Is this a pure imaginary number or real number?











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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







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    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















up vote
3
down vote

favorite












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




















  • 1




    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













up vote
3
down vote

favorite









up vote
3
down vote

favorite











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

4,23811140




4,23811140










asked Nov 30 at 22:33









Maske13

161




161








  • 1




    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














  • 1




    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








1




1




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

















up vote
3
down vote













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




























    up vote
    2
    down vote













    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




























      up vote
      1
      down vote













      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


















      up vote
      1
      down vote













      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











      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote













      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

























        up vote
        3
        down vote













        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























          up vote
          3
          down vote










          up vote
          3
          down vote









          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

          146k22117217




          146k22117217






















              up vote
              2
              down vote













              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

























                up vote
                2
                down vote













                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























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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






















                      up vote
                      1
                      down vote













                      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















                      up vote
                      1
                      down vote













                      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













                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      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

                      92.5k94495




                      92.5k94495












                      • ...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










                      up vote
                      1
                      down vote













                      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















                      up vote
                      1
                      down vote













                      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













                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      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

                      4,2221620




                      4,2221620












                      • 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


















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