×

# prove it!!!!!!!!!!!!!!!!(2)

Prove that a purely real number can never became equal to a purely imaginary number (Provided that the number is not 0+0i)

Note by Aman Sharma
3 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

A pure imaginary number is of the form $$bi$$ where $$b$$ is a [non-zero] real number and $$i^2=-1$$.

If $$a$$ real number were equal to $$bi$$, then we would have $$a^2=-b^2$$ which is impossible since the right hand side is negative while the left hand side is non-negative.

- 3 years, 2 months ago

Thanks for replying you are awsome in proof problems

- 3 years, 2 months ago

Thanks! But I wouldn't say I'm awesome.

- 3 years, 2 months ago

No, you realy are awsome, also your set ''proof problem of the day'' is realy cool one

- 3 years, 2 months ago

Actually, by the definition of complex numbers, a real number is also considered a complex number. Think of it as like, we can write x as x + 0i.

- 3 years, 2 months ago

I am editing it thanks for pointing out the mistake

- 3 years, 2 months ago

Sorry, all real numbers are complex numbers. If this is from NCERT, exercise and problem number please?

- 3 years, 2 months ago

Edited

- 3 years, 2 months ago