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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
3 years, 1 month ago

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\(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) 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 · 3 years, 1 month ago

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@Mursalin Habib 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! Lokesh Sharma · 3 years ago

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@Lokesh Sharma (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. Mursalin Habib · 3 years ago

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@Mursalin Habib

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. Lokesh Sharma · 3 years ago

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@Mursalin Habib 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. Lokesh Sharma · 3 years, 1 month ago

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@Lokesh Sharma Not that interesting if you know why that happens. The whole thing is because of the fact that \(\frac{1}{\sqrt{-1}}=-\sqrt{-1}\). Mursalin Habib · 3 years, 1 month ago

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@Mursalin Habib Yeah Lokesh Sharma · 3 years, 1 month ago

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@Mursalin Habib \(\sqrt{a/b} = \sqrt{a}/\sqrt{b}\) is true also if both a and b are negative integers Lokesh Sharma · 3 years, 1 month ago

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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 · 3 years ago

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The identity root a *root b =root ab is not applicable for complex no.s Kushagra Sahni · 3 years ago

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It would be in the 3rd step. Aman Jaiswal · 3 years, 1 month ago

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