I am sure that this has been discussed many times and the answer may be trivial. But I was confronted with the following issue and could not give a satisfactory answer: $$1=(-1^{6})^{\frac{1}{2}}=-1^{6\cdot\frac{1}{2}} = -1^{\frac{1}{2}\cdot 6}=(-1^{\frac{1}{2}})^6 = (\sqrt{-1})^6.$$  On the left handside of the chain, we have $1$, and on the right handside, we have the square root of a negative number, which is not defined with real numbers. Also moving towards complex numbers does not seem to help, as  $$(\sqrt{-1})^6 = (\sqrt{i^2})^6 = i^6 = (i^2)^3 = (-1)^3 = -1.$$ Where is the mistake or the step that is not allowed?
43762
I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
102749
$$(z+1)^6=z^6 $$ $$\frac{(z+1)^6}{z^6 }=1$$ $$\left(\frac{(z+1)}{z }\right)^6=1$$ $$\left(\frac{(z+1)}{z }\right)^6=\cos(2*\pi*k)+i\sin(2*\pi*k)$$ $$\frac{(z+1)}{z }=\left(\cos(2*\pi*k)+i\sin(2*\pi*k)\right)^{1\over6}$$ $$1+\frac{1}{z }=\cos({2*\pi*k\over6})+i\sin({2*\pi*k\over6})$$  $$\frac{1}{z }=(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3})$$ $$\frac{1}{(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3}) }=z$$ I dont know how else to solve this. A lot of help is appreciated. Also i know that k should go from 1-5.
43217
Where is the mistake? $$-1=(-1)^{2/2}=\left(\left (-1\right)\displaystyle ^2\right)^{1/2}=1^{1/2}=\sqrt 1=1$$
6893
this is something that came up when working with one of my students today and it has been bothering me since.  It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.  Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?  With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?
96688
My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
74621
As my question says,   $$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$  As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here?
74109
Let $F$ be a field.   And let $a \in F$ such that $a \cdot a = 1$.  Claim: $a = 1$ or $ a = -1$  Before, I begin with the proof I cannot understand what $ a = -1$ represents...  Is $a = -1$ representing the inverse of $a$?   I am capable of getting to $a = 1$ with no trouble, but how do you get the $-$ sign to show up in your proof?   Proof: This is my attempt at getting to $a=-1$.  If $a = 0$ then we have $0 \cdot 0 = 0 \neq 1$. Then this implies that $a \neq 0$ which implies that $ \exists b \in F,$ s.t. $a \cdot b = 1$, i.e. $b$ is the multiplicative inverse of $a$.  Knowing this fact prove as follows...:  $a \cdot a = 1 \implies$ $ (a \cdot a) \cdot b = 1 \cdot b \implies$ $a \cdot (a \cdot b) = 1 \cdot b \implies$ $a \cdot 1 = 1 \cdot b \implies$ $ 1 \cdot a =1 \cdot b \implies$ $a = b$.   But since $b$ is the multiplicative inverse of $a$ is this enough to say that $a = -1$?
7108
Does $\sqrt{-1}=\pm\sqrt{-1}$ since it has 2 solutions for roots? Note the link says there are two complex square roots for -1 which are $i$ and $-i$ .
83316
I am trying to solve a 2 part question, the first part is to expand $\sqrt{1-x}$ up to and including the term $x^2$ which I did.  This gives me $1-(0.5)x-(0.125)x^2 + ...$  However, the 2nd part of the question says that by taking $x=(1/64)$ in the expansion of part (i) -> which refers to the one above, deduce that $\sqrt 7 = 10837/4096$. How do i do this part?  EDIT (After Hint Given By SimpliFire):  $$ \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 $$ $$ =\frac83\sqrt{1-\frac1{64}}\;=\sqrt7 $$ $$ =\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)\; $$ $$ =\frac83\left(1-\frac1{128}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32511}{32768}\right)\; $$ $$ =\frac{260088}{98304} $$ $$ \frac{10837}{4096}\;\approx\sqrt7 $$  I'm not sure if I can present it like this, if it is okay i will post it as my answer :X
2779
I simplified to get $6x^4-25x^3+12x^2+25x+6=0$  Or $6 (x^2+1)^2+25x(1-x^2)=0$  Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$  I can't find a way to proceed further.
45673
What's wrong with this demonstration?:  $$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$  So finaly we get  $$1 = -1$$
498
So I am about halfway through complex analysis (using Churchill amd Brown's book) right now.  I began thinking some more about the nature and behavior of $i$ and ran into some confusion.  I have seen the definition of $i$ in two different forms; $i = \sqrt{-1} $ and $i^2 = -1$.  Now I know that these two statements are not equivalent, so I am confused as to which is the 'correct' definition.  I see quite frequently that the first form is a common mistake, but then again Wolfram Math World says otherwise.  So my questions are:   What is the 'correct' definition of $i$ and why?  Or are both definitions correct and you can view the first one as a principal branch? It seems that if we are treating $i$ as the number with the property $i^2 = -1$, it is implied that we are treating $i$ as a concept and not necessarily as a "quantity"? If we are indeed treating $i$ as a concept rather than a "quantity", how would things such as $i^i$ and other equations/expressions involving $i$ be viewed?  How would such an equation have value if we treat $i$ like a concept?   I've checked around on the various imaginary number posts on this site, so please don't mark this as a duplicate.  My questions are different than those that have already been asked.
109854
I am stuck at   $$(a+d)^2 - 4(ad-bc) &lt; 0$$ $$\implies (a+d)^2&lt;4(ad-bc)$$ $$\implies (a+d)&lt;2\sqrt{ad-bc}$$  where $a,b,c,d \ge 0$.  Is there a contradiction to derive here?   Also, the square root on the R.H.S. may not be legit, since I don't have any more information about the relation of $ad$ and $bc$ -- so let's say I took the complex square root, to be safe.    (This lack of information also makes it hard to use the AM-GM inequality.)  Any help is appreciated.  Thanks,
40496
What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!  So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?
44331
How to solve $x^3=-1$? I got following:  $x^3=-1$   $x=(-1)^{\frac{1}{3}}$  $x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
24665
Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number    I have tried to look at $$\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}$$  But did not get far
74047
Can you enter the rabbit hole recursively?  If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?