb{C} = 0## and a subfield cannot have another" provokes the question why? The reason is, that an equation ##1+ \ldots +1 = 0## in the subfield ##F## is also an equation in ##\mathbb{C}## and therefore it is impossible. I know it is kind of hair splitting, but it is not trivial, because it uses the fact, that the embedding ##F \subseteq \mathbb{C}## is injective. If you consider it trivial, then the entire exercise is.

Oh right I did not knew the proof of the that fact. I just took as stated thinking it would be very difficult to prove.

Is the converse also true ? that each field with ##\operatorname {char} 0## is a subfield of complex numbers ?
 
  • #11
Buffu said:
Oh right I did not knew the proof of the that fact. I just took as stated thinking it would be very difficult to prove.

Is the converse also true ? that each field with ##\operatorname {char} 0## is a subfield of complex numbers ?
No, this is wrong. The simplest example would be the field of rational functions with rational (or real) coefficients: ##\mathbb{Q}(x)##. It is the quotient field of the polynomial ring ##\mathbb{Q}[x]##, that is it contains all quotients of polynomials in ##x##. This field is obviously not contained in ##\mathbb{C}##. But all fields of characteristic ##0## contain the rational numbers, as you have just proven, because you haven't used any property of complex numbers beside the fact, that ##1+ \ldots +1 \neq 0## no matter how many ##1##'s we add.

Another not as simple example are the p-adic numbers with very unusual properties.
 
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  • #12
I think that worrying about cycles shows that you are taking a confusing approach to the proof. It may be just the wording that seems strange to me. You only need to show that F contains Q. So pick an arbitrary p/q number in Q and show that it is in F.
 
  • #13
FactChecker said:
I think that worrying about cycles shows that you are taking a confusing approach to the proof. It may be just the wording that seems strange to me. You only need to show that F contains Q. So pick an arbitrary p/q number in Q and show that it is in F.

Let ##p/q## be in ##\Bbb Q##. Since ##p/q = p\times 1/q = (1 + 1 + \stackrel{\text{p times}}{\cdots} + 1) \times 1/(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)##.

Since ##1,0## is there in ##F## and ##F## has a characteristic of ##0## therefore both ##(1 + 1 + \stackrel{\text{p times}}{\cdots} + 1)## and ##(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)## is in ##F##.

Done.
 
  • #14
Buffu said:
Let ##p/q## be in ##\Bbb Q##. Since ##p/q = p\times 1/q = (1 + 1 + \stackrel{\text{p times}}{\cdots} + 1) \times 1/(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)##.

Since ##1,0## is there in ##F## and ##F## has a characteristic of ##0## therefore both ##(1 + 1 + \stackrel{\text{p times}}{\cdots} + 1)## and ##(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)## is in ##F##.

Done.
For being finicky's sake:

Given ## p/q; q \neq 0 ##:
a)##p,q := 1+1+1+...## ( ##p, q## times respectively ; it is not 0 , since the Complexes have characteristic 0. )We have p,q
b) Every element in a field is invertible; in particular, q is invertible, so ## 1/q## is in ## F##
c) By closure of operations, ## p (1/q):=p/q ## is in ##F##.
 
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  • #15
@WWGD is not just being "finicky". If you are learning how to do math proofs, remember that proofs should be done in methodical "baby steps". Only the most trivial proofs can be done otherwise. Try to make every statement depend on one fundamental fact, that you state. You can start to skip and combine steps after you have a lot of experience.
 
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  • #16
FactChecker said:
@WWGD is not just being "finicky". If you are learning how to do math proofs, remember that proofs should be done in methodical "baby steps". Only the most trivial proofs can be done otherwise. Try to make every statement depend on one fundamental fact, that you state.
Thanks, it is just that one gets often "accused" of this when being (necessarily , I agree) careful.
 
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  • #17
Buffu said:
Since ##1,0## is there in ##F## and ##F## has a characteristic of ##0## therefore both ##(1 + 1 + \stackrel{\text{p times}}{\cdots} + 1)## and ##(1 + 1 + \stackrel{\text{q times}}{\cdots} + 1)## is in ##F##.
The fact that 1+1+...+1 (p times) is in F is due to closure under addition, not due to the characteristic. The sum is in F regardless of whether it cycled back to 0 or not. Similariy for q. You only need the characteristic to say that q≠0 and therefore 1/q is in F.
 
  • #18
FactChecker said:
The fact that 1+1+...+1 (p times) is in F is due to closure under addition, not due to the characteristic. The sum is in F regardless of whether it cycled back to 0 or not. Similariy for q. You only need the characteristic to say that q≠0 and therefore 1/q is in F.
Strictly speaking you also need the injectivity of ##F## in ##\mathbb{C}## to preserve the characteristic which is needed to get all rationals. So if it cycles back it is still in ##F## but that would not be enough. But you (both) are right: this is a good example to see that many properties of ##F## are actually used.
 
  • #19
fresh_42 said:
Strictly speaking you also need the injectivity of ##F## in ##\mathbb{C}## to preserve the characteristic which is needed to get all rationals. So if it cycles back it is still in ##F## but that would not be enough. But you (both) are right: this is a good example to see that many properties of ##F## are actually used.
When would you not have an injectivity, other than for a finite field?
 
  • #20
FactChecker said:
The fact that 1+1+...+1 (p times) is in F is due to closure under addition, not due to the characteristic. The sum is in F regardless of whether it cycled back to 0 or not. Similariy for q. You only need the characteristic to say that q≠0 and therefore 1/q is in F.

No I said about the characteristic of ##F## because if it cycles back to 0, then there is at least one rational number which is not in field.
Closure of addition is an axiom so I took it for granted.
 
  • #21
WWGD said:
When would you not have an injectivity, other than for a finite field?
Yes, but who said finite fields were impossible? I know it's hair splitting, but ##\{0,1\} \subset \mathbb{Z} \rightarrow \mathbb{Z}_2 = \{0,1\} \subset \mathbb{C}## would not work. That it is a subfield, which is an embedding (= injective), is important. Of course one could say subfields cannot have a distinct characteristic, but why?
 
  • #22
fresh_42 said:
Yes, but who said finite fields were impossible? I know it's hair splitting, but ##\{0,1\} \subset \mathbb{Z} \rightarrow \mathbb{Z}_2 = \{0,1\} \subset \mathbb{C}## would not work. That it is a subfield, which is an embedding (= injective), is important. Of course one could say subfields cannot have a distinct characteristic, but why?
Because if they are subfields, their properties are inherited from the top field? Isn't a subfield required to ?
 
  • #23
Buffu said:
No I said about the characteristic of ##F## because if it cycles back to 0, then there is at least one rational number which is not in field.
Closure of addition is an axiom so I took it for granted.
That is why I asked if you were supposed to be learning how to do proofs. We could take the entire thing for granted, but that is not a proof. We know that the axioms are true. If you are learning how to do math proofs, you have to refer to the things you know in writing where they apply and are used.
 
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  • #24
WWGD said:
Because if they are subfields, their properties are inherited from the top field? Isn't a subfield required to ?
Yes, of course, but subset isn't sufficient, so it formally takes the injectivity (and homomorphy) from the embedding to keep the characteristic in place.
 
  • #25
WWGD said:
Because if they are subfields, their properties are inherited from the top field? Isn't a subfield required to ?
The definition of the multiplication and addition operations of a subfield must be inherited from the field. It wouldn't make sense to say that the binary arithmetic operations can be used in a subfield of the rational numbers. I think that a complete proof of the OP would say that 1+1+...+1 q times is not 0 in the subfield because it is not 0 in the field.
 
  • #26
FactChecker said:
The definition of the multiplication and addition operations of a subfield must be inherited from the field. It wouldn't make sense to say that the binary arithmetic operations can be used in a subfield of the rational numbers.
Of course not, I never said nor believed this. Just that the subfield's properties are bound by those of the superfield.
 
  • #27
fresh_42 said:
Yes, of course, but subset isn't sufficient, so it formally takes the injectivity (and homomorphy) from the embedding to keep the characteristic in place.
But werebn't we talking about subfields?
 
  • #28
WWGD said:
Of course not, I never said nor believed this. Just that the subfield's properties are bound by those of the superfield.
I meant to be agreeing, not arguing with your statement.
 
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  • #29
FactChecker said:
I meant to be agreeing, not arguing with your statement.
Sorry, too much caffeine :).
 
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  • #30
WWGD said:
Sorry, too much caffeine :).
I just wasn't clear.
 
  • #31
WWGD said:
But werebn't we talking about subfields?
Yes, but we also got picky. ##\mathbb{Z}_7## is a field and ##\{0,1,2,3,4,5,6\} \subset \mathbb{C}##. And yes it is no subfield, because it cannot be embedded as a field into ##\mathbb{C}##. So in the same way you take this embedding for granted, @Buffu may take ##0,1 \in F##, the additional closure or other axioms for granted, which reduces the entire exercise to: Obvious. I only wanted to say, that this property in the meaning / definition of subfield is used same as the axioms of a field are used.
 
  • #32
FactChecker said:
That is why I asked if you were supposed to be learning how to do proofs. We could take the entire thing for granted, but that is not a proof. We know that the axioms are true. If you are learning how to do math proofs, you have to refer to the things you know in writing where they apply and are used.

I will mention every axiom used from next time when writing proofs :).
 
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  • #33
Buffu said:
I will mention every axiom used from next time when writing proofs :).
At least it will do no harm, except ...
Just a (not completely serious hint) in case you will turn in a paper for your master's degree or similar: Write it down in a way that you understand all steps. Make a copy for yourself. Then wipe out every second step and hand it over :wink:.
 
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  • #34
Buffu said:
I will mention every axiom used from next time when writing proofs :).
Good. The first few years of proofs should include all the steps. You will have a lot of time to learn which steps to skip. A thesis can skip some steps and an article for publication must skip a lot.
 
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1. Why do complex numbers have a copy of Q?

Complex numbers, denoted as C, are a mathematical set that includes all numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit. Q, or the set of rational numbers, is a subset of C. This means that every rational number can be expressed as a complex number. Thus, every subfield of complex numbers will also have a copy of Q.

2. How are complex numbers related to rational numbers?

As mentioned before, Q is a subset of C. This means that every rational number can be expressed as a complex number. Additionally, the set of rational numbers is closed under addition, subtraction, multiplication, and division, which are also operations in C. This shows a strong connection between complex and rational numbers.

3. Can complex numbers be used to represent all real numbers?

Yes, complex numbers can be used to represent all real numbers. This is because the set of real numbers is a subset of C. In fact, every real number can be expressed as a complex number with an imaginary part of 0. This is why complex numbers are often used in mathematics to solve problems involving real numbers.

4. Why are complex numbers important in mathematics?

Complex numbers are important in mathematics because they allow us to solve problems that cannot be solved using only real numbers. For example, complex numbers are used in solving polynomial equations, differential equations, and in the field of quantum mechanics. They also have many applications in engineering, physics, and other sciences.

5. Are there any practical applications of complex numbers?

Yes, there are many practical applications of complex numbers. For instance, they are used in electrical engineering to represent alternating current circuits, in signal processing to analyze signals, and in computer graphics to represent 3D objects. They are also used in economics, biology, and other fields to model complex systems and phenomena.

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