# Are complex numbers complex for Maths to handle too?

$-1\quad =\quad -1\\ \frac { 1 }{ -1 } \quad =\quad \frac { -1 }{ 1 }$ Taking square root both sides $\frac { \sqrt { 1 } }{ \sqrt { -1 } } \quad =\quad \frac { \sqrt { -1 } }{ \sqrt { 1 } }$ Cross multiplying both sides

$1\quad =\quad { i }^{ 2 }\\ 1\quad =\quad -1$

Where lies the Maths fallacy?

This puzzle is taken from a Maths book I read many years ago. I can't remember the name.

Note by Lokesh Sharma
6 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

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:

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ is true only when $a$ and $b$ have the same signs.

EDIT: I meant to say whenever $a$ is positive and $b$ is negative.

- 6 years, 7 months ago

I am replying after a long time. I was roaming in the garden and realised some funny thing happening here. $\sqrt{\frac{-2}{3}} = \frac{\sqrt{-2}}{3}$ So the expression $\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ is only invalid if $b < 0$ and $a > 0$.

Let's see the big picture now. You disproved the prove above in the post by claiming that the expression $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ is only invalid if $b < 0$ and $a > 0$. You actually said something else but I have reduced what you have said to this form because there was certain inaccuracies that I have talked about in the beginning.

Now the problem is that your statement needs to be proved in order to disapprove my given trick that $1 = -1$ and the proof of your statement in the end is the assumption that my trick uses some wrong methods.

If you are not able to understand then try the following example and you will understand what I mean:

• Try to disprove that $\sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}}$ The lesson here is that you can't do it unless you assume that $\frac{1}{-1} = \frac{-1}{1}$ but it you assume that then you are right back at my trick to prove that that $1 = -1$.

The challenged is reopened!

- 6 years, 6 months ago

(sigh) Here I go again...

Now the problem is that your statement needs to be proved in order to disapprove my given trick

No, my statement doesn't have to be proved. My statement makes the math consistent. You can claim that my statement is not true and continue doing inconsistent math. But what good would that do?

The point of posts like these is: "Hey, look! I'm doing seemingly correct mathematical operations and coming up with a crazy conclusion! Can you find what I'm doing wrong?"

Guess what? These 'crazy conclusions' make math inconsistent and no one wants that.

The correct way of looking at this is, $1=-1$ is wrong, so what I'm doing in between must be wrong. Not the other way around.

- 6 years, 6 months ago

No, my statement doesn't have to be proved. My statement makes the math consistent. You can claim that my statement is not true and continue doing inconsistent math. But what good would that do?

Here's the point my friend. The reason I am compelling you to prove your statement is to make sure that there isn't any other fallacy because of which the result 1 = -1 is inconsistent. OR you have to prove that the fallacy doesn't lie in any other step.

- 6 years, 6 months ago

$\sqrt{a/b} = \sqrt{a}/\sqrt{b}$ is true also if both a and b are negative integers

- 6 years, 7 months ago

It is very interesting to note that $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ is true even if at least one of them is positive while $\sqrt{a/b} = \sqrt{a}/\sqrt{b}$ is not.

- 6 years, 7 months ago

Not that interesting if you know why that happens. The whole thing is because of the fact that $\frac{1}{\sqrt{-1}}=-\sqrt{-1}$.

- 6 years, 7 months ago

Yeah

- 6 years, 7 months ago

It would be in the 3rd step.

- 6 years, 7 months ago

The identity root a *root b =root ab is not applicable for complex no.s

- 6 years, 6 months ago

It is in taking the roots. What the guy did was actually take the fourth root of 1 which is +i or - i (equal to the square roots you have taken of -1) and then equated them. That's where the catch lies! Good one tho...

- 6 years, 6 months ago