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This post requires familiarity with complex numbers, in particular \( i = \sqrt{-1}\), and \( i^2 = -1\).

This is taken from a Brilliant discussion, with slight edits.

\[ \begin{aligned} &\mbox{We know that} && 1 \times 1 = (-1) \times (-1) & &(1) \\ &\mbox{Dividing across, we get} && \frac {1}{-1} = \frac {-1}{1} & &(2)\\ &\mbox{Taking square roots, we get} && \frac {\sqrt{1} }{\sqrt{-1}} = \frac {\sqrt{-1}}{\sqrt{1}} &&(3) \\ &\mbox{Hence,} && \frac {1}{i} = \frac {i}{1} && (4)\\ &\mbox{Multiplying out denominators, we get} && 1 \times 1 = i \times i && (5)\\ &\mbox{Hence,} && 1 = 1 \times 1 = i \times i = -1 & &(6).\\ \end{aligned} \]

What went wrong? At which step did we go wrong? You can review the discussion to understand the subtleties of the proof.

Related to this is the following 'proof' that \( 1 = 3\).

\[ \begin{aligned} & \mbox{We know that} & & (-1)^1 = -1 = (-1)^3 & & (1) \\ & \mbox{Taking logarithms, we get} & & 1 \times \log (-1) = 3 \times \log (-1) & & (2)\\ & \mbox{Since }\log(-1) \neq 0, \mbox{ we can divide by} \log(-1) & & 1 = 3 & & (3) \\ \end{aligned}\]

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## Comments

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TopNewest\(\sqrt \frac {a}{b} = \frac {\sqrt{a}}{\sqrt{b}}\) is not valid for complex numbers

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the mistake in 1st proof is that we cant take root of negative numbers because no two same numbers when multiplied would give a -ve number and the mistake in 2nd is that log is not defined for -ve numbers because if in the case of natural log, the base is

ehence aseis +ve soeraised to any thing would be always be +ve .Log in to reply

You're right. I would like to add that when teacher teach about complex numbers they don't mention that -i is also a solution for \({x}^{2}+1=0\). In the complex numbers we also can't compare numbers so we don't really have a sign, (eg what's the square root of \({(2-i)}^{2}\) is it \(2-i\) or \(-2+1\)) that's why the square root can only be calculated for positive numbers.

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\( (i)In\quad first\quad proof,\\ \\ \cfrac { a }{ b } =\cfrac { p }{ q } \nRightarrow \sqrt { \cfrac { a }{ b } } =\sqrt { \cfrac { p }{ q } } \quad because\quad may\quad be\quad \sqrt { \cfrac { a }{ b } } =-\sqrt { \cfrac { p }{ q } } \\ \\ (ii)In\quad second\quad proof,\\ \\ log(x)\quad only\quad defined\quad for\quad x>0 \)

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\(\large \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) , only if a and b are both positive or both negative...

And if exactly one of a and b is positive, then

\(\large \sqrt{\frac{a}{b}}=i. \frac{\sqrt{|a|}}{\sqrt{|b|}}\).

You can say that i am trying to make a new rule , which would give its best to control some functionings of such expressions.

One rule I know is \(\sqrt{a}.\sqrt{b}=\sqrt{a.b}\) ; if and only if at least one of a and b is non-negative. And the second rule may be which i discussed above :P

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in second proof we cannot use negative numbers as it not in domain of logarithmic function

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i can't get the first proof.. base from my teacher's teaching, you should divide the same #.

So..

if...

1

1 = (-1)(-1)then,

(1*1)/(-1) = ((-1)(-1))/(-1)

And my teavher says.. YOU CAN'T DIVIDE ACROSS.. Joke.. justmy induction.. :)

second...

Log (base -1) (-1) = 1 -----> (Log -1)/(Log -1) = 1 ----> Log (-1) = (Log -1)

And:

Log (base -1) (1) = 3 ------> (Log -1)/(Log 1) = 3 ----> Log (-1) = 3 (Log 1)

By means of their similarity,

(Log -1) = 3 ( Log 1)

1.36437635 i = 3(0)

1.36437635 i ≠ 0

So...

the 1 = 3 is wrong.. :3

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What about something like this:

We know that \(\quad \quad \quad \qquad \qquad \qquad {i}^{ 2 }={i}^{6}=-1\) and \(i=\sqrt { -1 }\)

Combining, we get \(\quad \qquad \qquad \qquad i=\sqrt { i^{ 6 } } =i^{ \frac { 6 }{ 2 } }=i^{ 3 }\qquad hence,\qquad i=i^{ 3 }\quad\)

Since \(i\neq 0\) we can divide \(\qquad \qquad \frac { i^3 }{ i }=1\)

And \(\qquad \qquad \qquad \qquad \qquad \qquad \frac { i^{ 3 } }{ i } = i^{ 3-1 }= i^{ 2 }= -1\)

Therefore, we conclude \(\qquad \qquad \quad 1=-1\)

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We can not divide by an imaginary number.

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Why not? If \( i^2 = - 1 \), then we have \( \frac{ 1}{i} = - i \).

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What do you mean IF \( { i }^{ 2 }=-1 \)?! it CAN'T be -1! You might as well be saying if (idiotic BS)=-1, then \( \frac { 1 }{ (idiotic BS) } \) = -(idiotic BS). What is -1 x -1? Perhaps you need a calculator to figure that out. A negative Times a negative is a positive. A positive times a positive is a positive. There is NO WAY that ANY negative number can be squared to become a negative number. Since that is a mathematical impossibility it makes absolutely no sense to then use it in any equation as if it is not a mathematical impossibility in order to prove some other impossibility. No matter how much you may Claim to be able to fly, when you leap off a tall building you will still go splat!

But \( \sqrt [ 3 ]{ -1 } \)= -1, And \( \frac { \sqrt [ 3 ]{ -1 } }{ \sqrt { 1 } } =\frac { \sqrt { 1 } }{ \sqrt [ 3 ]{ -1 } } \), I checked on my calculator, they are both -1. That still doesn't mean that 1=-1, it only means that a Negative times a Positive is a Negative! Basically what this is saying is that -1/1=-1 AND 1/-1=-1. Why is that? Because a negative times a negative is a positive, AND a negative times a positive is a negative. Both rules are well known, and neither proves that 1=-1. However you Can say that |1|=|-1|, and be completely correct.

Just because 1/-1=-1/1 that does NOT mean that 1x1=-1x-1=-1, that would be very plainly wrong. -1x-1=1! 1/-1=-1, -1/1=-1, 1x1=1, -1x-1=1, so Yes, both sets of equations are true, but one set's answer is NOT the same as the other set's. 1 does NOT equal -1. Not by any so-called mathematical proof.

The logarithmic so-called proof is wrong as well, my calculator says \( 1 \times \log { (-1) } \) is a math error. No matter how much you manipulate mathematically impossible things you can never make them prove another mathematical impossibility. Logarithms have to do with multiplications too. The only way to get -1 is by multiplying \( -1 \times -1 \times -1 \), etc. etc. into infinity as long as it's an odd number of multiplications. But if I put in \( \log _{ -1 }({ -1 }) \) in my calculator, it gives me an error. Why? Because it's trying to find the exponent that gives the answer of -1, but Every Odd Exponent would do that! Can't be done on the calculator, too much to do. So what? No matter the exponent you choose, that still doesn't mean that 1=3!! BS! \( { -1 }^{ 1 } = -1 \), \( { -1 }^{ 3 } = -1 \), \( { -1 }^{ 5 } = -1 \), \( { -1 }^{ 7 } = -1 \), \( { -1 }^{ 9 } = -1 \), \( { -1 }^{ 11 } = -1 \), etc. etc. So.......... yah, 1 is not 3. Oh, and since \( 1 \times \log { (-1) } \) is giving an error and is therefore impossible to resolve to an actual number, you can not divide anything by it! That's like dividing something by infinity! And what, exactly, are you dividing by infinity in order to get 3? Since you can't multiply anything by infinity and get any kind of real number to work with, you might as well just say infinity equals infinity, which is true. 1=3 is not true. Period, end of discussion.

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ii = -1 * i = -iLog in to reply

@Aaron Ingebrigtsen

thats why i is called an imaginary number not a real numberLog in to reply