What does the following mean translated to English from mathematical language?












-2














What does this mean in English language translated from mathematical text?



$$S=left{ frac{1}{z} mid zin Rright}$$






complex-analysis complex-numbers







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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 '18 at 10:17








  • 1




    Fair question from a NEW contributor... Why downvotes? Why closing?
    – Robert Z
    Nov 25 '18 at 10:25












  • Stricto sensu this is undefined, as the inverse of $0$ is undefined.
    – Yves Daoust
    Nov 25 '18 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 '18 at 10:50


















-2














What does this mean in English language translated from mathematical text?



$$S=left{ frac{1}{z} mid zin Rright}$$






complex-analysis complex-numbers







share|cite|improve this question
























  • 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 '18 at 10:17








  • 1




    Fair question from a NEW contributor... Why downvotes? Why closing?
    – Robert Z
    Nov 25 '18 at 10:25












  • Stricto sensu this is undefined, as the inverse of $0$ is undefined.
    – Yves Daoust
    Nov 25 '18 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 '18 at 10:50
















-2












-2








-2







What does this mean in English language translated from mathematical text?



$$S=left{ frac{1}{z} mid zin Rright}$$






complex-analysis complex-numbers







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 10:31









Robert Z

93.3k1061132




93.3k1061132










asked Nov 25 '18 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 '18 at 10:17








  • 1




    Fair question from a NEW contributor... Why downvotes? Why closing?
    – Robert Z
    Nov 25 '18 at 10:25












  • Stricto sensu this is undefined, as the inverse of $0$ is undefined.
    – Yves Daoust
    Nov 25 '18 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 '18 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 '18 at 10:17








  • 1




    Fair question from a NEW contributor... Why downvotes? Why closing?
    – Robert Z
    Nov 25 '18 at 10:25












  • Stricto sensu this is undefined, as the inverse of $0$ is undefined.
    – Yves Daoust
    Nov 25 '18 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 '18 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 '18 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 '18 at 10:17






1




1




Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 '18 at 10:25






Fair question from a NEW contributor... Why downvotes? Why closing?
– Robert Z
Nov 25 '18 at 10:25














Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 '18 at 10:49




Stricto sensu this is undefined, as the inverse of $0$ is undefined.
– Yves Daoust
Nov 25 '18 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 '18 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 '18 at 10:50












3 Answers
3






active

oldest

votes


















2














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?






share|cite|improve this answer























  • 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 '18 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 '18 at 11:37






  • 1




    @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
    – Robert Z
    Nov 25 '18 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 '18 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 '18 at 12:04





















1














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.






share|cite|improve this answer





















  • This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
    – Bertrand Wittgenstein's Ghost
    Nov 25 '18 at 11:25



















0














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".






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    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?






    share|cite|improve this answer























    • 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 '18 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 '18 at 11:37






    • 1




      @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
      – Robert Z
      Nov 25 '18 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 '18 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 '18 at 12:04


















    2














    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?






    share|cite|improve this answer























    • 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 '18 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 '18 at 11:37






    • 1




      @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
      – Robert Z
      Nov 25 '18 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 '18 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 '18 at 12:04
















    2












    2








    2






    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?






    share|cite|improve this answer














    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?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 25 '18 at 11:49

























    answered Nov 25 '18 at 10:11









    Robert Z

    93.3k1061132




    93.3k1061132












    • 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 '18 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 '18 at 11:37






    • 1




      @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
      – Robert Z
      Nov 25 '18 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 '18 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 '18 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 '18 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 '18 at 11:37






    • 1




      @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
      – Robert Z
      Nov 25 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 11:37




    1




    1




    @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
    – Robert Z
    Nov 25 '18 at 11:51




    @BertrandWittgenstein'sGhost I hope that the proposer will clear up this mistery.
    – Robert Z
    Nov 25 '18 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 '18 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 '18 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 '18 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 '18 at 12:04













    1














    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.






    share|cite|improve this answer





















    • This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
      – Bertrand Wittgenstein's Ghost
      Nov 25 '18 at 11:25
















    1














    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.






    share|cite|improve this answer





















    • This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
      – Bertrand Wittgenstein's Ghost
      Nov 25 '18 at 11:25














    1












    1








    1






    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 '18 at 10:42









    Henno Brandsma

    105k346113




    105k346113












    • This is the best answer. The others lack colloquiality in translation: the set of non-zero real numbers.
      – Bertrand Wittgenstein's Ghost
      Nov 25 '18 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 '18 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 '18 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 '18 at 11:25











    0














    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".






    share|cite|improve this answer


























      0














      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".






      share|cite|improve this answer
























        0












        0








        0






        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".






        share|cite|improve this answer












        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".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 10:13









        CoffeeCrow

        578216




        578216






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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