Is this a pure imaginary number or real number?
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
edited Nov 30 at 23:15
user334732
4,23811140
asked Nov 30 at 22:33
Maske13
161
add a comment |
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
edited Nov 30 at 23:15
user334732
4,23811140
asked Nov 30 at 22:33
Maske13
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
add a comment |
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
edited Nov 30 at 23:15
user334732
4,23811140
asked Nov 30 at 22:33
Maske13
161
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
edited Nov 30 at 23:15
user334732
4,23811140
asked Nov 30 at 22:33
Maske13
161
edited Nov 30 at 23:15
user334732
4,23811140
asked Nov 30 at 22:33
Maske13
161
edited Nov 30 at 23:15
user334732
4,23811140
edited Nov 30 at 23:15
user334732
4,23811140
edited Nov 30 at 23:15
user334732
4,23811140
4,23811140
asked Nov 30 at 22:33
Maske13
161
asked Nov 30 at 22:33
Maske13
161
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
add a comment |
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
add a comment |
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.
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
add a comment |
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.
answered Nov 30 at 22:37
Alex R.
24.7k12352
add a comment |
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.
answered Nov 30 at 22:36
gimusi
92.5k94495
...and the absorbing element under multiplication.
– user334732
Nov 30 at 23:16
add a comment |
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).
answered Nov 30 at 22:45
badjohn
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
add a comment |
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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.
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
add a comment |
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.
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
add a comment |
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.
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
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.
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
answered Nov 30 at 22:37
José Carlos Santos
146k22117217
146k22117217
add a comment |
add a comment |
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.
answered Nov 30 at 22:37
Alex R.
24.7k12352
add a comment |
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.
answered Nov 30 at 22:37
Alex R.
24.7k12352
add a comment |
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.
answered Nov 30 at 22:37
Alex R.
24.7k12352
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.
answered Nov 30 at 22:37
Alex R.
24.7k12352
answered Nov 30 at 22:37
Alex R.
24.7k12352
answered Nov 30 at 22:37
Alex R.
24.7k12352
answered Nov 30 at 22:37
Alex R.
24.7k12352
24.7k12352
add a comment |
add a comment |
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.
answered Nov 30 at 22:36
gimusi
92.5k94495
...and the absorbing element under multiplication.
– user334732
Nov 30 at 23:16
add a comment |
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.
answered Nov 30 at 22:36
gimusi
92.5k94495
...and the absorbing element under multiplication.
– user334732
Nov 30 at 23:16
add a comment |
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.
answered Nov 30 at 22:36
gimusi
92.5k94495
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.
answered Nov 30 at 22:36
gimusi
92.5k94495
answered Nov 30 at 22:36
gimusi
92.5k94495
answered Nov 30 at 22:36
gimusi
92.5k94495
answered Nov 30 at 22:36
gimusi
92.5k94495
92.5k94495
...and the absorbing element under multiplication.
– user334732
Nov 30 at 23:16
add a comment |
...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
add a comment |
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).
answered Nov 30 at 22:45
badjohn
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
add a comment |
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).
answered Nov 30 at 22:45
badjohn
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
add a comment |
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).
answered Nov 30 at 22:45
badjohn
4,2221620
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).
answered Nov 30 at 22:45
badjohn
4,2221620
edited Dec 1 at 10:19
answered Nov 30 at 22:45
badjohn
4,2221620
answered Nov 30 at 22:45
badjohn
4,2221620
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
add a comment |
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
add a comment |
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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