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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.

Tom S
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4 Answers4

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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.

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If $y\neq 0$, 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.

Alex R.
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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.

user
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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).

badjohn
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  • 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 01 '18 at 00:00
  • I had intended to just add to the answers of others but I have expanded the answer to be self-sufficient. – badjohn Dec 01 '18 at 09:23