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set of natural numbers subset of the set of real numbers [closed]
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The natural numbers are said to be a subset of the real numbers but how is this possible since in the set of natural numbers division is not allowed.
elementary-number-theory
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
asked Jul 19 at 5:21
Rene Dongren
41
closed as unclear what you're asking by Shailesh, Morgan Rodgers, Peter, amWhy, Parcly Taxel Jul 24 at 15:25
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
-2
down vote
favorite
The natural numbers are said to be a subset of the real numbers but how is this possible since in the set of natural numbers division is not allowed.
elementary-number-theory
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
asked Jul 19 at 5:21
Rene Dongren
41
closed as unclear what you're asking by Shailesh, Morgan Rodgers, Peter, amWhy, Parcly Taxel Jul 24 at 15:25
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.
4
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
2
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
2
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
The natural numbers are said to be a subset of the real numbers but how is this possible since in the set of natural numbers division is not allowed.
elementary-number-theory
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
asked Jul 19 at 5:21
Rene Dongren
41
The natural numbers are said to be a subset of the real numbers but how is this possible since in the set of natural numbers division is not allowed.
elementary-number-theory
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
asked Jul 19 at 5:21
Rene Dongren
41
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
edited Jul 19 at 10:32
G Tony Jacobs
25.6k43482
25.6k43482
asked Jul 19 at 5:21
Rene Dongren
41
asked Jul 19 at 5:21
Rene Dongren
41
asked Jul 19 at 5:21
Rene Dongren
41
41
closed as unclear what you're asking by Shailesh, Morgan Rodgers, Peter, amWhy, Parcly Taxel Jul 24 at 15:25
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.
closed as unclear what you're asking by Shailesh, Morgan Rodgers, Peter, amWhy, Parcly Taxel Jul 24 at 15:25
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.
4
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
2
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
2
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02
add a comment |Â
4
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
2
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
2
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02
4
4
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
2
2
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
2
2
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
Real numbers are basically all numbers, decimal, whole, negative, and positive except for imaginary numbers. They include both rational and irrational numbers. A more formal definition is any value that can represent a distance along a line (-ve and positive denoting direction).
Natural numbers are just whole positive numbers. Since whole positive numbers can represent a distance along a line, they are a subset of real numbers.
I really don't understand what division has to do with this.
answered Jul 19 at 5:38
Rumi
615
add a comment |Â
up vote
1
down vote
That the set of natural numbers is a subset of the set of real numbers just means that all natural numbers are also real numbers. You may be thinking of the term subfield, which is a subset that is a field with respect to the same operations as the larger set. Since the set of natural numbers has no multiplicative inverses (or additive for that matter), it is indeed not a subfield of the set of real numbers.
answered Jul 19 at 5:39
apanpapan3
119110
add a comment |Â
up vote
1
down vote
A set $A$ is said to be a subset of a set $B$ if and only if every member of $A$ is a member of $B$.
Since every natural number is a real number, the set of natural numbers($Bbb N$) is a subset of the set of real numbers($Bbb R$). It should be clear from the definition that containment(subset) relationships have nothing to do with the possible binary operations defined on the set.
What you might be thinking of is a group, in which case the set of non-zero real numbers form a group under multiplication but the set of natural numbers do not form a subgroup because you cannot guarantee a multiplicative inverse for every element.
answered Jul 19 at 5:48
John Mitchell
19510
add a comment |Â
up vote
1
down vote
You mention that, "in the set of natural numbers, division is not allowed". That's not entirely true. We can do lots of division with natural numbers, e.g., $6div 2=3$. I think you mean that the set of natural numbers is not "closed under division". That's true. There are lots of division problems involving natural numbers whose solution is not a natural number.
The set of real numbers is closed under division, with the usual provision that we don't divide by $0$. The natural numbers $6$ and $5$ are also real numbers, since the naturals are a subset of the reals. Their quotient, $frac65$, is another real number, one which is not a natural number. There is no problem here.
Being a subset of the real numbers doesn't mean retaining all of the closure properties of real numbers. It's like this: the prime numbers are a subset of the natural numbers, and even though the natural numbers are closed under addition, the prime numbers are not: $3+5=8$. Here, the sum of two primes equals a natural that is not a prime, just like we saw the quotient of two naturals can equal a real that is not a natural.
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Real numbers are basically all numbers, decimal, whole, negative, and positive except for imaginary numbers. They include both rational and irrational numbers. A more formal definition is any value that can represent a distance along a line (-ve and positive denoting direction).
Natural numbers are just whole positive numbers. Since whole positive numbers can represent a distance along a line, they are a subset of real numbers.
I really don't understand what division has to do with this.
answered Jul 19 at 5:38
Rumi
615
add a comment |Â
up vote
2
down vote
Real numbers are basically all numbers, decimal, whole, negative, and positive except for imaginary numbers. They include both rational and irrational numbers. A more formal definition is any value that can represent a distance along a line (-ve and positive denoting direction).
Natural numbers are just whole positive numbers. Since whole positive numbers can represent a distance along a line, they are a subset of real numbers.
I really don't understand what division has to do with this.
answered Jul 19 at 5:38
Rumi
615
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Real numbers are basically all numbers, decimal, whole, negative, and positive except for imaginary numbers. They include both rational and irrational numbers. A more formal definition is any value that can represent a distance along a line (-ve and positive denoting direction).
Natural numbers are just whole positive numbers. Since whole positive numbers can represent a distance along a line, they are a subset of real numbers.
I really don't understand what division has to do with this.
answered Jul 19 at 5:38
Rumi
615
Real numbers are basically all numbers, decimal, whole, negative, and positive except for imaginary numbers. They include both rational and irrational numbers. A more formal definition is any value that can represent a distance along a line (-ve and positive denoting direction).
Natural numbers are just whole positive numbers. Since whole positive numbers can represent a distance along a line, they are a subset of real numbers.
I really don't understand what division has to do with this.
answered Jul 19 at 5:38
Rumi
615
answered Jul 19 at 5:38
Rumi
615
answered Jul 19 at 5:38
Rumi
615
answered Jul 19 at 5:38
Rumi
615
615
add a comment |Â
add a comment |Â
up vote
1
down vote
That the set of natural numbers is a subset of the set of real numbers just means that all natural numbers are also real numbers. You may be thinking of the term subfield, which is a subset that is a field with respect to the same operations as the larger set. Since the set of natural numbers has no multiplicative inverses (or additive for that matter), it is indeed not a subfield of the set of real numbers.
answered Jul 19 at 5:39
apanpapan3
119110
add a comment |Â
up vote
1
down vote
That the set of natural numbers is a subset of the set of real numbers just means that all natural numbers are also real numbers. You may be thinking of the term subfield, which is a subset that is a field with respect to the same operations as the larger set. Since the set of natural numbers has no multiplicative inverses (or additive for that matter), it is indeed not a subfield of the set of real numbers.
answered Jul 19 at 5:39
apanpapan3
119110
add a comment |Â
up vote
1
down vote
up vote
1
down vote
That the set of natural numbers is a subset of the set of real numbers just means that all natural numbers are also real numbers. You may be thinking of the term subfield, which is a subset that is a field with respect to the same operations as the larger set. Since the set of natural numbers has no multiplicative inverses (or additive for that matter), it is indeed not a subfield of the set of real numbers.
answered Jul 19 at 5:39
apanpapan3
119110
That the set of natural numbers is a subset of the set of real numbers just means that all natural numbers are also real numbers. You may be thinking of the term subfield, which is a subset that is a field with respect to the same operations as the larger set. Since the set of natural numbers has no multiplicative inverses (or additive for that matter), it is indeed not a subfield of the set of real numbers.
answered Jul 19 at 5:39
apanpapan3
119110
answered Jul 19 at 5:39
apanpapan3
119110
answered Jul 19 at 5:39
apanpapan3
119110
answered Jul 19 at 5:39
apanpapan3
119110
119110
add a comment |Â
add a comment |Â
up vote
1
down vote
A set $A$ is said to be a subset of a set $B$ if and only if every member of $A$ is a member of $B$.
Since every natural number is a real number, the set of natural numbers($Bbb N$) is a subset of the set of real numbers($Bbb R$). It should be clear from the definition that containment(subset) relationships have nothing to do with the possible binary operations defined on the set.
What you might be thinking of is a group, in which case the set of non-zero real numbers form a group under multiplication but the set of natural numbers do not form a subgroup because you cannot guarantee a multiplicative inverse for every element.
answered Jul 19 at 5:48
John Mitchell
19510
add a comment |Â
up vote
1
down vote
A set $A$ is said to be a subset of a set $B$ if and only if every member of $A$ is a member of $B$.
Since every natural number is a real number, the set of natural numbers($Bbb N$) is a subset of the set of real numbers($Bbb R$). It should be clear from the definition that containment(subset) relationships have nothing to do with the possible binary operations defined on the set.
What you might be thinking of is a group, in which case the set of non-zero real numbers form a group under multiplication but the set of natural numbers do not form a subgroup because you cannot guarantee a multiplicative inverse for every element.
answered Jul 19 at 5:48
John Mitchell
19510
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A set $A$ is said to be a subset of a set $B$ if and only if every member of $A$ is a member of $B$.
Since every natural number is a real number, the set of natural numbers($Bbb N$) is a subset of the set of real numbers($Bbb R$). It should be clear from the definition that containment(subset) relationships have nothing to do with the possible binary operations defined on the set.
What you might be thinking of is a group, in which case the set of non-zero real numbers form a group under multiplication but the set of natural numbers do not form a subgroup because you cannot guarantee a multiplicative inverse for every element.
answered Jul 19 at 5:48
John Mitchell
19510
A set $A$ is said to be a subset of a set $B$ if and only if every member of $A$ is a member of $B$.
Since every natural number is a real number, the set of natural numbers($Bbb N$) is a subset of the set of real numbers($Bbb R$). It should be clear from the definition that containment(subset) relationships have nothing to do with the possible binary operations defined on the set.
What you might be thinking of is a group, in which case the set of non-zero real numbers form a group under multiplication but the set of natural numbers do not form a subgroup because you cannot guarantee a multiplicative inverse for every element.
answered Jul 19 at 5:48
John Mitchell
19510
answered Jul 19 at 5:48
John Mitchell
19510
answered Jul 19 at 5:48
John Mitchell
19510
answered Jul 19 at 5:48
John Mitchell
19510
19510
add a comment |Â
add a comment |Â
up vote
1
down vote
You mention that, "in the set of natural numbers, division is not allowed". That's not entirely true. We can do lots of division with natural numbers, e.g., $6div 2=3$. I think you mean that the set of natural numbers is not "closed under division". That's true. There are lots of division problems involving natural numbers whose solution is not a natural number.
The set of real numbers is closed under division, with the usual provision that we don't divide by $0$. The natural numbers $6$ and $5$ are also real numbers, since the naturals are a subset of the reals. Their quotient, $frac65$, is another real number, one which is not a natural number. There is no problem here.
Being a subset of the real numbers doesn't mean retaining all of the closure properties of real numbers. It's like this: the prime numbers are a subset of the natural numbers, and even though the natural numbers are closed under addition, the prime numbers are not: $3+5=8$. Here, the sum of two primes equals a natural that is not a prime, just like we saw the quotient of two naturals can equal a real that is not a natural.
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
add a comment |Â
up vote
1
down vote
You mention that, "in the set of natural numbers, division is not allowed". That's not entirely true. We can do lots of division with natural numbers, e.g., $6div 2=3$. I think you mean that the set of natural numbers is not "closed under division". That's true. There are lots of division problems involving natural numbers whose solution is not a natural number.
The set of real numbers is closed under division, with the usual provision that we don't divide by $0$. The natural numbers $6$ and $5$ are also real numbers, since the naturals are a subset of the reals. Their quotient, $frac65$, is another real number, one which is not a natural number. There is no problem here.
Being a subset of the real numbers doesn't mean retaining all of the closure properties of real numbers. It's like this: the prime numbers are a subset of the natural numbers, and even though the natural numbers are closed under addition, the prime numbers are not: $3+5=8$. Here, the sum of two primes equals a natural that is not a prime, just like we saw the quotient of two naturals can equal a real that is not a natural.
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You mention that, "in the set of natural numbers, division is not allowed". That's not entirely true. We can do lots of division with natural numbers, e.g., $6div 2=3$. I think you mean that the set of natural numbers is not "closed under division". That's true. There are lots of division problems involving natural numbers whose solution is not a natural number.
The set of real numbers is closed under division, with the usual provision that we don't divide by $0$. The natural numbers $6$ and $5$ are also real numbers, since the naturals are a subset of the reals. Their quotient, $frac65$, is another real number, one which is not a natural number. There is no problem here.
Being a subset of the real numbers doesn't mean retaining all of the closure properties of real numbers. It's like this: the prime numbers are a subset of the natural numbers, and even though the natural numbers are closed under addition, the prime numbers are not: $3+5=8$. Here, the sum of two primes equals a natural that is not a prime, just like we saw the quotient of two naturals can equal a real that is not a natural.
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
You mention that, "in the set of natural numbers, division is not allowed". That's not entirely true. We can do lots of division with natural numbers, e.g., $6div 2=3$. I think you mean that the set of natural numbers is not "closed under division". That's true. There are lots of division problems involving natural numbers whose solution is not a natural number.
The set of real numbers is closed under division, with the usual provision that we don't divide by $0$. The natural numbers $6$ and $5$ are also real numbers, since the naturals are a subset of the reals. Their quotient, $frac65$, is another real number, one which is not a natural number. There is no problem here.
Being a subset of the real numbers doesn't mean retaining all of the closure properties of real numbers. It's like this: the prime numbers are a subset of the natural numbers, and even though the natural numbers are closed under addition, the prime numbers are not: $3+5=8$. Here, the sum of two primes equals a natural that is not a prime, just like we saw the quotient of two naturals can equal a real that is not a natural.
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
answered Jul 19 at 10:28
G Tony Jacobs
25.6k43482
25.6k43482
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
add a comment |Â
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
thank you for the comment the question came up when discussing Eulers formula eiÀ + 1 = 0. Some of the participants said that the 1 in Eulers formula does not belong to the set of natural numbers and is different from the 1 of the set of real numbers. What is your opinion
– Rene Dongren
Jul 20 at 7:20
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
I think I understand your question better now. You might want to see my answer here: math.stackexchange.com/questions/2525182/…
– G Tony Jacobs
Jul 20 at 17:34
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
Basically, the way we define complex numbers, in terms of sets, is very different from how we define reals, which is different from how we define naturals. The complex number $1$ is technically the ordered pair $(1,0)$, where both elements of the pair are real numbers. The real number $1$ is technically the set of all sequences of rational numbers converging to the rational number $1$. The rational number $1$ is technically the set of all ordered pairs $(z,z)$ where $z$ is any integer. The integer $1$ is technically the set of ordered pairs $(n+1,n)$ where $n$ is any natural number.... (cont)
– G Tony Jacobs
Jul 20 at 17:38
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
...and finally, the natural number $1$ is technically defined as the set containing the empty set. The thing is, we identify the natural number $1$ with the integer $1$, which we identify with the rational number $1$, which we identify with the real number $1$, which we identify with the complex number $1$. Nobody treats them as different objects, which is why we give them all the same name. When we're talking about the properties of the complex number $1$, or of the natural number $1$, then nothing about the differing set-theoretic definitions affects how they behave, arithmetically.
– G Tony Jacobs
Jul 20 at 17:41
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
Since the distinction amounts to nothing, practically speaking, unless we're doing nothing but set theory, we all tend to ignore these distinctions, and simply assume that $Bbb NsubsetBbb ZsubsetBbb QsubsetBbb RsubsetBbb C$. There is a subset of $Bbb C$ that "looks just like" $Bbb N$, so we just pretend it's $Bbb N$. Mathematical objects are typically defined "up to isomorphism" anyway, so there's no harm done.
– G Tony Jacobs
Jul 20 at 17:43
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4
I don't understand the question. What has division to do with subsethood?
– Lord Shark the Unknown
Jul 19 at 5:22
2
Operations don't change the underlying set. Subsethood only concerns members of the sets.
– Doug Spoonwood
Jul 19 at 5:33
2
What does division have to do with subsets? orange, apple, pear is a subset of banana, apple,orange, pineapple,pear but division of fruit is not allowed.
– fleablood
Jul 19 at 5:40
Smell and hearing are a subset of the five senses, but you couldn't tell a red painted ball by them alone.
– dxiv
Jul 19 at 6:02