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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}

We will be moving the discussion for this post to this Blog post discussion board, which will allow you to respond to comments made by other students, and also vote on the explanation that you think is correct.

From → Algebra, Level 3, Level 4, Math

1. [Comment moved to discussion board - Calvin]

[Comment moved to discussion board - Calvin]

log undefined for -ve nos

In that proof square root of one is treated as one but actually it’s plus or minus one

square root of 1 is |1| means mod(1) = +/- 1…
hear taken only +1.

log(-1) is not defined so our first step to take log of -1 is wrong.

• Bear in mind that $e^{i \pi} = -1$, so we can define $\log (-1)$.

Yeah but we have to put an absolute value when we put the logarithms so absolute value of e^iπ is one and log(1) is 0 isn’t it?

• What do you mean by put absolute value? $\log -1 \neq \log 1$.

7. You took the root of a negative 1 not possible. If you want that you have bring and i in so you will have i*sqrt(1)

• We took square root of negative 1, and had the value of $i$ in both places. I do not see how your argument is valid.

Using the logic of the second proof, 1^1 = 1^2 = 1^3 = …. could be used as a proof for 1 = 2 = 3 = …

This is invalid because for (-1)^1 = (-1)^3 to imply 1 = 3, you need to show that every power of -1 must yield a unique value. You don’t even have to go as far as worrying about logarithms of negative numbers to find a flaw.

• I disagree with your second paragraph. We only need the odd powers to yield a negative value. And the do, since $(-1)^{2k+1} = {[(-1)^2]}^k\times (-1) = -1$.

The first proof is wrong at line (3). Where the square root of both sides is taken but the assumption that sqrt(-1/1) = sqrt(-1)/sqrt(1) is made. This is not true for complex numbers.

i thought log(-1) does not exist?

• Bear in mind that $e^{i \pi} = -1$.

Log is defined from 0 to +oo, the square as well.

• Bear in mind that $e^{i \pi} = -1$. While you may have only seen logarithms defined for positive reals, it is possible to extend that definition.

12. You can’t say that i=/-1 because this value doesn’t exist. It means that can not be carried forward.

• Do you mean that “You can’t say $i = \sqrt{-1}$ because this value doesn’t exist”? What do you mean by it doesn’t exist? It exists as a complex number. Why can’t we do addition and multiplication with it? If we can’t, does that mean that all the complex numbers that we learnt are all useless?

awesome

there was a confusion with log(-1) . Otherwise its really surprising !

15. we know that the natural logarithm is inverse function of the exponential, but a function does have an inverse only if it was bijective for instance : e^(i pi) = e^(3 i pi), the exponential function is not bijective over all the complex numbers.

• Yes, you’re getting close to the reason why the proof breaks down. The logarithm function is actually a multivalued function, and one of the values of $\log -1$ is $i \pi$, while the other possible value is $3 i \pi$, and the latter is thrice of the former.

I think the problem is at line (6) because here you have done “i*i=-1″. But i don’t agree with this step as according to the rule of multiplication of complex numbers, a complex no. is always multiplied with it’s complex conjucate.So, it has to be multiplied with “-i” and “i*(-i)=+1″. and this proves that 1 is not equal to -1 but it equals to +1. And to prove “i*i can’t be equal to -1″…we know, exp((i*pi)/2)=i. and if we multi ply exp((i*pi)/2) by exp((i*pi)/2) using the method of multilication of line (6) and we don’t use the complex conjugate of this then it would give”ccos²(љ/2)+ i²sin²(љ/2)” here if we put “i*i=-1 according to the multiplication of line(6) then it will give cos²(љ/2)+ sin²(љ/2)= -1 (as,sin(љ/2)=1) which is not possible because we know, “cos²( Ө)+ i²sin²(Ө)=1″. so this is satisfied only when we follow the rule of multiplication of complex number and it gives “i*(-i)= -i²=+1″and it proves”1=1″.

I think the problem is at line (6) because here you have done “i*i=-1″. But i don’t agree with this step as according to the rule of multiplication of multiplication a complex no. is always multiplied with it’s complex conjucate.So, i has to be multiplied with “-i” and “i*(-i)=+1″. and this proves that 1 is not equal to -1 but it equals to +1. And to prove “i*i can’t be equal to -1″…we know, exp((i*pi)/2)=i. and if we multi ply exp((i*pi)/2) by exp((i*pi)/2) using the method of multilication of line (6) and we don’t use the complex conjugate of this then it would give”ccos²(љ/2)+ i²sin²(љ/2)” here if we put “i*i=-1 according to the multiplication of line(6) then it will give cos²(љ/2)+ sin²(љ/2)= -1 (as,sin(љ/2)=1) which is not possible because we know, “cos²( Ө)+ i²sin²(Ө)=1″. so this is satisfied only when we follow the rule of multiplication of complex number and it gives “i*(-i)= -i²=+1″and it proves”1=1″.

• I don’t understand why we can only multiply a complex number with it’s complex conjugate. We certainly have $(1+2i) \times (1 + 3i) = 1 + 2i + 3i + 6i^2 = 5i - 5$. Multiplying by the complex conjugate is done to obtain the norm of the complex number.

$i^2 = -1$ is the definition of the imaginary unit, and taken as a fact / axiom.

Also, in your 2nd last line, $\cos^2 \theta + i^2 \sin ^2 \theta \neq 1$. Instead, the trigonometric identity that you’re thinking of is $\cos^2 \theta + \sin^2 \theta = 1$.

Yeah log(-1)=iπ right? But i think the identity is {log(a)}^k=k*log(|a|) That is what I mean

• I don’t think that is the identity. Be very careful, especially if you have not seen logs of negative (or even complex) numbers before.

well sqrt(1/-1) is not the same as sqrt(1)/sqrt(-1).

(-1)^1 = (-1)^3
so antilog(1*log-1) = antilog(3*log-1)
but this doesn’t mean 1*log-1 = 3*log-1, because antilog is not a one-one function.

• Yup! Though it could depend on what you call ‘antilog’, and what the equality means to you. But you have to right ideas.

I COULDN’T UNDERSTAND YOUR 6 TH STEP, HOW YOU CONSIDER i*i=-1

• That is the definition of the imaginary unit, specifically $i^2 = -1$.

In line (3) a square root of 1 and -1 has been taken but square root gives both + and -ve value but here only + ve has been considered for next all the steps but it can be considered only when we are taking a mod value which always gives a positive quantity. So in line (5) Mod(±√1)=+1 and Mod(±√-1)=Mod(±i)=i
that’s why Mod(i).Mod(i)= (Mod(i)) ² = (i*).(i)= (-i).(i)= – i²=+1.So, in line (6) it gives,, 1=Mod(i).Mod(i)= (Mod(i)) ² = (i*).(i)= – i²=+1. So, 1=+1.
And I’m Sorry! because i said “according to the rule of multiplication of complex nos” it is actually the rule of finding the mod value of a complex no.
for example, (Mod(a+ib))²=(a+ib)*.(a+ib)=(a-ib).(a+ib)=a²-(ib)²=a²-i²b²=a²+b².
So, Mod(a+ib)=√(a²+b²).

• I’m not sure what kind of modulus you are working with, esp to say that $Mod(\pm i) = i$. As far as I know, modulus always returns a positive quantity (like you mentioned), that is real valued. What would you say is $Mod(1 - 2i)$?

I believe that you are on the right track, but you should crystallize your thoughts, and be certain that what you’re writing is exactly what you’re thinking.

P.S. In future, please reply to your previous comments, so that it is easier to keep track of what you have been saying. I.e. this is not a new comment, but a continuation.

The 4th line is wrong because root over 1 has two solutions one is 1 and other is -1.

(root of -1) * (root of -1) wil b equal to -1 nt 1
so proof is surely wrong

25. I think the mistake is in third step root(a/b) is equal to root(a)/root(b) only if both a and b are positive or negative.

to prove 1=-1 we have to select only real numbers not the imaginary numbers

if x+2=3 on taking x=1 equation is satisfied but if 1=3 and on putting 3 in the equation the equation is not satisfied its mean 1 is not equal to 3.

Log of a number less than zero is not possible. Any real number to any power is always greater than zero, if only by a negligible amount. (e.g. Two to the power of minus infinity).
I’m thinking it’s step 3 in the 1=-1 part. (it breaks between stage 2 and 4.)

• Why is it not possible? Because you haven’t seen it before? Is $\sqrt{-1}$ possible? It is if you allow for complex numbers. Likewise, we have $e^{\pi i} = -1$. So, can’t we take logarithms on both sides?

29. Good introduction to high school and complex math for a 14 year old? I feel confused by half of the things on this site but feel as if I learn some higher concept every time I visit. I actually understand some of this post! Keep them coming, Calvin.

there is no log-1 log is only ++++ numpers
no 0 no – numpers Lol

• By Euler’s Theorem, $e^{i \pi} = -1$. This suggests that we can take the logarithm of -1.

That’s good but proof(1) can be done without using i=√(-1 ) in the following way:-
We know,
1 X 1 = (-1) X (-1)
Or, 1^2 = (-1)^2
Or, 1 = (-1) [Taking out square-root]

Isn’t it?

32. $$\log(m^n) = n \log (m)$$ does not hold for $$m < 0$$.

a^2-a^2=a^2-a^2
=> (a+a)(a-a)=a(a-a)
=> a+a=a
=> 2a=a
=> 2=1 ….(1)
now
3=1+2
=> 3=1+1 (from-1)
=> 3=2
=> 3=1 (from-1)
hence any no. can be equated to any no.