Is this a pure imaginary number or real number?
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3
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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
New contributor
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
New contributor
1
isnt that number just $0$?
– Jorge Fernández
6 hours ago
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.
– Robert Frost
5 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago
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
New contributor
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
New contributor
New contributor
edited 5 hours ago
Robert Frost
4,1961039
4,1961039
New contributor
asked 6 hours ago
Maske13
161
161
New contributor
New contributor
1
isnt that number just $0$?
– Jorge Fernández
6 hours ago
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.
– Robert Frost
5 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago
add a comment |
1
isnt that number just $0$?
– Jorge Fernández
6 hours ago
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.
– Robert Frost
5 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago
1
1
isnt that number just $0$?
– Jorge Fernández
6 hours ago
isnt that number just $0$?
– Jorge Fernández
6 hours ago
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.
– Robert Frost
5 hours ago
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.
– Robert Frost
5 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago
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.
add a comment |
up vote
2
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.
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
add a comment |
up vote
1
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.
add a comment |
up vote
1
down vote
$0$ being purely real and purely imaginary need not be more surprising than, as a real number, it is neither positive nor negative.
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
4 hours ago
add a comment |
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.
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.
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.
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 6 hours ago
José Carlos Santos
142k20112208
142k20112208
add a comment |
add a comment |
up vote
2
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.
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
add a comment |
up vote
2
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.
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
add a comment |
up vote
2
down vote
up vote
2
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.
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 6 hours ago
gimusi
88.5k74394
88.5k74394
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
add a comment |
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
...and the absorbing element under multiplication.
– Robert Frost
5 hours ago
add a comment |
up vote
1
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.
add a comment |
up vote
1
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.
add a comment |
up vote
1
down vote
up vote
1
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.
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 6 hours ago
Alex R.
24.6k12352
24.6k12352
add a comment |
add a comment |
up vote
1
down vote
$0$ being purely real and purely imaginary need not be more surprising than, as a real number, it is neither positive nor negative.
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
4 hours ago
add a comment |
up vote
1
down vote
$0$ being purely real and purely imaginary need not be more surprising than, as a real number, it is neither positive nor negative.
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
4 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
$0$ being purely real and purely imaginary need not be more surprising than, as a real number, it is neither positive nor negative.
$0$ being purely real and purely imaginary need not be more surprising than, as a real number, it is neither positive nor negative.
answered 5 hours ago
badjohn
4,2101620
4,2101620
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
4 hours ago
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
4 hours ago
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
4 hours ago
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
4 hours ago
add a comment |
Maske13 is a new contributor. Be nice, and check out our Code of Conduct.
Maske13 is a new contributor. Be nice, and check out our Code of Conduct.
Maske13 is a new contributor. Be nice, and check out our Code of Conduct.
Maske13 is a new contributor. Be nice, and check out our Code of Conduct.
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1
isnt that number just $0$?
– Jorge Fernández
6 hours ago
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.
– Robert Frost
5 hours ago
You left out another possibility: when $y=0$, the expression is undefined.
– amd
4 hours ago