\[-1\quad =\quad -1\\ \frac { 1 }{ -1 } \quad =\quad \frac { -1 }{ 1 }\] 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.

## Comments

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

EDIT: I meant to say whenever \(a\) is positive and \(b\) is negative. – Mursalin Habib · 2 years, 9 months ago

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invalidif \( b < 0\) 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

invalidif \( b < 0\) 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:

The challenged is reopened! – Lokesh Sharma · 2 years, 8 months ago

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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

correctway of looking at this is, \(1=-1\) is wrong, so what I'm doing in between must be wrong. Not the other way around. – Mursalin Habib · 2 years, 8 months agoLog in to reply

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. – Lokesh Sharma · 2 years, 8 months ago

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– Lokesh Sharma · 2 years, 9 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.Log in to reply

– Mursalin Habib · 2 years, 9 months ago

Not that interesting if you know why that happens. The whole thing is because of the fact that \(\frac{1}{\sqrt{-1}}=-\sqrt{-1}\).Log in to reply

– Lokesh Sharma · 2 years, 9 months ago

YeahLog in to reply

– Lokesh Sharma · 2 years, 9 months ago

\(\sqrt{a/b} = \sqrt{a}/\sqrt{b}\) is true also if both a and b are negative integersLog in to reply

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... – Kalyan Pakala · 2 years, 8 months ago

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The identity root a *root b =root ab is not applicable for complex no.s – Kushagra Sahni · 2 years, 8 months ago

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It would be in the 3rd step. – Aman Jaiswal · 2 years, 9 months ago

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