On a test I got a question asking to simplify $\sqrt{-25}\times \sqrt{-3}$ I answered $5\sqrt 3$ but apparently it's only $-5\sqrt 3$ because of some rule of imaginary numbers, could someone please explain to me why $\sqrt{-25}\times \sqrt{-3} \ne \pm 5\sqrt 3$
64753
I've been following a series on understanding about how imaginary numbers came to be, and in the series, it mentions that imaginary numbers mostly follows the algebra rules for real numbers, such as adding or multiplying by real numbers. However, it specifically mentions this "inconsistency(?)" about multiplying square roots of imaginary numbers do not follow the rule for multiplying square roots of real numbers, namely $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$. For example, why can't I do this: $\sqrt{-2} \times \sqrt{-3} = \sqrt{(-2)(-3)} = \sqrt{6}$, which I know is the wrong answer. In the past, I've just memorized to factor out the imaginary parts first, and that $i^2 = -1.$ However, can anyone show me an explanation for this, or tell me where I could learn more about it?
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If w = $\frac {\sqrt2+\sqrt3}{\sqrt5-\sqrt3} $  Express the following in the simplest form (with a rational denominator)  i) w  ii) $\frac {1}{w} $  It must be like this right ?  w = $\frac {\sqrt2+\sqrt3}{\sqrt5-\sqrt3} $ x $\frac {\sqrt5+\sqrt3}{\sqrt5+\sqrt3} $  Please help me with this question coz I don't get it at all :/
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Introduction:  Recently I found out that $i^i \approx 0.20788$ has no imaginary part. I got interested and then wanted to know whether there are other $n$ for which $\underbrace {i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}$ has no inaginary part.  So I wrote  (I'm a beginner at python, could be very bad code :) which plots what I call the $i-Tower\ up\ to\ n = 100$. It looks like this:   Question:  Let's use this convention: ${}^ni = \underbrace{i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}$.   Why is the sequence ${}^ni\ |\ n \in \mathbb{N}$ converging?   ${}^{20}i \approx 0.48770 + 0.41217i$   ${}^{60}i \approx 0.437584 + 0.360535i$   ${}^{100}i \approx 0.43829+ 0.36059i$  What I already noticed is that the angle between the lines you can draw from ${}^ni$ over ${}^{n+1}i$ to ${}^{n+2}i$ is $&lt;90°$.   Is that angle equal for all $n$? Is there a way to give the exact value of $x = {}^ni \ $with$\ {n \rightarrow \infty}$? What is the connection between $i$ and $x$. Is $x$ a special complex number that is maybe already known or comes up in other places?
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I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:  $$\sqrt{x+5} = x - 1$$  So I used this logic:  $$ \begin{align} \sqrt{x+5} &amp;= x - 1 \\ x + 5 &amp;= (x-1)^2 \\ x + 5 &amp;= (x-1)(x-1) \\ x + 5 &amp;= x^2 - 2x + 1 \\ 0 &amp;= x^2 - 3x - 4 \\ 0 &amp;= (x-4)(x+1) \\ \end{align} $$  Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.  But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:  $$ \begin{align} \sqrt{4 + 5} &amp;= 4 - 1 \\ \sqrt{9} &amp;= 3 \\ 3 &amp;= 3 \end{align} $$  So using 4 works as expected, but when using $-1$:  $$ \begin{align} \sqrt{-1 + 5} &amp;= -1 - 1 \\ \sqrt{4} &amp;= -2 \\ 2 &amp;\ne -2 \end{align} $$  At what stage am I going wrong?  And according the , the solution is $x = 4$.
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Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.  Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.  If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.  And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
7382
Problem: prove $\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}&lt;3$ by induction. I tried some, but stopped in $\sqrt[2^n]{n+1}$. Also tried with $2\sqrt{3\cdots}&lt;3^2$ and so on.
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It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
21646
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational   My attempt:-     Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$ Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$ and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
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Are they equal?      -5 = $\sqrt{(-5)^2}$
80000
An old challenge problem I saw asked to prove that  $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} &lt; 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C &lt; 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} &lt; C$?
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Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?
72082
I have tried this on my Windows 7 calculator with $\sqrt9 -3 $ it too gives some weird answer- ie$1.1546388020691628168216106791278e-37$. And so for any $n$(positive) $\sqrt n^2-n= wierd_ .answer$ Why does this happen?
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Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
94937
From a previous question, we have that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$, and I am assuming that we would need to use this to justify the answer to the question.   Would it be right to use some idea of a splitting field here? Or is it something to do with $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ not containing a primitive 3rd root of unity?   This is a past exam paper question worth 8 marks, so it doesn't seem like it's a few lined answer, but I am a little stuck on where to go.
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How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?  I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$  and  $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$  How to Proceed further?
12243
I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!
86086
I want to show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$.  I think it would be easier to prove it using the following: $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{3})$. Then $\sqrt{2}\notin\mathbb{Q}(\sqrt{3})$ so $\sqrt{2}\in \mathbb{Q}(\sqrt[4]{3})\Longleftrightarrow \sqrt{2}=a+b\sqrt[4]{3}$, with $a,b\in\mathbb{Q}(\sqrt{3})$.   I tried simplifying the above equation, but I did not get anything.