What does the following mean translated to English from mathematical language?
up vote
-2
down vote
favorite
What does this mean in English language translated from mathematical text?
$$S=left{ frac{1}{z} mid zin Rright}$$
complex-analysis complex-numbers
edited Nov 25 at 10:31
Robert Z
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
add a comment |
up vote
-2
down vote
favorite
What does this mean in English language translated from mathematical text?
$$S=left{ frac{1}{z} mid zin Rright}$$
complex-analysis complex-numbers
edited Nov 25 at 10:31
Robert Z
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
1
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
What does this mean in English language translated from mathematical text?
$$S=left{ frac{1}{z} mid zin Rright}$$
complex-analysis complex-numbers
edited Nov 25 at 10:31
Robert Z
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
What does this mean in English language translated from mathematical text?
$$S=left{ frac{1}{z} mid zin Rright}$$
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Nov 25 at 10:31
Robert Z
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
edited Nov 25 at 10:31
Robert Z
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
edited Nov 25 at 10:31
Robert Z
91.8k1058129
edited Nov 25 at 10:31
Robert Z
91.8k1058129
edited Nov 25 at 10:31
Robert Z
91.8k1058129
91.8k1058129
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
asked Nov 25 at 10:07
Riccardo CAIULO 12C02S
7
7
Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
1
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50
add a comment |
Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
1
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50
Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
1
1
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
Since you have chosen "complex analysis" as a tag I guess that here we should see $z$ as a complex number (although this is not explicitly written).
Having said that, I guess that $S={frac{1}{z};|;zinmathbb{R}}$ is the set of all the complex numbers $1/z$ such that $z$ is real number, that is the imaginary part of the complex number $z$ is zero.
In order to find a simpler description of the elements of the set $S$, note that, when $znot=0$, if the imaginary part of $z$ is zero then also imaginary part of $1/z$ is zero. Why? Is the converse true?
answered Nov 25 at 10:11
Robert Z
91.8k1058129
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
add a comment |
up vote
1
down vote
For all real numbers $z$ we compute the number $frac{1}{z}$, whenever possible (so when $z neq 0$) and put all these (and only these) in the set $S$. Note that the reciprocal of a real is also real, and every non-zero real can be written this way so $S$ is really the set of all nonzero real numbers, $mathbb{R}setminus{0}$, written in a weird way.
answered Nov 25 at 10:42
Henno Brandsma
103k345112
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
add a comment |
up vote
0
down vote
This is standard set notation. In English it is read as "The set of numbers $frac{1}{z}$ such that $z$ is a real number".
answered Nov 25 at 10:13
CoffeeCrow
578216
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since you have chosen "complex analysis" as a tag I guess that here we should see $z$ as a complex number (although this is not explicitly written).
Having said that, I guess that $S={frac{1}{z};|;zinmathbb{R}}$ is the set of all the complex numbers $1/z$ such that $z$ is real number, that is the imaginary part of the complex number $z$ is zero.
In order to find a simpler description of the elements of the set $S$, note that, when $znot=0$, if the imaginary part of $z$ is zero then also imaginary part of $1/z$ is zero. Why? Is the converse true?
answered Nov 25 at 10:11
Robert Z
91.8k1058129
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
add a comment |
up vote
2
down vote
Since you have chosen "complex analysis" as a tag I guess that here we should see $z$ as a complex number (although this is not explicitly written).
Having said that, I guess that $S={frac{1}{z};|;zinmathbb{R}}$ is the set of all the complex numbers $1/z$ such that $z$ is real number, that is the imaginary part of the complex number $z$ is zero.
In order to find a simpler description of the elements of the set $S$, note that, when $znot=0$, if the imaginary part of $z$ is zero then also imaginary part of $1/z$ is zero. Why? Is the converse true?
answered Nov 25 at 10:11
Robert Z
91.8k1058129
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
add a comment |
up vote
2
down vote
up vote
2
down vote
Since you have chosen "complex analysis" as a tag I guess that here we should see $z$ as a complex number (although this is not explicitly written).
Having said that, I guess that $S={frac{1}{z};|;zinmathbb{R}}$ is the set of all the complex numbers $1/z$ such that $z$ is real number, that is the imaginary part of the complex number $z$ is zero.
In order to find a simpler description of the elements of the set $S$, note that, when $znot=0$, if the imaginary part of $z$ is zero then also imaginary part of $1/z$ is zero. Why? Is the converse true?
answered Nov 25 at 10:11
Robert Z
91.8k1058129
Since you have chosen "complex analysis" as a tag I guess that here we should see $z$ as a complex number (although this is not explicitly written).
Having said that, I guess that $S={frac{1}{z};|;zinmathbb{R}}$ is the set of all the complex numbers $1/z$ such that $z$ is real number, that is the imaginary part of the complex number $z$ is zero.
In order to find a simpler description of the elements of the set $S$, note that, when $znot=0$, if the imaginary part of $z$ is zero then also imaginary part of $1/z$ is zero. Why? Is the converse true?
answered Nov 25 at 10:11
Robert Z
91.8k1058129
edited Nov 25 at 11:49
answered Nov 25 at 10:11
Robert Z
91.8k1058129
answered Nov 25 at 10:11
Robert Z
91.8k1058129
answered Nov 25 at 10:11
Robert Z
91.8k1058129
91.8k1058129
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
add a comment |
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
How did you read complex numbers into the set? It is the set of all real numbers 1/z, given z $in mathbb R$ or simply, the set of of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:21
1
1
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
@BertrandWittgenstein'sGhost The proposer choose "complex analysis" as a tag (see below the question). This is not explicitly said in the question, but it is plausible otherwise the exercise is quite trivial.
– Robert Z
Nov 25 at 11:37
1
1
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
@BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
– Robert Z
Nov 25 at 11:51
1
1
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
@BertrandWittgenstein'sGhost I edited my answer. I am not going to write the full answer. The downvote is fine. But if you ask for an explanation, please wait for it and then downvote...
– Robert Z
Nov 25 at 12:03
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
Even if we assume the set $S subseteq mathbb C$, the way you have defined it is not consistent, since $zneq 0$. Put this somewhere in your answer: The set of non-zero complex numbers $(a+bi)(b=0 land aneq 0)$. I upvoted it. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 25 at 12:04
add a comment |
up vote
1
down vote
For all real numbers $z$ we compute the number $frac{1}{z}$, whenever possible (so when $z neq 0$) and put all these (and only these) in the set $S$. Note that the reciprocal of a real is also real, and every non-zero real can be written this way so $S$ is really the set of all nonzero real numbers, $mathbb{R}setminus{0}$, written in a weird way.
answered Nov 25 at 10:42
Henno Brandsma
103k345112
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
add a comment |
up vote
1
down vote
For all real numbers $z$ we compute the number $frac{1}{z}$, whenever possible (so when $z neq 0$) and put all these (and only these) in the set $S$. Note that the reciprocal of a real is also real, and every non-zero real can be written this way so $S$ is really the set of all nonzero real numbers, $mathbb{R}setminus{0}$, written in a weird way.
answered Nov 25 at 10:42
Henno Brandsma
103k345112
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
add a comment |
up vote
1
down vote
up vote
1
down vote
For all real numbers $z$ we compute the number $frac{1}{z}$, whenever possible (so when $z neq 0$) and put all these (and only these) in the set $S$. Note that the reciprocal of a real is also real, and every non-zero real can be written this way so $S$ is really the set of all nonzero real numbers, $mathbb{R}setminus{0}$, written in a weird way.
answered Nov 25 at 10:42
Henno Brandsma
103k345112
For all real numbers $z$ we compute the number $frac{1}{z}$, whenever possible (so when $z neq 0$) and put all these (and only these) in the set $S$. Note that the reciprocal of a real is also real, and every non-zero real can be written this way so $S$ is really the set of all nonzero real numbers, $mathbb{R}setminus{0}$, written in a weird way.
answered Nov 25 at 10:42
Henno Brandsma
103k345112
answered Nov 25 at 10:42
Henno Brandsma
103k345112
answered Nov 25 at 10:42
Henno Brandsma
103k345112
answered Nov 25 at 10:42
Henno Brandsma
103k345112
103k345112
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
add a comment |
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
– Bertrand Wittgenstein's Ghost
Nov 25 at 11:25
add a comment |
up vote
0
down vote
This is standard set notation. In English it is read as "The set of numbers $frac{1}{z}$ such that $z$ is a real number".
answered Nov 25 at 10:13
CoffeeCrow
578216
add a comment |
up vote
0
down vote
This is standard set notation. In English it is read as "The set of numbers $frac{1}{z}$ such that $z$ is a real number".
answered Nov 25 at 10:13
CoffeeCrow
578216
add a comment |
up vote
0
down vote
up vote
0
down vote
This is standard set notation. In English it is read as "The set of numbers $frac{1}{z}$ such that $z$ is a real number".
answered Nov 25 at 10:13
CoffeeCrow
578216
This is standard set notation. In English it is read as "The set of numbers $frac{1}{z}$ such that $z$ is a real number".
answered Nov 25 at 10:13
CoffeeCrow
578216
answered Nov 25 at 10:13
CoffeeCrow
578216
answered Nov 25 at 10:13
CoffeeCrow
578216
answered Nov 25 at 10:13
CoffeeCrow
578216
578216
add a comment |
add a comment |
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Welcome to MSE! Please edit your question and replace the image with the code S={frac{1}{z};|;zinmathbb{R}} between dollars. See math.stackexchange.com/help/notation
– Robert Z
Nov 25 at 10:17
1
Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 at 10:25
Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 at 10:49
$S=mathbb Rsetminus{0}$, that means that $S$ is the set of all non-zero real numbers, in plain English.
– Michael Hoppe
Nov 25 at 10:50