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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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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