The square root of i is (1 + i)/sqrt (2). [Try it out my multiplying it by itself.] It has no special notation beyond other complex numbers; in my discipline, at least, it comes up about half as often as the square root of 2 does --- that is, it isn't rare, but it arises only because of our prejudice for things which can be expressed using small integers.

2014年6月24日 · Hi guys I just want to ask how to solve this. The given is:

So, my friend show me prove that $1=-1$ by using this way: $$1=\\sqrt{1}=\\sqrt{(-1)\\times(-1)}=\\sqrt{-1}\\times\\sqrt{-1}=i\\times i=i^2=-1$$ At first sight, I ...

2017年7月22日 · The square root symbol does not make sense when applied to a negative number. Granted, people often write it that way for convenience. i i is defined by the equation i2 = −1 i 2 = − 1. Yes, if there is a field containing a root of x2 + 1 x 2 + 1 then if i i is such a root so is −i − i. When we define a field such as C C we are implicitly choosing a root of x2 + 1 x 2 + 1.

I thought of this a while ago, but can't make up a proof or a counterexample. Does anyone know more about this? $$\sqrt {a+b} \le \sqrt a + \sqrt b , \forall a,b \in \mathbb R_+$$ Moreover, what ha...

2017年2月2日 · An alternative solution. Assume that α = 2–√ + 3–√ + 5–√ = a b ∈ Q. By quadratic reciprocity, there is some prime p> b such that 3 and 5 are quadratic residues (mod p) while 2 is not. That implies that α is an algebraic number over K = Fp with degree 2, since 2–√ does not belong to K but belongs to a quadratic extension of K. …

2015年1月16日 · I'm trying to figure Taylor series for $\sqrt {x}$. Unfortunately all web pages and books show examples for $\sqrt {x+1}$. Is there any particular reason no one shows Taylor series for exactly $\sqrt...

It's just notation most likely. Yes, (−2)2 = 4 (− 2) 2 = 4, but often the 4–√ 4 symbol is reserved for the positive square root, so 4–√ = 2 4 = 2. If you want the negative square root, that would be − 4–√ = −2 − 4 = − 2. Both −2 − 2 and 2 2 are square roots of 4 4, but the notation 4–√ 4 corresponds to only the positive square root.

Consider the sequence defined by a1 = 2–√ a 1 = 2, a2 = 2 + 2–√− −−−−−√ a 2 = 2 + 2, so that in general, an = 2 +an−1− −−−−−−√ a n = 2 + a n − 1 for n> 1 n> 1. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least ...

2013年11月17日 · Prove that the function $\sqrt x$ is uniformly continuous on $\ {x\in \mathbb {R} | x \ge 0\}$. To show uniformly continuity I must show for a given $\epsilon > 0$ there exists a $\delta>0$ suc...