Is this a pure imaginary number or real number?












3














Is $dfrac{0}{2yi}$ a pure imaginary number or a real number?



I'm debating, $0$ is a real number but if you divide by $i$, it's imaginary.






complex-numbers







share|cite|improve this question




















  • 2




    isnt that number just $0$?
    – Jorge Fernández
    Nov 30 at 22:33










  • Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
    – user334732
    Nov 30 at 23:16












  • You left out another possibility: when $y=0$, the expression is undefined.
    – amd
    Dec 1 at 0:06
















3














Is $dfrac{0}{2yi}$ a pure imaginary number or a real number?



I'm debating, $0$ is a real number but if you divide by $i$, it's imaginary.






complex-numbers







share|cite|improve this question




















  • 2




    isnt that number just $0$?
    – Jorge Fernández
    Nov 30 at 22:33










  • Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
    – user334732
    Nov 30 at 23:16












  • You left out another possibility: when $y=0$, the expression is undefined.
    – amd
    Dec 1 at 0:06














3












3








3







Is $dfrac{0}{2yi}$ a pure imaginary number or a real number?



I'm debating, $0$ is a real number but if you divide by $i$, it's imaginary.






complex-numbers







share|cite|improve this question















Is $dfrac{0}{2yi}$ a pure imaginary number or a real number?



I'm debating, $0$ is a real number but if you divide by $i$, it's imaginary.






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 23:15









user334732

4,24311240




4,24311240










asked Nov 30 at 22:33









Maske13

161




161








  • 2




    isnt that number just $0$?
    – Jorge Fernández
    Nov 30 at 22:33










  • Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
    – user334732
    Nov 30 at 23:16












  • You left out another possibility: when $y=0$, the expression is undefined.
    – amd
    Dec 1 at 0:06














  • 2




    isnt that number just $0$?
    – Jorge Fernández
    Nov 30 at 22:33










  • Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
    – user334732
    Nov 30 at 23:16












  • You left out another possibility: when $y=0$, the expression is undefined.
    – amd
    Dec 1 at 0:06








2




2




isnt that number just $0$?
– Jorge Fernández
Nov 30 at 22:33




isnt that number just $0$?
– Jorge Fernández
Nov 30 at 22:33












Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
– user334732
Nov 30 at 23:16






Hi, and welcome to MSE. This is a great help on here: math.meta.stackexchange.com/questions/5020 Look what I did to your post! Click edit and you can see the MathJax that does this.
– user334732
Nov 30 at 23:16














You left out another possibility: when $y=0$, the expression is undefined.
– amd
Dec 1 at 0:06




You left out another possibility: when $y=0$, the expression is undefined.
– amd
Dec 1 at 0:06










4 Answers
4






active

oldest

votes


















2














You told us nothing about $y$ but, assuming that $y$ is a non-zero complex number, then $dfrac0{2yi}=0$, which is both a real number and a pure imaginary number. It's actually the only complex number with both properties.






share|cite|improve this answer





























    2














    If $yneq 0$, in the complex plane, by definition, $0=0+0i$. Since the imaginary and real parts are 0, 0 is purely real and imaginary. However, it's also member of the complex numbers.






    share|cite|improve this answer





























      1














      Well, provided that $y neq 0$, you are multiplying by $frac{1}{2y}$. (Of course, if $y = 0$ then it is undefined.) So, I will talk of multiplication rather than division. I am guessing that $y$ is intended to be real but that does not actually affect the answer.



      A complex number is real if the imaginary component is zero. Conversely, it is imaginary if the real component is zero. Most complex numbers e.g. $1 + i$ are neither. $0$ is special in that it is both.



      Generally multiplying by $i$ will flip real numbers to imaginary and vice versa. Since $0 times i = 0$ multiplying by $i$ does change it, it goes from both to both.



      Something a little like this occurs with the more familiar real numbers. Multiplying by $-1$ flips positive and negative. It leaves $0$ alone, so is $0$ positive or negative? The usual answer is that $0$ is neither but considering it as both could work. The Bourbaki (Wikipedia) school proposed this. I found a reference thanks to posting a question here: Bourbaki and zero (this site).






      share|cite|improve this answer























      • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
        – max_zorn
        Dec 1 at 0:00










      • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
        – badjohn
        Dec 1 at 9:23



















      0














      We have that



      $$frac{0}{2yi}=0$$



      which is an integer, a rational, a real and a complex number.



      Notably it indicates the neutral element with respect to addition.






      share|cite|improve this answer





















      • ...and the absorbing element under multiplication.
        – user334732
        Nov 30 at 23:16











      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020734%2fis-this-a-pure-imaginary-number-or-real-number%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      You told us nothing about $y$ but, assuming that $y$ is a non-zero complex number, then $dfrac0{2yi}=0$, which is both a real number and a pure imaginary number. It's actually the only complex number with both properties.






      share|cite|improve this answer


























        2














        You told us nothing about $y$ but, assuming that $y$ is a non-zero complex number, then $dfrac0{2yi}=0$, which is both a real number and a pure imaginary number. It's actually the only complex number with both properties.






        share|cite|improve this answer
























          2












          2








          2






          You told us nothing about $y$ but, assuming that $y$ is a non-zero complex number, then $dfrac0{2yi}=0$, which is both a real number and a pure imaginary number. It's actually the only complex number with both properties.






          share|cite|improve this answer












          You told us nothing about $y$ but, assuming that $y$ is a non-zero complex number, then $dfrac0{2yi}=0$, which is both a real number and a pure imaginary number. It's actually the only complex number with both properties.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 22:37









          José Carlos Santos

          150k22120221




          150k22120221























              2














              If $yneq 0$, in the complex plane, by definition, $0=0+0i$. Since the imaginary and real parts are 0, 0 is purely real and imaginary. However, it's also member of the complex numbers.






              share|cite|improve this answer


























                2














                If $yneq 0$, in the complex plane, by definition, $0=0+0i$. Since the imaginary and real parts are 0, 0 is purely real and imaginary. However, it's also member of the complex numbers.






                share|cite|improve this answer
























                  2












                  2








                  2






                  If $yneq 0$, in the complex plane, by definition, $0=0+0i$. Since the imaginary and real parts are 0, 0 is purely real and imaginary. However, it's also member of the complex numbers.






                  share|cite|improve this answer












                  If $yneq 0$, in the complex plane, by definition, $0=0+0i$. Since the imaginary and real parts are 0, 0 is purely real and imaginary. However, it's also member of the complex numbers.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 at 22:37









                  Alex R.

                  24.7k12352




                  24.7k12352























                      1














                      Well, provided that $y neq 0$, you are multiplying by $frac{1}{2y}$. (Of course, if $y = 0$ then it is undefined.) So, I will talk of multiplication rather than division. I am guessing that $y$ is intended to be real but that does not actually affect the answer.



                      A complex number is real if the imaginary component is zero. Conversely, it is imaginary if the real component is zero. Most complex numbers e.g. $1 + i$ are neither. $0$ is special in that it is both.



                      Generally multiplying by $i$ will flip real numbers to imaginary and vice versa. Since $0 times i = 0$ multiplying by $i$ does change it, it goes from both to both.



                      Something a little like this occurs with the more familiar real numbers. Multiplying by $-1$ flips positive and negative. It leaves $0$ alone, so is $0$ positive or negative? The usual answer is that $0$ is neither but considering it as both could work. The Bourbaki (Wikipedia) school proposed this. I found a reference thanks to posting a question here: Bourbaki and zero (this site).






                      share|cite|improve this answer























                      • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                        – max_zorn
                        Dec 1 at 0:00










                      • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                        – badjohn
                        Dec 1 at 9:23
















                      1














                      Well, provided that $y neq 0$, you are multiplying by $frac{1}{2y}$. (Of course, if $y = 0$ then it is undefined.) So, I will talk of multiplication rather than division. I am guessing that $y$ is intended to be real but that does not actually affect the answer.



                      A complex number is real if the imaginary component is zero. Conversely, it is imaginary if the real component is zero. Most complex numbers e.g. $1 + i$ are neither. $0$ is special in that it is both.



                      Generally multiplying by $i$ will flip real numbers to imaginary and vice versa. Since $0 times i = 0$ multiplying by $i$ does change it, it goes from both to both.



                      Something a little like this occurs with the more familiar real numbers. Multiplying by $-1$ flips positive and negative. It leaves $0$ alone, so is $0$ positive or negative? The usual answer is that $0$ is neither but considering it as both could work. The Bourbaki (Wikipedia) school proposed this. I found a reference thanks to posting a question here: Bourbaki and zero (this site).






                      share|cite|improve this answer























                      • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                        – max_zorn
                        Dec 1 at 0:00










                      • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                        – badjohn
                        Dec 1 at 9:23














                      1












                      1








                      1






                      Well, provided that $y neq 0$, you are multiplying by $frac{1}{2y}$. (Of course, if $y = 0$ then it is undefined.) So, I will talk of multiplication rather than division. I am guessing that $y$ is intended to be real but that does not actually affect the answer.



                      A complex number is real if the imaginary component is zero. Conversely, it is imaginary if the real component is zero. Most complex numbers e.g. $1 + i$ are neither. $0$ is special in that it is both.



                      Generally multiplying by $i$ will flip real numbers to imaginary and vice versa. Since $0 times i = 0$ multiplying by $i$ does change it, it goes from both to both.



                      Something a little like this occurs with the more familiar real numbers. Multiplying by $-1$ flips positive and negative. It leaves $0$ alone, so is $0$ positive or negative? The usual answer is that $0$ is neither but considering it as both could work. The Bourbaki (Wikipedia) school proposed this. I found a reference thanks to posting a question here: Bourbaki and zero (this site).






                      share|cite|improve this answer














                      Well, provided that $y neq 0$, you are multiplying by $frac{1}{2y}$. (Of course, if $y = 0$ then it is undefined.) So, I will talk of multiplication rather than division. I am guessing that $y$ is intended to be real but that does not actually affect the answer.



                      A complex number is real if the imaginary component is zero. Conversely, it is imaginary if the real component is zero. Most complex numbers e.g. $1 + i$ are neither. $0$ is special in that it is both.



                      Generally multiplying by $i$ will flip real numbers to imaginary and vice versa. Since $0 times i = 0$ multiplying by $i$ does change it, it goes from both to both.



                      Something a little like this occurs with the more familiar real numbers. Multiplying by $-1$ flips positive and negative. It leaves $0$ alone, so is $0$ positive or negative? The usual answer is that $0$ is neither but considering it as both could work. The Bourbaki (Wikipedia) school proposed this. I found a reference thanks to posting a question here: Bourbaki and zero (this site).







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 1 at 10:19

























                      answered Nov 30 at 22:45









                      badjohn

                      4,2421620




                      4,2421620












                      • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                        – max_zorn
                        Dec 1 at 0:00










                      • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                        – badjohn
                        Dec 1 at 9:23


















                      • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                        – max_zorn
                        Dec 1 at 0:00










                      • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                        – badjohn
                        Dec 1 at 9:23
















                      This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                      – max_zorn
                      Dec 1 at 0:00




                      This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                      – max_zorn
                      Dec 1 at 0:00












                      I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                      – badjohn
                      Dec 1 at 9:23




                      I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient.
                      – badjohn
                      Dec 1 at 9:23











                      0














                      We have that



                      $$frac{0}{2yi}=0$$



                      which is an integer, a rational, a real and a complex number.



                      Notably it indicates the neutral element with respect to addition.






                      share|cite|improve this answer





















                      • ...and the absorbing element under multiplication.
                        – user334732
                        Nov 30 at 23:16
















                      0














                      We have that



                      $$frac{0}{2yi}=0$$



                      which is an integer, a rational, a real and a complex number.



                      Notably it indicates the neutral element with respect to addition.






                      share|cite|improve this answer





















                      • ...and the absorbing element under multiplication.
                        – user334732
                        Nov 30 at 23:16














                      0












                      0








                      0






                      We have that



                      $$frac{0}{2yi}=0$$



                      which is an integer, a rational, a real and a complex number.



                      Notably it indicates the neutral element with respect to addition.






                      share|cite|improve this answer












                      We have that



                      $$frac{0}{2yi}=0$$



                      which is an integer, a rational, a real and a complex number.



                      Notably it indicates the neutral element with respect to addition.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 30 at 22:36









                      gimusi

                      1




                      1












                      • ...and the absorbing element under multiplication.
                        – user334732
                        Nov 30 at 23:16


















                      • ...and the absorbing element under multiplication.
                        – user334732
                        Nov 30 at 23:16
















                      ...and the absorbing element under multiplication.
                      – user334732
                      Nov 30 at 23:16




                      ...and the absorbing element under multiplication.
                      – user334732
                      Nov 30 at 23:16


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020734%2fis-this-a-pure-imaginary-number-or-real-number%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      AnyDesk - Fatal Program Failure

                      Image
                      up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
                      Read more

                      How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                      Image
                      up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
                      Read more

                      QoS: MAC-Priority for clients behind a repeater

                      Image
                      1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
                      Read more