# Tag Info

## Hot answers tagged proof-writing

4

Your line is, after multiplying by 6 and moving things to the left, $3x+2y-6=0$, whose distance to the origin is $|-6|/\sqrt{3^2+2^2}=6/\sqrt{13}\approx 1.6641.$ See this page for the distance from point to line formula, with several proofs of it. Added: to proceed via the contrapositive, assume in fact that $x^2+y^2 \le 1$. Then we have $|x|\le 1$ and ...

4

Generally, derangements come in pairs: the inverse of a derangement is a derangement. The exceptions are the fixed-point-free involutions, i.e., the permutations in which every cycle is a $2$-cycle. Thus the parity of the number of derangements is the same as the parity of the number of fixed-point-free involutions. If $n$ is odd, there are no ...

4

The simplest way of deriving/proving this is through Euler's formula: Take $e^{i(\theta+\alpha)}$ which, through basic exponent rules we know is equal to $e^{i\theta +i\alpha} = e^{i\theta}\cdot e^{i\alpha}\ .$ Now expanding this (using Euler's formula) we get: \begin{align} e^{i\theta}\cdot e^{i\alpha} &= ... 4 Let's see: We have that 2^n>n for all n, and 2^n>n^2 for n\ge 5. So 4^n=2^n2^n>n^3 for n\ge 5, and we just need to prove the cases n\le 4 by hand, and this is straightforward. 3 Well, if you buy or have access to (through previous learning) e^{i\theta} = \cos \theta + i \sin \theta, then setting \theta = a - b one has e^{i(a - b)} = \cos(a - b) + i\sin(a - b). \tag{1} But also, e^{i(a - b)} = e^{ia}e^{-ib} = (\cos a + i \sin a)(\cos b - i\sin b). \tag{2} The imaginary part of the product on the right of (20) is \sin a ... 3 It is possible to find the parity from the formula\sum_{i=0}^n (-1)^i \frac{n!}{i!}=\sum_{i=0}^n (-1)^i n(n-1)\cdots(i+1).$$Working modulo 2: If n is even, then there is one non-zero contribution to the sum is when i=n, so we have an odd number. If n is odd, then there are two non-zero contributions to the sum, when i=n and when i=n-1, so ... 3 I don't think there is a reasonable general answer to this very genuine phenomenon: that's just the way the (mathematical) world is. In a related area, most of modern cryptography (and thus of modern economics: think banking) relies on the fact that it is very easy to multiply two huge prime numbers p, q but extremely difficult, given just their product ... 3 Hint: Try the other way: x/2+y/3=1, so, \frac32x+y=3, yielding y=3-\frac32 x. Then compute x^2+y^2. Or, even better: try geometrically: the set of points (x,y) on the plane that satisfy x/2+y/3=1 is a line. This line contains (2,0) and (0,3). Draw it and draw also the disk x^2+y^2<1. 2 For x\in(0,\frac\pi 2) we have$$\cos x+\sin x-1=\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)-1\\=-2\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)=2\sin\left(\frac x2\right)\left(\underbrace{\cos\left(\frac x2\right)-\sin\left(\frac ...

2

Hint $\ {\rm mod}\ m\!:\,\ ax\equiv b\,\overset{\large {\rm times}\ c}\Rightarrow\,cax\equiv cb.\,$ Conversely, by Bezout, $\,\gcd(c,m) = 1\,\Rightarrow\, c^{-1}\,$ exists, thus the opposite direction follows by multiplying $\ cax\equiv cb\,$ by $\,c^{-1}$ to cancel $\,c.$ Remark $\$ Generally, scaling an equation by a unit (invertible) yields an ...

2

In my experience, there is no plan of making a proof bidirectional. Really, you just try to prove $P\implies Q$ and see what happens. When you have your proof that $P$ implies $Q$, you take a good look at it and try to reverse every step of it. The usual case is that you can reverse some of the steps, but not all of them, but sometimes you get lucky and can ...

2

$$e^{i\alpha}=\cos \alpha + i\sin \alpha$$ $$e^{i\theta}=\cos \theta + i\sin \theta$$ \begin{align}e^{i(\alpha+\theta)}&=e^{i\alpha}\cdot e^{i\theta} \text{ (law of exponents)}\\ \cos(\alpha+ \theta) + i\sin (\alpha+\theta)&=(\cos\alpha+i\sin\alpha)(\cos \theta + i\sin \theta) \end{align} Multiply, and compare $\Re(z)$ and $\Im(z)$ of both ...

2

With vectors there is a geometric formula for the dot product $$\vec{u}\cdot\vec{v}=\left\|\vec{u}\right\|\left\|\vec{v}\right\|\cos\theta$$ where $\theta$ is the angle between the two vectors. Take $\vec{u}=\langle\cos(a),\sin(a)\rangle$ and $\vec{v}=\langle-\sin(b),\cos(b)\rangle$ and this formula gives $$\sin(a)\cos(b)-\cos(a)\sin(b)=\cos\theta$$ ...

2


1

Here's one way to look at this: Suppose $\nabla$ is the unique covariant derivative operator associated with the metric tensor $g$ on the surface $S$, i.e. it is the unique symmetric connection on $S$ with $\nabla g = 0$. It is well know that the geodesics $\gamma(t)$ of $g$ satisfy $\nabla_{\mathbf w}\mathbf w = 0, \tag{1}$ where $\mathbf w$ is the ...

Only top voted, non community-wiki answers of a minimum length are eligible