What I'm missing in $p=3^{q}\cdot 2^{r}$? Given positive integer $p,q,r$ with $p=3^{q}\cdot 2^{r}$ and $100<p<1000$. The difference between maximum and minimum values of $(q+r)$ is?

My tries

It's clear that $p=6k$ for some positive integer $k$

Hence, $6*17=102\neq 3^{q}\cdot 2^{r}$, on trying next one $6*18=108=3^3\cdot2^2$
.Hence min $(q+r)=5$

Similarly, $972=6*162=3^52^2$, max $(q+r)=7$

difference $=2$, but didn't matched with answer given.

please help.

You're finding the largest number of the form $2^p 3^q$, but that doesn't maximize $p+q$. As $2<3$ getting at high exponent on $2$ makes a smaller number than getting a high exponent of $3$. Trying the powers of $2$, we quickly find that $256=2^8$ is the highest that we can multiply by three and stay below $1000$.

Similar considerations might be relevant to find $\min(p+q)$.

And note that I haven't in any way proved that you'll find the $\min/\max$ this way, that's left for you.

Yoou don'h have to take minimum and maximum values of $p$, but of $q+r$, which is the number of prime factors in the expanded notation. To minimize  $q+r$, we need larger factors appearing more. hence we search for numbers where $r=1$. For $162=2\cdot3^4$. Similarly, for maximum, $768=3\cdot2^8$. this gives difference 4


Existential quantifier I've been doing a bit of independent research on existential quantification, and I was wondering if anyone could tell me if I have the hang of it. Here's a statement regarding the intermediate value theorem of calculus:

If $f$ is a function of $x$ that is continuous on the interval $[a,b]$, then $$\exists\left( p\in [a,b]\right)\,\left( f(p)\in [f(a),f(b)]\right)$$ is true.

If it helps, I typed this as
If $f$ is a function of $x$ that is continuous on the interval $[a,b]$, then $$\exists\left( p\in [a,b]\right)\,\left( f(p)\in [f(a),f(b)]\right)$$ is true.

If you are having trouble with existential quantifiers, there are just four rules that govern its usage in a mathematical proof (with each quantifier explicitly restricted to a set):

From $\exists x\in A: P(x)$, we can infer that $y\in A$ and $P(y)$ where $y$ is a free variable not previously used in that proof. (Existential specification or instantiation.)
(EDIT) From $y\in A$ and $P(y)$, we can infer that $\exists y\in A: P(y)$, or that $\exists x\in A: P(x)$ where $x$ is not a bound variable in $P(y)$, and where $P(x)$ is formed replacing some or all occurrences of $y$ in $P(y)$ with $x$. (Existential generalization)
From $\exists x\in A: P(x)$, we can infer that $\neg \forall x\in A:\neg P(x)$. (Changing quantifier, $\exists \to \forall$)
From $\forall x\in A:P(x)$, we can infer that $\neg\exists x\in A: \neg P(x)$. (Changing quantifier, $\forall \to \exists$)

where $A$ is a set and $P$ is a unary predicate.

Note that: $$\exists x\in A: P(x) \equiv \exists x: [x\in A \land P(x)]$$  $$\forall x\in A: P(x) \equiv \forall x: [x\in A \implies P(x)]$$

Avoid the construct:

$$\exists x: [x\in A \implies P(x)]$$

Very strange things can happen otherwise! (See "Drinker's Paradox" at Wikipedia and at my blog.)


Evaluate $\lim_{n\to{\infty}}\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}=c$ If $$\lim_{n\to{\infty}}\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}=c$$

  where $c$ is a non zero constant. Then evaluate $12(c-\alpha)$.

My Attempt:

I know that $$\lim_{n\to{\infty}}\bigg(1-\frac{1}{n}\bigg)^n=\frac{1}{e}$$

i.e. as $$\lim_{n\to{\infty}}{e\bigg(1-\frac{1}{n}\bigg)^n-1}=0$$

And as the question says limit is non zero, we can conclude that $\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}$ is an $\frac{0}{0}$ form.

i.e. $$\lim_{n\to{\infty}}n^{\alpha}=0$$

From this I can conclude that $\alpha\lt{0}$.

But I am not able to solve this further. I know that the exact value of $\alpha$ can be found using the fact that the limit exists and is a finite non zero constant. But I don't know how to approach this? I tried using L-Hopital but it wasn't of much help. How do I proceed?

Thank you.

Hint:

By the Binomial theorem,
$$\left(1-\frac1n\right)^n=1-\frac nn+\frac{n(n-1)}{2n^2}-\frac{n(n-1)(n-2)}{3!n^3}+\cdots$$

and this should be made close to

$$\frac{1+cn^\alpha}e.$$

Develop the RHS in terms of $1$ and $n^{-1}$.


How do I show that $\ell^{\infty}$ is a normed linear space? How do I show that $\ell^{ \infty}$ is a normed linear space,
where $\ell^{ \infty}$ is define as $$\|\{a_n\}_{n=1}^{\infty}\|_{\ell^\infty}=\sup_{1 \leq k \leq \infty} |a_k|?$$
There are three properties that I need to check in order for this to be a normed linear space. Nonnegativity, positive homogeneity, and the triangle inequality.
I am having trouble working with the supremum. Any idea will be greatly appreciated thanks

For the triangle inequality, suppose that $\{a_n\},\{b_n\}\in \ell^{\infty}$. For each index $n$ we have
$$|a_n+b_n|\leq |a_n|+|b_n|\leq \sup_{m}|a_m|+\sup_m|b_m|=\|a\|_{\infty}+\|b\|_{\infty}.$$
Then taking the supremum over all $n$ shows that
$$ \|a+b\|_{\infty}\leq \|a\|_{\infty}+\|b\|_{\infty}. $$

Let $(a_n),(b_n) \in \ell^\infty$ Then we have for each $k$:
$$|a_k+b_k| \le |a_k|+|b_k| \le ||(a_n)||_\infty+||(b_n)||_\infty.$$
This shows that $(a_n+b_n) \in \ell^\infty$ and that
$$||(a_n+b_n)||_\infty \le ||(a_n)||_\infty+||(b_n)||_\infty.$$


Prove that $f(x)=x^3 −x−1$ has at least one real root. How would I go about proving this?
Would I try finding a value for $x$ that will make $f(x) = 0$?

It's continuous, odd powered polynomial, which means it goes from $-\infty$ to $+\infty$, which in turn means that it crosses $x$ axis somewhere in between.

So, $\space$the$\space$ $f(x) = x^3 - x - 1$

Then,$\space$ $x$ $(x^2 - 1) - 1$, by pulling out an $x$.

$\rightarrow$ $x - 1$ is equal to $x = 1$

$\rightarrow$ $x^2 - 1$ is equal to $x^2 = 1$, then sqrt both sides, $x = 1$

However, If you apply the limit test of $lim_x\rightarrow_\infty$ of the $f(x) = x^3 -x -1$, then the $f(x) =\infty$ so the function is continuous.


Do quotient groups contain the empty set? I was reading Foote & Dummit's book on abstract algebra and this question arose. The authors start their discussion on quotient groups by first defining the quotient group $G/K$ where $K$ is the kernel of some homomorphism $\varphi$ defined on $G$. Let me introduce some of their definitions and terminology.
Let $G$ and $H$ be groups and $\varphi: G \to H$ a group homomorphism with kernel $K$. A fiber in $G$ over $h$ is the pre-image the singleton $\{h\} \subset H$, i.e. $\varphi^{-1}(\{h\})$. The quotient group $G/K$ is defined to be the set of all such fibers with group operation defined as follows. If $X = \varphi^{-1}(\{h\})$ and $Y = \varphi^{-1}(\{s\})$, then $XY := \varphi^{-1}(\{hs\})$.
Now, I might be wrong but as far as I know, the homomorphism $\varphi$ is not assumed to be surjective. Thus, in principle the empty set can be an element of $G/K$.
However, the authors discuss a more natural way to define $G/K$ in which the explicit use of the homomorphism $\varphi$ is not required. First, they prove that if $X$ is any fiber and $u \in X$ then $X = uK :=\{uk: \hspace{0.1cm} k \in K\}$. Then, they proceed to prove that $G/K$ is precisely the set of all such left cosets $gK$, $g \in K$. At this point, I prefer not to follow the argumentation of the book. My thoughts are the following. If $X$ is a nonempty fiber on $G/K$, then $X = uK$ for some $u \in X \subseteq G$, so that every nonempty fiber is a left coset of the form $gK$, $g \in G$. Conversely, if $g \in G$, there exists some fiber $X$ such that $g \in X$ and $X = gK$. However, the reasoning here is not taking the empty set into account and, in addition, the empty set cannot be expressed as a left coset since $K$ is a subgroup of $G$.
Question: Does the quotient group $G/K$ contain the empty set or not?
Note: I know that the problem can be solved by taking $H = \text{Im}\varphi$ instead, which is a subgroup of $H$, but there is no mention of this fact on Foote & Dummit's book and I got tricked by it.

No, the empty set is not an element of the quotient group $G / K$. This appears to be a mistake in Dummit and Foote: they should define $G / K$ to be set of nonempty fibers of $\varphi : G \to H$.
One explanation for defining quotient groups in this way is that we want the first isomorphism theorem to hold: if $\varphi : G \to H$ is a group homomorphism then we want $G / \ker \varphi \cong \operatorname{im} \varphi$, however if $\varphi$ is not surjective and hence with the incorrect definition $G / \ker \varphi$ contains the empty set (and $G$ is finite), then these two sets will not even have the same size.

The quotient group cannot be empty because it contains the identity element. It may be the only element but it's still at least a singleton.


Why is $nlog(n)=\Theta(n^2)$ not true? I want to demonstrate this relation. I know that $f(n)=\Theta(g(n))$ when $\exists c_1>0, c_2>0, n_o\in \mathbb{N} \mid \forall n \geq n_0$ for which:
$$c_1g(n) \leq f(n) \leq c_2g(n)$$
For the first part of the relation, I have $c_1n^2 \leq nlog_{10}(n)$. How can I now demostrate that these $c_1$ and $n$ exist (or better, don't exist, since I know that the initial equation is wrong)?

EDIT: Sorry, I forgot to say. I can't use logarithms, only the definition of $\Theta$

If you can use calculus, it is fairly simple. Suppose that there exist $c_1>0$ such that $c_1\le \frac{n\log_{10}(n)}{n^2}$ for all $n$ large enough. Take the limit:
$$c_1\le \lim_{n\rightarrow \infty}\frac{n\log_{10}(n)}{n^2}=\lim_{n\rightarrow \infty}\frac{\log_{10}(n)}{n}=0$$
contradiction.

In other words you have proved that $n\log_{10}(n)=o(n^2)$, so $n\log_{10}(n)\ne\Theta(n^2)$

A little example ,make a sense . $$n=100 \to nlogn=200  ,n^2=10^4\\n=10^5 \to nlogn=5\times 10^5 ,n^2=10^{10}$$ does it make a sense ?


Is the interior of this closure in the union of the closures? 2 Hi everyone: Suppose $V$ is an open set, and $(F_{n})$ is an increasing sequence (so $F_{n}\subset F_{n+1}$) of closed sets, all of them in $\mathbb{R}^{k}$, $k\geq2$. If
$V$ is included into the union of the interiors of the $F_{n}$'s,   can we say that the interior of the closure of $V$ is included into the union of the $F_{n}$'s?

The closure of $V$ is just $V$ union boundary. Boundary points are not interior. Also a set is open iff it is equal to its interior. So i think this should make the closure of the interior of $V$ just $V$. As $V$ is in the union of the sets $F_n$, then the interior of the closure should be in the union of $F_n$.


very stupid confusion in probablity or combination? i confuse very much on these when any question from probability is asked for example
suppose you have three beggars let they be x,y,z and 2 biscuits and you have to distribute between them so total number of ways should be the numbers of solution of equation x+y+z=2 which is 6 and i know it i am certain of that but if we look at this way

consider first biscuit it has three options and the other one it also has three options so it should be 3*3 that is 9 but it isn't i know it isn't but i am still very confused what it means mathematically and practically

to explain what i am talking about consider these question i am not making these up they are in my textbooks

1 find the probability that a particular beggar gets exactly 1 biscuit

2 number of ways of distributing biscuits between beggars

now on the first one the solution uses 9 as total number of ways but in second its 6 which is according to my intuition and which i feel should be used in first too ... what is going on here can someone explain me please examples of both cases with proper logic would be much appreciated

The number of non-negative integer solutions of the system:

$\begin{cases}x_1+x_2+\dots+x_r=n\\
0\leq x_1\\
\vdots\\
0\leq x_r\end{cases}$

is known to be $\binom{n+r-1}{r-1}$ and can be proven using a stars and bars argument.

Assuming the biscuits are considered identical and may not be broken apart, the number of ways of distributing the biscuits is then $\binom{2+3-1}{3-1}=\binom{4}{2}=6$ as you correctly found.

However: assuming each biscuit is given to one of the beggars uniformly at random, note that these six possibilities are not equiprobable.  It is twice as likely to arrive at the scenario $(1,1,0)$ than you are to arrive at the scenario $(2,0,0)$ (where the numbers correspond to the number of biscuits received by the respective beggar).

To answer the question of the probability, again assuming that the biscuits are distributed uniformly at random (this really should have been specified in the problem.  Perhaps in reality the shortest man is more likely to receive the biscuits since he looks more pathetic), imagine for a moment that the biscuits are in fact different colors.

You have nine different ways of distributing the biscuits now, and each of these are equiprobable.  The probability that the first beggar receives exactly one biscuit is then $\frac{1}{3}\cdot \frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{9}$  (he either receives the red biscuit but not the blue one or he doesn't receive the red biscuit but does receive the blue one).

Remember that the scenarios $(\color{red}{1},\color{blue}{1},0)$ and $(\color{blue}{1},\color{red}{1},0)$ are considered different in this case, whereas in the other problem where the biscuits are considered identical.

To calculate probability: (no. of favorable outcomes)/(total no. of outcomes).

Probability that beggar gets exactly 1 biscuit is calculated as follows:

Total no. of favorable outcomes are : 6 (3 * 2 first biscuit can      be
distributed in 3 ways while second biscuit can be distributed in 2 ways
since  none beggar can get more than one)

Total no. of outcomes : 9( 3 * 3 since first biscuit can be distributed in
3 ways and second biscuit can be distributed in 3 ways and each beggar can
get more than one biscuit )

So probability : 6/9


Find the remainder when $ p(x) = (x+2)^{101} + (x+3)^{200}$ is divided by $ x^2 +5x + 6 $. Find the remainder when $p(x) = (x+2)^{101} + (x+3)^{200}$ is divided by $ x^2 +5x + 6 $.

HINT.   Set $P(x)=Q(x)(x^2+5x+6)+ax+b$ and choose two suitable values of $x$ to make a simultaneous equation for $a$ and $b$

Set the polynomial x^2 + 5x + 6 to zero then find x by quadratic formula. and put that x in the first polynomial.


Extension of Euclid's lemma This is a somewhat obvious fact that is intuitively obvious to me, but I haven't been able to construct a proof of it. Euclid's lemma states for for $p$ a prime and $ab$ a product of integers (let's take everything to be positive for simplicity), if $p \mid ab$, then $p \mid a$ or $p \mid b$. This is clear, and I know how to prove it. Let's extend it somewhat. Suppose that $a$ and $b$ are two relatively prime integers, and we have $a \mid bc$ for some other integer $c$. Then $a \not \mid b$, so it must divide $c$. This fact is obvious to me, but I can't figure out how to prove it.
Does anyone have any hints or advice? Do I need the assumption of positivity? (For my purposes at the moment, I only need them to be positive, but there is value in having the most general result possible).
EDIT: Updated attempt:
We have that $a,b$ are relatively prime, so there exist $r,s \in \mathbb{Z}$ such that $ar + bs = 1$ by Bézout's lemma. Multiply through by $c$ to get $arc + bsc = c$. Then $a \mid arc$ and $a \mid bsc$, so $a \mid c$.
How is that?

We have $m|n \iff \gcd(m,n)=m$.
We have $\gcd(a,bc)=\gcd(a,c)$, because $\gcd(a,b)=1$.
So, $a=\gcd(a,c)$, therefore $a|c$.


Determining $\arg(u-\sqrt{3})$ in exact form. Let $u$ be the solution to $z^5=-9\sqrt{3}i$ so that $\frac{\pi}{2}\le arg(u) \le \pi$.
Determine $\arg(u-\sqrt{3})$ in exact form.
How would I go about completing this question?

Let $z=r(\cos\theta+i\sin\theta)$. We know $-9i\sqrt3=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2} )$
This means (using De Moivre's Theorem)
$$r^{5}(\cos5\theta+i\sin5\theta)=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2})=9\sqrt3(\cos(-\frac{\pi}{2}+2k\pi)+i\sin(-\frac{\pi}{2}+2k\pi)$$
So $r^{5}=9\sqrt3$ so $r=\sqrt3$.
We also have $5\theta=-\frac{\pi}{2}+2k\pi$ so $\theta=-\frac{\pi}{10}+\frac{2k\pi}{5}$
For different values of $k$:
$$k=0: \theta=-\frac{\pi}{10}$$
$$k=1: \theta=-\frac{\pi}{10}+\frac{2\pi}{5}=\frac{3\pi}{10}$$
$$k=-1: \theta=-\frac{\pi}{10}-\frac{2\pi}{5}=-\frac{\pi}{2}$$
$$k=2: \theta=-\frac{\pi}{10}+\frac{4\pi}{5}=\frac{7\pi}{10}$$
$$k=-2: \theta=-\frac{\pi}{10}-\frac{4\pi}{5}=-\frac{9\pi}{10}$$
Ignore all values of $\theta<\frac{\pi}{2}$ as this is what you specified. This leaves us with : $$\theta= \frac{7\pi}{10}$$
So (using Euler's relation, that $e^{i\theta}=\cos\theta+i\sin\theta)$:
$$ u=\sqrt3 e^{\frac{7i\pi}{10}} $$
Can you do it from there?
Hope that helped :)

The specified root is
$$u=\sqrt3\text{ cis}\frac{7\pi}{10}.$$
Then
$$u-\sqrt3=\sqrt3\left(\cos\frac{7\pi}{10}-1+i\sin\frac{7\pi}{10}\right)=\sqrt3\left(-2\sin^2\frac{7\pi}{20}+2i\sin\frac{7\pi}{20}\cos\frac{7\pi}{20}\right)$$
and
$$\arg(u-\sqrt3)=\frac{7\pi}{20}-\frac\pi2.$$


What is special about $x^T P y$? We know that if a matrix $P$ is positive definite, then for any vector $x \ne 0$, $x^T P x > 0$. But what about $x^T P y$ where $x \ne y$?

$$\rm x^{\top} P \, y = \frac 12 \begin{bmatrix} \mathrm x\\ \mathrm y\end{bmatrix}^{\top} \begin{bmatrix} \mathrm O & \mathrm P\\ \,\mathrm P^{\top} & \,\mathrm O\end{bmatrix} \begin{bmatrix} \mathrm x\\ \mathrm y\end{bmatrix}$$

The characteristic polynomial of the block matrix above is

$$\det \begin{bmatrix} s \mathrm I & -\mathrm P\\ -\mathrm P^{\top} & s\mathrm I\end{bmatrix} = \det( s^2 \mathrm I - \mathrm P \mathrm P^{\top} )$$

where $\mathrm P \mathrm P^{\top}$ is positive semidefinite. The spectrum of the block matrix contains the positive and negative square roots of the (nonnegative) eigenvalues of $\mathrm P \mathrm P^{\top}$. Hence, the block matrix

$$\begin{bmatrix} \mathrm O & \mathrm P\\ \,\mathrm P^{\top} & \,\mathrm O\end{bmatrix}$$

is indefinite, and so is the bilinear form $\rm x^{\top} P \, y$.


Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$ Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$
Then I think its $x=0, y=0$ and $z=0$ and the surface $z=4-x^2-y^2$
I think the integral is:
$\int_0^{2}\int_0^{\sqrt{{-x^2}+4}}4-x^2-y^2dydx$
is right?

Your integral approach is correct. An easy way to do it is using polar coordinates. @Rezha Adrian Tanuharja said cylindrical coordinates, but you use them for triple integrals which is not the case (you can also solve this problem with $\displaystyle\iiint_E dV$).
Let $x=rcos(\theta)$, $y=rsin(\theta)$.
Your integration region is the circle in the first quadrant:

Then, we can see that $\theta$ should go from 0 (starting from point $C$) to B, and that angle is $90$ degrees. Then $0\leq\theta\leq \frac{\pi}{2}$. It is also easy to see that the radius of the cirlce goes from $0$ to $2$ (because $2^2=x^2+y^2$).
You should also remember that the jacobian when using polar coordinates is $r$, and we set up the following integral:
$\displaystyle\int_0^{{\pi/}{2}}\int_0^2(4-r^2)rdrd\theta=2\pi$ (this is an easy to solve integral, just a polynomial).

Use cylindrical coordinate: $z=4-r^{2}\implies r^{2}=4-z$
$$
\begin{aligned}
V&=\frac{1}{4}\int_{0}^{4}\pi r^{2}dz\\
&=\frac{\pi}{4}\int_{0}^{4}(4-z)dz\\
&=2\pi
\end{aligned}
$$


for $z,w\in\mathbb{C}$, $\sqrt{zw} = \sqrt{z}\sqrt{w}$? for $z,w\in\mathbb{C}$, $\sqrt{zw} = \sqrt{z}\sqrt{w}$

I started by writing $z$ and $w$ in polar coordinates, and writing it out, giving another form for the question:
$$
\begin{align}
y &= r_1r_2cos(\theta_1 + \theta_2) +  r_1r_2sin(\theta_1 + \theta_2) i\\
y &= a + bi
\end{align}
$$
so
$$
\sqrt{zw} = \sqrt{r_1(cos(\theta_1)+isin(\theta_2))}\sqrt{r_2(cos(\theta_2)+isin(\theta_2))}
$$
but how to prove the original problem without just saying: square both sides?

Another correct formulation.
A complex number may have two square-roots.  If you take any square-root of $w$ and multiply it by any square-root of $z$, the product will be a square-root of $zw$.

This is very dangerous...

Even if you choose a consistent definition (e.g. $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2}$ with $\theta\in[0,2\pi]$), what happens when $z=i$ and $w=-i$?


How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$ I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.

So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.

Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.

How would I start with this integration?

Here's something you might try. Note that the length of that arc will be the same as the length of the same arc, reflected over the line $y=x$. That is, the arc $y=x^2/4$, from $x=0$ to $x=2$.

Notice that the integrand is a differential binomial, and then you may apply "Integration of differential binomial" (P. L. Chebyshev) and you're immediately done. See here.


Justify the given locus Let $\square ABCD$ be a convex quadrilateral, and $\overline{\rm CD}=2$. $N$ is the midpoint of $\overline{\rm DB}$. Determine the locus of $N$ when $D$ moves.

What I know is that the locus is the circumference shown(centre F, radius 1), but I don't know why.

Let F’ be the midpoint of BC. We further let $D_1$ and $D_2$ be two different locations of D. Correspondingly, we have $N_1$ and $N_2$ as the midpoints.

Note that $CD_1 = CD_2$. By midpoint theorem, $N_1F’ = (\frac 12)CD_1 = (\frac 12)CD_2 = N_2F’$.

This means F’ is the center of the circle passing through $N_1$ and $N_2$ with radius = half of $CD$.

(Too long for a comment.)

That's an oddly worded question, either carelessly or perhaps deliberately so.

At face value, point $\,A\,$ doesn't figure anywhere, so the quick answer is to take the homothety centered at $\,B\,$ with ratio $\,1/2\,$ which sends $\,D\,$ to $\,N\,$ and $\,C\,$ to the midpoint $\,F\,$ of $\,BC\,$, determine that $\,NF = CD /2 = 1\,$, then conclude that the locus is the circle of radius $\,1\,$ centered at $\,F\,$.

It is certainly true that the locus is a subset of that circle, but on the other hand the problem also states that the quadrilateral $\,ABCD\,$ is convex, and $\,D\,$ "moves". This could be interpreted either as the other three vertices staying put in place, or as the side lengths staying constant with $\,B,C\,$ fixed and $\,A\,$ moving along with $\,D\,$ like in an articulated mechanism. Each interpretation would lead to different answers as to what arcs of the full circle satisfy the condition, neither entirely trivial.

Otherwise put, not enough information for a definitive answer.


Show that function is injective and surjective using given condition on $f$ Consider $f$ a continuous function from $\mathbb{R}$ to $\mathbb{R}$

It is given that $$|f(x)-f(y)|\geq \frac{1}{2}|x-y|\tag{1}$$ for all $x,y$ in $\mathbb{R}$

I want to show that $f$ is one-one and onto

My efforts

Injectivity

Suppose $a,b\in \mathbb{R}\text{ and } a\neq b$ such that $f(a)=f(b).$

Then consider the following fraction $\frac{f(b)-f(a)}{b-a}$ which is equal to zero as $f(b)=f(a)$ and $b\neq a$

But that is a contradiction as given condition on $f$ says that $|f(b)-f(a)|\geq \frac{1}{2}|b-a|$ and that would actually show that $0\geq \frac{1}{2}|b-a|$ and right hand side is strictly greater than zero.

Therefore function is one one.

Surjectivity

Let $x_0\in \mathbb{R}$ be any arbitrary point and WLOG assume $x_0\geq 0$.

If I put $y=0$ condition (1) says $f(x)\geq \frac{1}{2}x+f(0)$ if $x\geq 0$ (also using the fact that WLOG that function is increasing).

Take $x=2x_0$,

we have $f(2x_0)\geq x_0 +f(0)$

Take $x=-2x_0$ and then we have $f(-2x_0)\leq x_0-f(0)$

So we have $f(-2x_0)\leq x_0-f(0)\leq x_0\leq x_0+f(0)\leq f(2x_0)$

Now Intermediate Value theorem works like a magic and we are done!!!!

Am I correct?

Edit: If people are confused, why I can assume that function is increasing in surjectivity part, here is the answer A continuous, injective function $f:\mathbb{R}\rightarrow \mathbb{R}$ is either strictly increasing or strictly decreasing.

The proof doesn't seem to be right, for example, f(x)=-x will make the proof fail.

The conclusion is correct, though: since R is connected, so is the image of f(x). In other words, the image fills a chuck of R. The only thing we need to prove is f(x) is unbounded, which follows immediately from the inequality with y=0.


Minimizing the surface area of a rectangular prism with volume $50000$L. A town wants to build a water tank to hold 50000L of water. They want to minimize the amount of material used to build the tank by minimizing the surface area. If they make the water tank into a closed rectangular prism, what would the dimensions be?

Volume of the cube is $50 kL$. Hence, $lbh=50000$. If the surface area is minimum, then $lb+2h(l+b)$ has to be minimized.
Using  $h=\frac{50000}{lb}$ we get $SA=lb+100000\frac{l+b}{lb}$.
By argument of symmetry, $l=b$.
Minimizing $l^2+\frac{200000}{l}$ gives $l=10\sqrt[3]{100}$ and $h=5\sqrt[3]{100}$

A cube, $a=\sqrt[3]{50}\approx3.68 m=368 cm$, $A=6a^²≈81.22 m^²$
If a shape of water tank is arbitrary, then it's a sphere , $R=\sqrt[3]{\frac{3*50}{4\pi})}\approx2.28 m, A_1=4\pi R^2\approx 65.52 m^2$


Number of elements in the ring $\mathbb Z [i]/\langle 2+2i\rangle$ The question is :

Show that $I=\langle 2+2 i\rangle$ is not a prime ideal of $\mathbb Z[i]$. Also find the number of elements in $\mathbb Z[i]/I$ and its characteristic.

My try:

I started with elements $2, 1+i \in \mathbb Z[i]$. Clearly, the product is $0$ in the coset representation.

The part where i am struck at is the number of elements and the characteristic.

Well, $2+2i+\langle 2+2i\rangle=0+\langle 2+2i\rangle$

Can we directly write $i^2=-1$ in the coset representation?

The following is not the most elementary approach, but it puts the question in a nice perspective (in my humble opinion):

There is an isomorphism $\Bbb{Z}[i]\cong\Bbb{Z}[X]/(X^2+1)$, given by $i\ \mapsto\ X$, yielding an isomorphism
$$\Bbb{Z}[i]/(2+2i)\cong\Bbb{Z}[X]/(X^2+1,2+2X).$$
The characteristic of the latter the least positive integer in the ideal $(X^2+1,2+2X)\cap\Bbb{Z}$ of $\Bbb{Z}$. Its number of elements is precisely the index of the ideal $(X^2+1,2+2X)$ in $\Bbb{Z}[X]$, which is the same as the resultant of $X^2+1$ and $2+2X$, which is in itself a nice fact to prove.

I usually solve it by drawing the square generated by $2+2i$.

....|....
..xbbbx..
..aoooa..
--aoooa--
..aoooa..
..xbbbx..
....|....

I count 16 points:

the 9 points marked o are distinct
the 3 points marked a on the left and the right are congruent mod $(2+i)$
the 3 points marked b on the top and bottom are congruent mod $(2+i)
the 1 points marked x are the same


Fastest prime generating algorithm What is the fastest known algorithm that generates all distinct prime numbers less than n?

Is it faster than Sieve of Atkin?

What is the fastest known algorithm that generates all prime numbers?

I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p \le n$ ? Currently it is the Sieve of Atkin.

  And what is the fastest known algorithm that generates any infinite subset of the prime numbers?

Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!

  And is there a theoretical lowest possible O(n) of such programs?

Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p \le n$ , the Sieve of Atkin has time complexity $O(n/\log \log n)$ . So no.

Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.

Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.

Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.

A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.


Prove polynomial is irreducible? Suppose we have a monic polynomial $p(x)\in \mathbb{Z}[x]$ of degree greater than 1 with all even integer coefficients except for the leading term such that the coefficient on the linear term is an odd multiple of 2. We know that the polynomial does not have an integer root. How then can we prove that the polynomial is irreducible over the integers?

So far, I've proven that not having an integer root means that the polynomial does not have a root in the rationals either by using the fact that it's monic and reproving the rational root theorem in this case, but I don't see how to use this information to prove that the polynomial is completely irreducible over the integers.

To clear up some confusion, the claim is:
If $$p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 \in \mathbb{Z}[x],$$
with $2\mid a_0,a_1,\dots,a_{n-1}$, $4 \nmid a_1$ and $p(x)$ has no integer root, then $p(x)$ is irreducible.
Following generalization of this claim is proven in article A mild generalization of Eisenstein’s criterion by Steven H. Weintraub (see copy at https://www.lehigh.edu/~shw2/preprints/eisenstein.pdf):

Theorem. Let $f(x)=a_nx^n+\dots+a_0 \in \mathbb{Z}[x]$ be a polynomial and suppose there is a prime $p$ such that $p$ does not divide $a_n$, $p$ divides $a_i$ for $i=0,\dots,n-1$, and for some $k$ with $0 \leq k \leq n-1$, $p^2$ does not divide $a_k$. Let $k_0$ be the smallest such value of $k$. If $f(x)=g(x)h(x)$, a factorization in $\mathbb{Z}[x]$, then $\min (\deg(g(x)),\deg(h(x))) \leq k_0$.
Proof. Suppose we have a factorization $f(x)=g(x)h(x)$. Let $g(x)$ have degree $d_0$ and $h(x)$ have degree $e_0$. Let $d$ be the smallest power of $x$ whose coefficient in $g(x)$ is not divisible by $p$, and similarly for $e$ and $h(x)$. Then $g(x)=x^dg_1(x)+pg_2(x)$ and $h(x)=x^eh_1(x)+ph_2(x)$ for polynomials $g_1(x),g_2(x),h_1(x),h_2(x)\in \mathbb{Z}[x]$, with the constant terms of $g_1(x)$ and $h_1(x)$ not divisible by $p$. Then
$$
f(x)=g(x)h(x)=x^{d+e}g_1(x)h_1(x)+p(x^eh_1(x)g_2(x)+x^dh_2(x)g_1(x))\\+p^2g_2(x)h_2(x).
$$
The condition that all of the coefficients of $f(x)$ except $a_n$ be divisible by $p$ forces $d+e=n$ and hence $d=d_0$ and $e=e_0$. Thus $g(x)=b_{d_0}x^{d_0}+pg_2(x)$ and $h(x)=c_{e_0}x^{e_0}+ph_2(x)$, in which case
$$
f(x)=g(x)h(x)=a_nx^n+ph_2(x)b_{d_0}x^{d_0}+pg_2(x)c_{e_0}x^{e_0}\\+p^2g_2(x)h_2(x),
$$
and so $k_0\geq \min(d_0,e_0)$.
$\square$

In your case $p=2$ and $k_0=1$ (if smallest $k$ with $p^2\nmid a_k$ was $k=0$, you can use Eisenstein criterion directly, hence we can assume $k_0=1$). Then the statement says that if the polynomial was reducible, it must have at least one of its factors with degree $\leq k_0=1$. It cannot be a degree $0$ since your polynomial is primitive, so it must have degree one factor. Since $p(x)$ is monic, this implies an integer root, a contradiction. So $p(x)$ is irreducible.

HINT.-$$p(x)=x^n+2n_1x^{n-1}+2n_2x^{n-2}_2+\cdots+2(2m+1)x+2a_n\\p(x)=x(x^{n-1}+2n_1x^{n-2}+2n_2x^{n-3}_2+\cdots+2(2m+1))+2a_n$$ Then $$p(x)=xq(x)+2a_n$$ where $q(x)$ is irreducible by Einsenstein so the most general form of $q(x)$ is
$$q(x)=\prod_ j(x-r_j)\prod(x-(s_k+it_k))(x-(s_k-it_k))=\prod_j(x-r_j)\prod_k(x^2-2s_kx+(s_k^2+t_k^2))$$ where each $r_j$ is irrational real.

It follows that if $p(x_0)=0$ then
$$x_0\prod_j(x_0-r_j)\prod_k(x_0^2-2s_kx_0+(s_k^2+t_k^2))+2a_n=0$$
If $a_n=0$ then $p(x)$ has the root $x=0$ which is contrary to the problem ($p(x)$ has no integer root)  so we assume that $a_n\ne0$. But in this case $x_0$ is irrational because $q(x_0)\ne0$. We are done.


Find the unknown area by different methods this problem has many ways of calculating, could provide some solutions, particularly using proportions

EDIT : A it's not midpoint

If we view $AEB$ as having base $AE$ and triangle $FEB$ as have base $FE$, then the both have the same height--- the altitude of line $AF$ to point $B$.

Since triangles $AEB$ and $FEB$ both have the same height then their bases are in proportion to their areas.  So $EF = \frac 32 AE$.

Triangle $DEF$ is similar $AEB$ and the sides are that $EF$ corresponds to $AE$ and so the sides are $\frac 32$ as long.  So the area of the $DEF$ (white) area is $2cm^2 *(\frac 32)^2 = 4.5 cm^2$.

So the area of the white plus the yellow is $4.5cm^2 + 3cm^2 = 7.5 cm^2$.

As this is a rectangle triangles $DCB$ and $DFB$ are congruent and have the same area.

And so the area of the blue plus the green is $7.5 cm^2$.  So the area of the blue is $7.5 cm^2 - 2cm^2 = 5.5 cm^2$

Area of the triangle $ABF=5$.

And $A$ is the midpoint of the side $CB$.

So the area of rectangle $BCDF$= 4 × area  of triangle$ ABF$
$=4×5= 20$.

Thus area of the triangle $BDC$ is half of the area of the rectangle =$10$.

Thus area of the blue reigion= $10-2=8$.


Hint on proving $\sum_{k=0}^n {n \choose k} 2^k (-1)^k = (-1)^n$ I am trying to prove that this series : $$\sum_{k=0}^n {n \choose k}2^k (-1)^k = (-1)^n$$

However, i just honestly dont know where to begin. I tried expanding the LHS out and trying to use the fact that $n\choose k$ = $n \choose n-k$ however the $2^k$ in there is ruining things cancelling out nicely.

When i do a couple of examples on paper when n = 2 and n =3 , i seem to be getting things like:

$n = 2: 1 + (4 - 4) = 1$

$n = 3: 1 + (6 - 12 + 8) = 1 - 2 = -1$

So i assume that $\sum_{k=1}^n {n \choose k} 2^k(-1)^k$ = 0 or -2 depending on even or odd respectively, Then once we add the first term of ${n \choose 0}2^0(-1)^0$ we'll get our 1 or -1

Im looking for a hint on how to approach this.

Thanks.

Hint: The binomial identity states
$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}$$
For any real $a,b$ and nonnegative integer $n$. With $a=2$, $b=-1$, this almost becomes the sum you have, except that the exponent of $b$ is $n-k$ instead of $k$. But this is easily fixed, since the value of $(-1)^m$ only depends on the parity of $m$. Can you fill in the details?

Hint: do a binomial expansion on $(2-1)^n$


Is minimal polynomial of field of rational numbers and field of real numbers are same? Why? Is minimal polynomial of field of  rational numbers and field of real numbers are same?    Why?

Note that over the field $\mathbb{R}$ of the real numbers irreducible polynomials have degree $\leq2$.
On the other hand over the field $\mathbb{Q}$ of rational numbers you can find irreducible polynomials of any degree, meaning that the degree of irreducible polynomials is not bounded. For instance, if $p$ is a prime number the polynomial
$$
1+X+X^2+\cdots+X^{p-1}
$$
is notoriously irreducible.
This answers your questions because irreducible polynomials are the minimal polynomials of their roots.

Depends.
The minimal polynomial of the imaginary unit $i\in\Bbb C$ is $x^2+1$, which is a polynomial over the rationals and the reals.
But the minimal polynomial of $\sqrt 2\in\Bbb R$ is $x^2-2$ over the rationals and $x-\sqrt 2$ over the reals.


Poisson Distribution of sum of two random independent variables $X$, $Y$ $X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.

This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$:
\begin{align*}
  P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\
    &= \sum_{i=0}^k P(Y = k-i , X =i)\\
    &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\
    &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\
   &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\
   &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\
   &= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)}
\end{align*}
Hence,  $X+ Y \sim \mathcal P(\mu + \lambda)$.

Here's a much cleaner solution:

Consider a two Poisson processes occuring with rates $\lambda$ and $\mu$, where a Poisson process of rate $r$ is viewed as the limit of $n$ consecutive Bernoulli trials each with probability $\frac{r}{n}$, as $n\to\infty$.

Then $X$ counts the number of successes in the trials of rate $\lambda$ and $Y$ counts the number of successes in the trials of rate $\mu$, so the total number of successes is the same as if we had each trial succeed with probability $\frac{\lambda + \mu}{n}$, where we take $n$ to be large enough so that the event where the $i$th Bernoulli trial in both processes are successful has a negligible probability.
Then we are done.


Intutively, why does $x^{\frac{1}{\ln x}} = e$? For any $x \gt 0$, we have this odd identity:

$$x^{\frac{1}{\ln x}} = e$$.

You can see this by using the fact that $x = e^{\ln x}$.

I'm wondering if there's a good intuitive explanation for this one, given that $x^{\frac{1}{k}}$ is the operation that inverts raising $x$ to the $k$th power and $\ln x$ is the inverse of the exponential function. Is there some compelling intuitive or geometric argument that makes this identity more obvious than algebraic rearrangement?

The expression $\log x$ describes a number that is an exponent:  specifically, it is the exponent when, to which $e$ is raised, yields $x$.  When we speak of $1/\log x$, this is in some sense an inverse exponent:  for example, if I write $$a^b = c,$$ then $$c^{1/b} = a.$$  So $1/b$ is the exponent to which $c$ is raised to "recover" $a$.  Similarly, then, as $\log x$ is the exponent to which $e$ is raised to yield $x$, we must then have $1/\log x$ as the exponent to which $x$ must be raised to yield $e$.

Of course, all of the above is nothing more than a more prosaic restatement of the algebraic rules that you already wrote in your question.

$$x^{\displaystyle\left(\frac 1{\ln x}\right)}=e\iff \left(x^{\displaystyle\left(\frac 1{\ln x}\right)}\right)^{\ln x}=e^{\ln x}\iff x^1=x^{\ln e}=x^1\equiv{\bf T}$$


Alternative definition of complex number, showing it is equivalent to the tradidional one. The author of a book makes an alternative definition of the complex numbers, later he shows that this definition is equivalent to the ordinary definition where we define $i^2=-1$.

Here is his definition:

Here he shows that they are equivalent:

Now comes my questions:

Has he really shown that they are equivalent. It seems to me that maybe he has shown that starting with the ordered pair definition he gets the tradisional definition of $i^2=-1$. But isn't that only an implication and not an equivalence. Should he have also shown it the other way around, starting with the original definiton, and showed that it implies the ordered pair definition?
Why is it enough to show what he did. If I were asked to show it I would have thought that I must also show that both definitions are equivalent when you add or multiply. Why hasn't he showed that? Or has he showed that implicitly?

They are the same thing, just written in an other way! Complex numbers in the plane can be represented as an ordered pair, and there they are homeomorphic to $R^2$. If in $R^2$ we define the addition and multiplication of two vectors, as in the definition given, than $R^2$ and complex numbers will be algebraically the same, so they would be equivalent.

P.S. the other way of the given proof is trivial: $i=0+1i=(0,1)$.


Need help with proving equivalence relation on Z x Z I need help with a problem from my discrete math course. I'm sure this problem is rather simple but I just can't figure out how to start it. I know I need to prove it is reflexive, symmetric, and transitive but I don't have any similar examples.

Let $\operatorname R$ be a relation on $\mathbb Z \times\mathbb Z$ such that $(a,b)\operatorname R(c,d) \iff a+b^3=c+d^3$.
Prove that $\operatorname R$ is an equivalence relation.

To prove R is an equivalence relation by definition is to prove it is reflexive, symmetric, and transitive

-Reflexivity: $(a,b)R(a,b) \iff a+b^3=a+b^3$ this one is trivial.

-Symmetry:  $(a,b)R(c,d) \implies (c,d)R(a,b)$ this one is also trivial since it's the same equality.

-Transitivity:   $(a,b)R(c,d) \land (c,d)R(e,f) \implies (a,b)R(e,f)$

$$\begin{align}(a,b)R(c,d)\land(c,d)R(e,f) &\iff (a+b^3=c+d^3) \land (c+d^3=e+f^3) \\&\iff a+b^3=e+f^3 \implies (a,b)R(e,f)\end{align}$$

Reflexivity: $\forall (a,b)\in\Bbb Z\times\Bbb Z : (a,b)\operatorname R(a,b)$ which means:

$$\forall (a,b)\in\Bbb Z\times\Bbb Z : a+b^3=a+b^3$$

Q: Does $a+b^3=a+b^3$ for every possible pair?  Are there any possible exceptions?

A: Yes and no (resp.).  So relation $\operatorname R$ is reflexive.

Symmetry: provide your definition.   Does this hold for all two pairs? Are there any exceptions?

Transitivity: repeat the drill.


Why is radian so common in maths? I have learned about the correspondence of radians and degrees so 360° degrees equals $2\pi$ radians. Now we mostly use radians (integrals and so on)

My question: Is it just mathematical convention that radians are much more used in higher maths than degrees or do radians have some intrinsic advantage over degrees?

For me personally it doesn't matter if I write $\cos(360°)$ or $\cos(2\pi)$. Both equals 1, so why bother with two conventions?

The reasons are mostly the same as the fact that we usually use base $e$ exponentiation and logarithm. Radians are simply the natural units for measuring angles.

The length of a circle segment is $x\cdot r$, where $x$ is the measure and $r$ is the radius, instead of $x\cdot r\cdot \pi/180$.
The power series for sine is simply $\sin(x)=\sum_{i=0}^\infty(-1)^i{x}^{2i+1}/(2i+1)!$, not $\sin(x)=\sum_{i=0}^\infty(-1)^i(x\cdot \pi/180)^{2i+1}/(2i+1)!$.
The differential equation $\sin$ (and $\cos$) satisfies is $f+f''=0$, not $f+f''\pi^2/(180)^2=0$.
$\sin'=\cos$, not $\cos\cdot 180/\pi$.

You could add more and more to the list, but I think the point is clear.

The answer is simple, it's a distance measure. When you move in a straight line you use inches or metres, in circles it is radians.
If you are at Disneyland and ask how far it was to Anaheim Stadium [go, Angels!] and I tell you that from my house it's about 45º, you are probably not going to be happy.
You want the distance traveled, at 1 mile out from my house, from one point to another. This distance is pi/4 * 1 mi = about 0.8 miles. Enjoy the game.


If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.

Let
$$f(x,y)= 21x^2 -6xy +29y^2 +6x-58y-151 $$
$$f_x’(x,y)= 42x-6y+6$$
$$f_y’(x,y)= 6x+58y-58$$
The center $(0,1)$ is obtained via $f’_x=f’_y=0$
and by matching the normal vectors at  vertexes, i.e.
$$f’_x:f’_y=(x-0):(y-1)$$
the equations for the major and minor axes are obtained
$$3y-x-3=0,\>\>\>\>\>y+3x-1=0$$
Then, substitute  them into $f(x,y)= 0$ to get the major vertexes
$(\pm\frac9{\sqrt{10}},1\pm\frac3{\sqrt{10}})$ and the minor vertexes
$(\mp\sqrt{\frac35},1\pm3\sqrt{\frac35})$, and in turn their respective lengths $2a=6$ and $2b=2\sqrt6$. Thus, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}= \frac1{\sqrt3}$.

The fact that there is an $xy$ term means that the axes of the conic are rotated from the coordinate axes.  So the first thing I would do is rotate to new $x'y'$ at some angle $\theta$:
$$x= x'\cos(\theta)- y'\sin(\theta), \quad y= x'\sin(\theta)+ y'\cos(\theta)$$
\begin{align}
21x^2&= 21(x'^2\cos^2(\theta)- 2x'y'\sin(\theta)\cos(\theta)+ y'^2\sin^2(\theta))\\
-6xy&= -6(x'^2\cos(\theta)\sin(\theta)+ x'y'(\cos^2(\theta)- \sin^2(\theta))- y'^2\sin(\theta)\cos(\theta))\\
29y^2&= 29(x'^2\sin^2(\theta)+ 2x'y'\sin(\theta)\cos(\theta)+y'^2\cos^2(\theta)\\\\
21x^2- 6xy+ 29y^2&= (21\cos^2(\theta)- 6\cos(\theta)\sin(\theta)+ 29\sin^2(\theta))x'^2\\&\qquad+ (-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta))x'y'\\
&\qquad+ (21\sin^2(\theta)+ 6\sin(\theta)\cos(\theta)+ \cos^2(\theta)y'^2
\end{align}
The point is to eliminate the coefficient of $x'y'$ so we must have $$-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta)= 0$$ or
$$6\sin^2(\theta)+ 69\sin(\theta)\cos(\theta)- 6\cos^2(\theta)= 0$$
We can think of that as the quadratic equation $6X^2+ 69XY- 6Y^2= 0$.  By the quadratic formula $$X= \frac{-69Y\pm\sqrt{4761Y^2+ 144Y^2}}{12}= \frac{-69Y\pm Y\sqrt{4905}}{12}= \frac{1.03}{12}Y.$$
So $\sin(\theta)= \dfrac{1.03}{12}cos(\theta)$ and $\tan(\theta)= \dfrac{1.03}{12}$.


Correct notation for "for all positive real $c$" I am wondering how to notate "for all positive real value $c$"

Is there a correct notation among the following?
$$
\forall c \in \mathbb{R} > 0\\
\forall c \left( \in \mathbb{R} \right) > 0\\
\forall c > 0 \in \mathbb{R}\\
\forall c > 0 \left( \in \mathbb{R} \right)\\
$$

My ultimate goal is notating the following sentence.

"$o(g(n))=\{f(n):$ For any constant positive real value $c$, there is a constant $n_0$ such that $0 \le f(n) \lt cg(n)$ for all $n \ge n_0\}$"

My trial is
$$
o(g(n))=\{f(n):\forall c>0(c\in\mathbb R), \exists n_0\in\mathbb{N} \ \ \ \ s.t.\  \forall n>n_0,\ \ 0 \le f(n) \lt cg(n)\}
$$

I want to correct this part: $\forall c>0(c\in\mathbb R)$

If you would like to have mercy on your reader, please avoid squeezing too many relations together. "$\forall c \in \mathbb{R} > 0$", for example, is readable, but it is not logically precise.

There are at least two ways out; the first one is to say "for all $c \in \mathbb{R}$ such that $c > 0$", and the other is to define the set of all reals $> 0$ and say "for all $c$ in the set ".

You may also use "for all positive $c \in \mathbb{R}$", but this is risky if you do not specify in the first place what your "positive" means; for people may interpret "positive" differently.

In sum, the precise and safe way seems to be "for all $c \in \mathbb{R}$ such that $c > 0$".

Taking advantage of the fact that a variable has a left side and a right side, and also, adding some parenthesis and brackets: can produce a logical and intuitive arguement: $$\forall(c:0<c\in\mathbb R )$$
I would normally just write $\forall(0<c\in \mathbb R)$.


Find $\lim_\limits{x\to\infty} \frac{1+\sqrt2+\sqrt3+...+\sqrt{n}}{n\sqrt{n}} $ Evaluate the limit:

$$\lim_{x\to \infty} \frac{1+\sqrt2+\sqrt3+...+\sqrt{n}}{n\sqrt{n}} $$

Am I right in saying that we can apply Stolz-Cesaro to find limit of $\lim_{x\to\infty} \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1} - n\sqrt{n}}$ which is $\lim_{x\to\infty} \frac{1}{(n+1) + n\sqrt\frac{n}{n+1}}$ = $0$ ?
My bad. $\lim_{x\to\infty} \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1} - n\sqrt{n}}$ =
$\lim_{n \to \infty} \dfrac{\sqrt{n+1}[(n+1)\sqrt{n+1}+n\sqrt n]}{(n+1)^3-n^3}$
= $\lim_{n \to \infty} \dfrac{(n+1)^2+n\sqrt{n(n+1)}}{3n^2+3n+1}$
=$\lim_{n \to \infty} \dfrac{n^2+2n+1 + n^2\sqrt{1+\frac{1}{n}}}{3n^2+3n+1}$
=$\frac{2}{3}$

Yes that's fine, for the last step it is better to add this one also
$$ \dfrac{n^2+2n+1 + n^2\sqrt{1+\frac{1}{n}}}{3n^2+3n+1}= \dfrac{1+\frac2n+\frac1{n^2} + \sqrt{1+\frac{1}{n}}}{3+\frac3n+\frac1{n^2}} \to \frac{1+1}{3}=\frac23$$


How does the dot product convert a matrix into a scalar? I am learning linear algebra, and I am a bit confused by the dot product and how the answer to the process turns out to be a scalar rather than a matrix.

For $2$ vectors with $2$ components, I learned that dot product is equivalent to a $1 \times 2$ row vector left multiplied by a $2 \times 1$ column vector.   The result of such a multiplication should result in a $ 1 \times 1 $ vector.   But I am learning that the dot product somehow transforms the result into a scalar, rather than a $ 1 \times 1 $ vector.

Maybe I am missing something but the difference between a $ 1 \times 1 $ vector and a scalar seems important, because you can multiply a scalar by a matrix of any size, but you can only left-multiply a $ 1 \times 1 $ vector by another matrix with $1$ row.

Thanks for any help in understanding this.

Sometimes, we just say that a $1\times 1$ matrix is the same as a scalar.  Afterall, when it comes to addition and multiplication of $1\times 1$ matrices vs addition and multiplication of scalars, the only difference between something like $\begin{bmatrix}3\end{bmatrix}$ and $3$ is some brackets.  Consider $$(3+5)\cdot 4 = 32 \\ (\begin{bmatrix} 3\end{bmatrix} + \begin{bmatrix} 5\end{bmatrix})\begin{bmatrix} 4\end{bmatrix} = \begin{bmatrix} 32\end{bmatrix}$$ The algebra works out exactly the same.  So sometimes it's not ridiculous to think of $1\times 1$ matrices as just another way of writing scalars.

But if you do want to distinguish the two, then just think of the formula $a\cdot b = a^Tb$ as a way of finding out which scalar you get from the dot product of $a$ and $b$ and not literally the dot product value itself (which should be scalar).  That is, we calculate the dot product of $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4\end{bmatrix}$ by using the formula $$\begin{bmatrix} 1 \\ 2\end{bmatrix}^T\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 1 & 2\end{bmatrix}\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 11\end{bmatrix}$$ and then say that this tells us that the dot product is really $11$.  So the formula $a^Tb$ is just an algorithm we use to find the correct scalar.

You can view it either way.  It doesn't really make a difference.

A dot product is not really a scalar, but it behaves just like one.   In math we call that an ISOMORPHISM.  For every dot product result, there is a corresponding real number that you get by simply removing the brackets.   All of the operations you do with the 1x1 matrix correspond to the same operations done with a real number.  Multipliction of a matrix by a 1x1 matrix is defined as multiplcation by the corresponding scalar.

Recommend you look up "isomorphism" in a math dictionary.   To do this, you may need to brush up on your German language skills.


Linear algebra. Question about symmetric matrix Prove that if

$(Au)\cdot v = u\cdot (Av)$ for every $u$ and $v$

then $A$ is symmetric.

I tried using a $2\times2$, but didn't help.

Look at $ u,v$ being standard unit vectors $e_i, e_j$. Then, what are the quantities in question?


What is negation of the following sentence? What is negation of All birds can fly.

The question seems bit funny but i don't know which of the following two sentences is correct:

Some birds can not fly
There is at least one bird which can not fly.

Both the sentence seems almost logically same. But which of the following is true.

In book correct option out of four is sentence 1.

$B(x):$ x is a bird.

$F(x):$ x can fly.

All birds can fly: $$\forall x(B(x)\rightarrow F(x))$$

Negation of the above $$\lnot \forall x(B(x)\rightarrow F(x))\equiv \exists x \Big(\lnot\big(\lnot B(x) \lor F(x)\big)\Big) \equiv \exists x (B(x) \land \lnot F(x)$$

You can certainly translate the statement as you did, and your translations are equivalent: mathematically, some is equivalent to at least one. So you may as well stick with "some".

An alternative translation of the negation would be simply the negation symbolized at the very left: "Not all birds can fly."

I would say only 2 is the negation of All birds can fly.

All birds can fly is only true if all birds can do it.

So if not all birds can fly it would be a negation.

There is at least one bird which can not fly, is equal to not all birds can fly


How many numbers of 4 digits contains at least one "5"? How many numbers of $4$ digits contains at least one "$5$"?

My suggestion is:

First I count all the numbers of type 5xxx (where x can be a number between 0 and 9). There is 999.

Then I add all the numbers of type y5xx (where y can be 1, 2, 3, 4, 6, 7, 8, 9). There is 8*99

Then I add all the numbers of type yz5x (where z can be 0, 1, 2, 3, 4, 6, 7, 8, 9). There is 8*9*9

Then I add all the numbers of type yzz5. There is 8*9*9

Total: 999+8*99+8*9*9+8*9*9 = 3087

My problem is, I'm not sure if this is correct OR if there exist a smarter method.

HINT: Compute how many 4 digit numbers contain no 5, and subtract that from the total count of 4 digit numbers.

If first digit is 5 the rest can't be five
1*9*9*9

If the second digit is five the rest can't be five and the first can't be zero
8*1*9*9

If the third digit is five the rest can't be five and the first can't be zero
8*9*1*9

If the fourth digit is five the rest can't be five and the first can't be zero
8*9*9*1

So in conclusion,
(1*9*9*9) + (8*1*9*9) + (8*9*1*9) + (8*9*9*1) = 2673


Does the series $\sum_{n=1}^\infty 1/(n+n \cos(n))$ converge or diverge? Does the series $\sum_{n=1}^\infty 1/(n+n \cos(n))$ converge or diverge? How can I use Direct Comparison Test for this problem?

Notice that $\sum_{n=1}^{\infty}\frac{1}{n+ncos(n)}≥ \sum_{n=1}^{\infty}\frac{1}{n+n}$ (Why?)

Now:

  $\sum_{n=1}^{\infty}\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}$ diverges (Why?)

$a_n=\dfrac{1}{n+n \cos{n} }$ and let $b_n=\dfrac{1}{n}$

$\lim_{n \to \infty}\dfrac{a_n}{b_n}=\dfrac{1}{1+cosn}=finite $

$\sum_n a_n $ and $\sum_n b_n$ converge or diverge together.$\sum b_n$ is a divergent or convergent series ?


Can a bivariate polynomial with only algebraic coefficients have monic polynomial factors with non-algebraic coefficients? Can a bivariate polynomial with only algebraic coefficients have monic polynomial factors with non-algebraic coefficients?

Are there literature references with lists of bivariate polynomials that are irreducible over the algebraic numbers?

I already know that a bivariate polynomial can have solutions $(z_1,z_2)\in\mathbb{C}^2$.

The quick answer is no:

Anytime you factor a polynomial of one or several variables $f \in K[x_1, \ldots, x_n]$
you can rescale the factors so that the coefficient of each factor is algebraic over the
field generated by the coefficients of $f$.  The reason is that each polynomial can be factored in (essentially) finitely many ways into two factors.


Solving $v_{t}+v(x,t)v_{x}=0$ with initial condition This problem comes from an undergraduate course in PDE.

The first question of the problem was to solve the following PDE: $v_{t}+v(x,t)v_{x}=0$ with the following initial condition:
$v(x,0)=5x$ where $v(x,t)$ is an unknown function of the two independent variables: $x$, and $t$. $v_{t}$ and $v_{x}$ refer to the partial derivatives of $v(x,t)$ with respect to $t$ and $x$ respectively.

The second question of the problem given the PDE above, but with the following initial condition: $$v(x,0)= 5\ \ if\ \ -3<x<3$$$$v(x,0)=0\  otherwise$$

The question was to find the time $S$ when the rarefaction wave catches up with the shock wave, and then to find $v(x,t)$ for the case:$0<t<S$

I solved the first question of the problem, and I found $v(x,t)=\frac{5x}{1+5t}$.
However, I am completely stuck when it comes to solving the second question. I have no idea how to find that time $S$, and this is the first I see a problem where the initial condition is defined piecewise. Any help is appreciated!

You used Method of Characteristics for the first question to get $v = \phi(x-vt)$.

However, you can't use Method of Characteristics for the second question. For the characteristics will cross each other. Consider initially at $x=3^-$, the velocity of the wave is $v=5$, however at $x=3^+$, the velocity is $0$. Physically the wave front is propagating faster, thus creating a shock.

By talking rarefaction and shock wave, we are referring to the weak solution. Rewriting the equation as
$$
v_t + f(v)_x = 0, \quad \text{where }f(v) = \frac{v^2}{2}.
$$
At $t=0^+$, any $x = x_0$, when
$$v_l:= \lim\limits_{x\to x_0^-} v(x,0) > v_r:= \lim\limits_{x\to x_0^+} v(x,0),$$
we get a shock wave. This is happening on the right end of the interval $(-3,3)$. Using Rankine–Hugoniot formula, equation (10) on the Wikipedia entry, we have the shock wave is propagating at a constant speed of
$$
v_s = \frac{v_l^2/2 - v_r^2/2}{v_l - v_r} = \frac{v_l+v_r}{2} = 5/2
$$
moving to the right starting from $x=3$. And the weak solution is:
$$
v(x,t) = \begin{cases}
v_l = 5 & \text{when } x -3 < v_s t,
\\
v_r = 0 & \text{when } x -3 \geq  v_s t.
\end{cases}\tag{1}
$$

When $v_l<v_r$, you get a rarefaction wave (this is obtained by vanishing viscosity method). This is on the left end of the interval $(-3,3)$. Using the formula here (in the example of Korteweg-de Vries Equation, it starts from $x=0$, we modify $x=-3$ as the starting point), we have on the left end the rarefaction wave is :
$$
v(x,t) = \begin{cases}
0 & \text{when } x +3 <0,
\\
(x+3)/t & \text{when } 0\leq x +3 \leq 5t,
\\
5 & \text{when }   x +3 > 5t.
\end{cases}\tag{2}
$$
This is the vanishing viscosity solution, which is obtained by solving $v_t + vv_x = \epsilon v_{xx}$, then letting $\epsilon \to 0$, you will get a continuous solution. This tells us that the rarefaction wave is propagating to the right at a constant speed of $5$.

Eventually, combining the shock wave (1) on the left end with the rarefaction wave (2) on the right end will give you the answer: before the rarefaction wave catches up the shock wave, $0<t<S$, the solution is:
$$
v(x,t) = \begin{cases}
0 & \text{when } x +3 <0,
\\
(x+3)/t & \text{when } 0\leq x +3 \leq 5t,
\\
5 & \text{when }   x +3 > 5t \;\text{ and }\; x-3 < 5t/2,
\\
0 & \text{when } x-3 \geq  5t/2.
\end{cases}
$$
So now computing when the rarefaction wave (propagating at a speed of $5$ from $x=-3$) will catch up with the shock wave (propagating at a speed of $5/2$ from $x=3$) is not a difficult task.

Some update: to answer doraemonpaul's concern, the solution above is more of a physical solution than mathematical, $v(x,t)$'s behavior when $t\to 0^+$ and $x\to -3^+$ can be viewed like a line with a slope $1/t$ approaching infinity to describe the discontinuity at $t=0, x=-3$. Notice on the left end, $x+3\to 5t$ from the right (spacial and temporal), we have:
$$
\lim_{\substack{x\to -3^+\\ t\to 0^+}} v(x,t) =
\lim_{\substack{x\to -3^+\\ t\to 0^+}} \frac{x+3}{t} = \lim_{\substack{x\to -3^+\\ t\to 0^+}} \frac{5t}{t} = 5, \;\text{ and }\; \lim_{\substack{x\to -3^-\\ t\to 0^+}} v(x,t)  = 0.
$$
For reference, please view formula (22.31) and explanation here.

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dx}{ds}=v=v_0$ , letting $x(0)=f(v_0)$ , we have $x=v_0s+f(v_0)=vt+f(v)$ , i.e. $u=F(x-vt)$

$v(x,0)=\begin{cases}5&\text{when}-3<x<3\\0&\text{when}~x\leq-3~\text{or}~x\geq3\end{cases}$ :

$\therefore v=\begin{cases}5&\text{when}-3<x-vt<3\\0&\text{when}~x-vt\leq-3~\text{or}~x-vt\geq3\end{cases}=\begin{cases}5&\text{when}-3<x-5t<3\\0&\text{when}~x\leq-3~\text{or}~x\geq3\end{cases}=\begin{cases}5&\text{when}~5t-3<x<5t+3\\0&\text{when}~x\leq-3~\text{or}~x\geq3\end{cases}$

Hence $v(x,t)=\begin{cases}5&\text{when}~5t-3<x<5t+3\\0&\text{when}~x\leq-3~\text{or}~x\geq3\\c&\text{otherwise}\end{cases}$


Why can't we remove the sqrt from rms? In chemistry, we define the root-mean-square speed as

$\sqrt{\bar{u^2}}$ = $\sqrt{\frac{3\text{RT}}{\text{M}}}$

A student asked me why we can't just remove the square root symbol. And aside from "because this is how we define it", I didn't actually have a reason.

So, I'm hoping someone can shed some light on why the above equation is used and not:

$\bar{u^2} = \frac{3\text{RT}}{\text{M}}$

In case it is important, we use this equation to determine the rms speed of a gas. It depends on the temperature (T) and the molecular mass of the gas (M). R is a constant value. I understand we don't just use the average because in a set of gases, they move in a random direction so the average is 0. But, by squaring isn't that issue resolved, without the square root?

$T$ has units of Kelvin (K).  The gas constant $R$ has units of Joule/mole/K.  The molecular mass $M$ has units of kg/mole.  Also remember that a Joule is $\mathrm{kg.m^2/s^2}$.  So the units of $3RT/M$ are
$$\mathrm{\frac{kg\,m^2\,K}{s^2\,mole\,K\,kg\,mole^{-1}}=\frac{m^2}{s^2}}$$
which is a velocity squared.  So taking the square root gives the correct units for a velocity.

Because if you remove 'Square Root', its unit will no longer be speed ( in this case ) i.e ${m^2}$/$sec^2$

This term  $\bar{u^2}$ represents 'Mean of Squared Component', aka 'ms'  from 'rms'. Hence to get the 'root mean square' value, you need to take square of 'ms'.


How to derive the 3D equation of a torus? I'm doing a presentation on 3D surfaces for college and one of the equations I am using is a Torus. I know that the equation is

$$z^2 = 25 - \left(10 - \sqrt{x^2 + y^2}\right)^2$$

For a torus with radius 5, with the inner circle 10 units away from the z axis but I was wondering how this is derived from a circle of the form

$$x^2 + y^2 = 15^2$$

Being the centre for the other circles in the $oxz$ and $oxy$ planes (15 being the distance from the $z$ axis + the radius.

Thanks, Any help will be appreciated

Let's select one point $p = (x, y, 0)$. Distance from point p to origin is $\sqrt{x^2 + y^2}$ Distance from point $p$ to the circle with radius $r$ in $xy$-plane is $d = \lvert r - \sqrt{x^2 + y^2} \rvert$. Point $(x, y, z)$ is on the surface of torus if it satisfies $d^2 + z^2 = h^2$ ie. $z^2 = h^2 - \left(r - \sqrt{x^2 + y^2} \right)^2$

First of all, to understand how the equation for a torus relates to that of a circle you must become familiar with the object: $\sqrt{x^2+y^2}$. As you can probably guess, this takes any point $p$ on the $x$-$y$ plane and generates $p$'s distance from the origin. The useful aspect of this tool is that it treats any point within the same distance of the origin as the same. Therefore, in order to create objects that are rotationally symmetric about the $z$-axis, it is essential to treat $z$ as a function of radius $r$ and then later substitute $r$ with $\sqrt{x^2+y^2}$.
Since we know that the torus is rotationally symmetric, we can simplify it by finding the equation to its cross section (by any origin intersecting plane perpendicular to the $x$-$y$ plane). We then get the graph of a circle translated horizontally.
Keep in mind that the horizontal axis is radius $r$ and the vertical axis is $z$. Now you may now see how the equation for a circle arises-- it's just the graph of $\pm\sqrt{a^2-(r-c)^2}=z$. Variable c is the length from the origin to the center of the circle while a is the radius of the circle.
Now we can expand this into three dimension by replacing the r with $\sqrt{x^2+y^2}$.
By then you have two parameters
$a$ = radius of the "tubing"
$c$ = distance from the center of the torus to the middle of the "tubing"
$z=\pm\sqrt{a^2-(\sqrt{x^2+y^2}-r)^2}$


Maximum sum of the number of saturdays and sundays in a leap year Question is to find maximum sum of the number of Saturdays and Sundays in a leap year...

I do not have much to show  off but I guess in a leap year there would be maximum of $53$ Saturdays/$53$ Sundays but i am not sure if it is possible to have $53$ Saturday and $53$ Saturdays..

So, I guess maximum sum would be $105$.

Could some one tell me if this is correct

I would like to learn something more of this kind and i would be so thankful if some one wants to say something more than this.

A regular year has exactly $52$ full weeks, and $1$ day. A leap year has $52$ full weeks, and $2$ days. Were such a year to start on a Saturday, then it would have exactly $53$ full weekends.


Is it faster to count to the infinite going one by one or two by two? A child asked me this question yesterday:

  Would it be faster to count to the infinite going one by one or two by two?

And I was split with two answers:

In both case it will take an infinite time.
Skipping half of the number should be really faster.

Which brings me this question:

Could an infinite be greater than another one?

YES and NO.

Galileo made the "discovery", the so-called Galileo's paradox, that if you start with the succession of natural numbers:

  $1, 2, 3, ...$

and you map it into the succession of even numbers :

  $2, 4, 6, ...$

you may map (i.e.associate) every number into its double (today, we call it one-to-one mapping).

So, you have the same "number" of numbers and of even numbers.

Modern set theory (from Cantor on) solved the paradox extending the "counting" process to infinite sets, but proving that the euclidean common notion that "The whole is greater than the part" [see Euclid, The Elements, trans T.L.Heath, Dover, Common notions 5] will not hold for an infinite set.

According to modern set theory, the two above sets can be put in one-to-one correspondence, so they have the same cardinal number, and their "type" of infinity is called denumerable (a set is called denumerable exactly when it can be put in one-to-one correspondence with the set of natural numbers).

But, again by a result of Cantor, not all infinite sets can be put in one-to-one correspondence: there are infinite sets "more infinite" than another. A set of "a larger kind" of infinity is the set of real numbers; it is not denumerable, in the meaning defined above.

As for counting "faster": of course, if you count two by two, after a while you will be way "ahead" of your friend that is counting by one. The only problem is that you will not "end" before him, because there is no final goal to be reached!

You can form a bijection from the set of all positive integers to a set of all even integers. If infinity were a quantity, then infinity would map to infinity. Mathematically then you'd reach infinity at the same time.


Natural Deduction stuck $$\vdash (( p → q) →  p) →  p$$ how would I go about solving this with no premises. to get the implication i know i need $$(p → q) →  p$$ as my first assumption but im stuck to where to go from there

First hint: Prove ($\lnot$p$\rightarrow$(p$\rightarrow$q)) as a lemma.

More detailed analysis:

What is the goal?

The goal is to produce to (((p $\rightarrow$ q) $\rightarrow$ p) $\rightarrow$ p).

What evidence do you have so far?

You have evidence that if you assume ((p$\rightarrow$q)$\rightarrow$p) and produce p, that a solution can get constructed.

If you assume something, what is the structure of that assumption?

The assumption ((p$\rightarrow$q)$\rightarrow$p), has the structure such that (p$\rightarrow$q) is the antecedent and p is the consequent.  Thus, if you can produce (p$\rightarrow$q), by detachment ("modus ponens"), you can deduce p.

Can you assume anything and get rid of that assumption?

Suppose you assume ((p$\rightarrow$q)$\rightarrow$p).  You want 'p'.  But, let's say your enemy says that obtaining 'p' is impossible.  So, let's assume your enemy correct, and thus assume that $\lnot$p makes sense.  Can you then obtain 'p' under that assumption, with the analysis of 3. in mind?

I just did it with a truth table. Set p to each of 0 and 1 (false and true), and set q to each of 0 and 1 as well. That gives you 4 combinations of the two. Then, work your way from the inside out.

  Case 1: $p = 0$, $q = 0$

$p→q = 1$

$1→p=0$

$0→p=1$

  Case 2: $p = 0$, $q = 1$

$p→q = 1$

$1→p=0$

$0→p=1$

  Case 3: $p = 1$, $q = 0$

$p→q = 0$

$0→p=1$

$1→p=1$

  Case 4: $p = 1$, $q = 1$

$p→q = 1$

$1→p=1$

$1→p=1$

Therefore the statement holds true for every case.


Branch Cuts of $f(z) + g(z) = \sqrt{p(z)} + \sqrt{q(z)}$ How does one find the branches of

$$f(z) + g(z) = \sqrt{p(z)} + \sqrt{q(z)}$$

where $p$ & $q$ are second degree polynomials?

It would be very nice to see this general method applied to, say,

$$f(z) + g(z) = \sqrt{2z^2 + 3z - 1} + \sqrt{z^2 + 4z - 2}$$

or something better that's nice though similar.

This is apparently an impossible problem, evading just about everybody in our class, though it looks so simple.

Each quadratic has two zeros.  A branch cut starts and ends at these zeros.
Take one path between the zeros of the first quadratic, and another path between the zeros of the second quadratic.
You can have curves instead of straight lines to make them avoid each other.
A path that avoids these cuts can't go around one zero without going around the quadratic's other zero.  Both square-roots change by a factor of $e^{i\pi}$ as the path goes around both zeros.  Their product changes by a full $e^{2i\pi}$.  That is why the square-root of the quadratic is well-defined on the cut plane.


Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.

I would start by using the substitution $u=\frac{1-x}{1+x}$. This is a handy substitution because it is its own inverse:

$$u=\frac{1-x}{1+x}\implies x=\frac{1-u}{1+u},~\mathrm{d}x=\frac{-2}{(1+u)^2}\,\mathrm{d}u.$$

Then,

$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=\int\left(\frac{1+u}{1-u}\right)\sqrt{u}\frac{(-2)}{(1+u)^2}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{(1-u)(1+u)}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u.\\
\end{align}$$

Another substitution such as $u=t^2$ would transform this to an integral of a rational function:

$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u\\
&=-2\int\frac{t}{1-t^4}\left(2t\,\mathrm{d}t\right)\\
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t.\\
\end{align}$$

From there, partial fraction decomposition will break the integral down into elementary integrals, and back-substitution will yield the desired anti-derivative. PFD gives us,

$$-\frac{4t^2}{1-t^4}=\frac{2}{1+t^2}-\frac{1}{1+t}-\frac{1}{1-t},$$

so,

$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t\\
&=2\int\frac{\mathrm{d}t}{1+t^2}-\int\frac{\mathrm{d}t}{1+t}-\int\frac{\mathrm{d}t}{1-t}\\
&=2\tan^{-1}{\left(t\right)}-\ln{\left(1+t\right)}+\ln{\left(1-t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(t\right)}-2\tanh^{-1}{\left(t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{u}\right)}-2\tanh^{-1}{\left(\sqrt{u}\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}-2\tanh^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}+\color{grey}{constant}.~\blacksquare\\
\end{align}$$

Solution

$$\text{substitute x with tan }\theta \text{ as pointed out by abel}$$

Now

$$\int\frac{1}{\tan \theta}\sqrt\frac{1 -\tan\theta}{1+\tan \theta}\ dx = \int \cot \sqrt{\frac{(1-\tan \theta)^2}{(1+\tan\theta) (1 -\tan\theta)}}\sec^2\theta
$$

Then

$$ \int \cot\sec^2 \frac{1-\tan\theta}{\sec\theta} = \int{\cot\sec (1-\tan\theta}) = \int\frac{\sec (1-\tan\theta)}{\tan\theta} = \int\frac{\sec\theta}{\tan\theta} -1$$

Now you can find integral of this.


Conjugate and dual representation Came by from this link https://en.wikipedia.org/wiki/Complex_conjugate_representation

I am wondering if $\Pi$ a representation, then $\bar{\Pi}$ and dual of $\Pi$ are isomorphic representations.

I can see $\Pi$ and $\bar{\Pi}$ is naturally isomorphic as vector space, hence $\bar{\Pi}$ and the dual is isomorphic as vector space. However, it seems like the natural isomorphism is not a representation isomorphism. How do we deduce this?

Let $G$ be a group, let $\Pi$ be a representation of $G$ on a finite-dimensional $\mathbf{C}$-vector space $V$. We will write simply $gv=\Pi(g)v$ for the action of $G$ on $V$ via $\Pi$. Suppose $V$ is equipped with a non-degenerate Hermitian form $(\cdot,\cdot)$. That is, $(v_1,v_2) \in \mathbf{C}$ is linear in $v_2$, satisfies $(v_2,v_1)=\overline{(v_1,v_2)}$, and for each $v_1 \in V$ there is some $v_2 \in V$ with $(v_1,v_2) \neq 0$. If in addition the form is compatible with the action of $G$ on $V$ in the sense that
$$(g v_1,g v_2)=(v_1,v_2) \quad \hbox{for all $g \in G$ and $v_1,v_2 \in V$}$$ then the map $v \mapsto (v,\cdot)$ defines a $G$-equivariant conjugate-linear isomorphism of $V$ onto the dual space $V^*$, or in other words an isomorphism from the conjugate of $\Pi$ to the dual of $\Pi$.

Here is one situation in which such a form is guaranteed to exist: if $G$ is a finite group (or more generally a compact topological group, where sums are replaced by integrals with respect to a Haar measure) then you can start with any positive-definite Hermitian form $(\cdot,\cdot)'$ on $V$ and produce a $G$-invariant positive definite Hermitian form by averaging over $G$,
$$(v_1,v_2):= \sum_{g \in G} (g v_1,gv_2)'.$$ Here we use that a sum of positive-definite forms is positive definite and hence non-degenerate.

The dual representation of $\rho$ is generally not isomorphic to $\rho$. For instance, for each integer $n$ the group of complex numbers of absolute value $1$ has a one-dimensional representation $\mathrm e^{\mathrm i\phi}\to\mathrm e^{\mathrm in\phi}$, whose dual is the representation $\mathrm e^{\mathrm i\phi}\to\mathrm e^{-\mathrm in\phi}$. These two representations are not isomorphic unless $n=0$.


Picking two random points on a disk I try to solve the following:

Pick two arbitrary points $M$ and $N$ independently on a disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2 \leq 1\}$ that is unformily inside. Let $P$ be the distance between those points $P=d(M,N)$. What is the probabilty of $P$ being smaller than the radius of the disk?

Picking a specific point $s$ as the first point $M$ on a disk has probabilty $P(M=s)=0$ since we are looking at the real numbers. My intention, that is probably wrong, is that chosing the second point $N$ such that the distance between them is bigger than the radius of the disk, is as likely as being smaller.

I have absolutely no idea how to solve this.

Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is
$$P(R\le r) = \begin{cases}
\frac{\pi r^2}{\pi\cdot 1^2} = r^2, & 0\le r\le 1\\
1, &r>1\\
0, &\text{otherwise}
\end{cases}$$
The probability density function of $R$ is
$$f_R(r) = \frac{d}{dr}P(R\le r) = \begin{cases}
2r, & 0< r< 1\\
0, &\text{otherwise}
\end{cases}$$

If $OM = d$, $0\le d\le 1$, then the area of the intersection of the unit circles centred at $O$ and $M$ respectively is

$$\begin{align*}
A(d) &= 4\left(\frac12\cos^{-1}\frac d2\right) - 2\left(\frac12d\sqrt{1^2-\frac {d^2}{2^2}}\right)\\
&= 2\cos^{-1}\frac d2 - \frac 12 d\sqrt{2^2-d^2}\\
\frac{A(d)}{\pi\cdot 1^2}&=\frac{2}{\pi}\cos^{-1}\frac d2 - \frac{1}{2\pi} d\sqrt{2^2-d^2}
\end{align*}$$
The last line is the probability that $N$ is within a unit distance of $M$, as a function of distance $d = OM$.

Combining with the probability density function $f_R$ above, the required probability is
$$p = \int_0^1\left(\frac{2}{\pi}\cos^{-1}\frac r2 - \frac{1}{2\pi} r\sqrt{2^2-r^2}\right)\cdot 2r\ dr \approx 0.58650$$

(WolframAlpha)

$$\begin{align*}
I_1 &= \int_0^1 r\cos^{-1}\frac r2\ dr\\
&= \int_{\pi/2}^{\pi/3} 2\theta\cos\theta\cdot(-2)\sin\theta \ d\theta && (r = 2\cos \theta)\\
&= -2\int_{\pi/2}^{\pi/3}\theta\sin 2\theta \ d\theta\\
&= \int_{\pi/2}^{\pi/3}\theta\ d\cos2\theta\\
&= \left[\theta\cos2\theta\right]_{\pi/2}^{\pi/3} - \int_{\pi/2}^{\pi/3}\cos2\theta\ d\theta\\
&= \left[\theta\cos2\theta - \frac 12 \sin2\theta\right]_{\pi/2}^{\pi/3}\\
&= -\frac{\pi}{6} - \frac{\sqrt3}{4} +\frac\pi2\\
&= \frac\pi 3 - \frac{\sqrt3}{4}\\
\frac4\pi I_1 &= \frac 43 - \frac{\sqrt3}{\pi}
\end{align*}$$

$$\begin{align*}
I_2 &= \int_0^1 r^2\sqrt{2^2-r^2}\ dr\\
&= -16\int_{\pi/2}^{\pi/3} \cos^2\theta\sin^2\theta\ d\theta&&(r=2\cos \theta)\\
&= -4\int_{\pi/2}^{\pi/3} \sin^2 2\theta\ d\theta\\
&= -4\int_{\pi/2}^{\pi/3} \frac{1-\cos4\theta}{2}\ d\theta\\
&= -2\left[\theta-\frac{1}{4}\sin4\theta\right]_{\pi/2}^{\pi/3}\\
&= -2\left(\frac\pi3+\frac{\sqrt3}8-\frac\pi2\right)\\
&= \frac\pi3-\frac{\sqrt3}{4}\\
\frac1\pi I_2 &= \frac13-\frac{\sqrt3}{4\pi}
\end{align*}$$

$$p = 1-\frac{3\sqrt3}{4\pi}$$

Unless you are given a specific sample space, the answer is 1/2, as there are uncountably many pairs of points distance r apart and uncountably many not. Without a sample space, the addition rule for probabilities no longer applies, so you have a 1/2 probability of the distance being each of less than, equal to, and greater than r.


Recurrence relation problem If $a$ is a sequence defined recursively by $a_{n+1} = \frac{a_n-1}{a_n+1}$ and $a_1=1389$ then can you find what $a_{2000}$ and $a_{2001}$ are?

it would be really appreciated if you could give me some pointers on how to solve this mathematically.

(Obviously using the formula to find the values one by one is possible but is extremely tedious, actually i wrote a simple piece of software to calculate it and the answers were a little odd, $a_{2000}=-1.0014409$ and $a_{2001}=1389.0381$)

By the way here is the code i used for calculating the answers (it's in java) :

        float answer = 1389;

        for (int i = 2; i <= 2001; i++) {

            answer = (answer - 1) / (answer + 1);

        }

(i ran it two times the first time i set the loop condition to 2000.)

The fixed points of the recursion $$a\to\frac{a-1}{a+1},$$ are $\mathrm i$ and $-\mathrm i$. Even if the whole sequence $(a_n)$ is real valued, this suggests to look at the dynamics of the modified (complex) variable $$b_n=\frac{a_n-\mathrm i}{a_n+\mathrm i}.$$ Behold! It happens that $$b_{n+1}=-\mathrm i\,b_n,$$ hence $b_{n+1}=(-\mathrm i)^nb_1$ for every $n$. Furthermore, $(-\mathrm i)^4=1$ hence the sequences $(a_n)$ and $(b_n)$ have period $4$.

In particular, $a_{2001}=a_1=1389$. It remains to note that $$\frac{a_4-1}{a_4+1}=a_5=a_1,$$ to deduce $$a_{2000}=a_4=-\frac{a_1+1}{a_1-1}=-\frac{695}{694}.$$

Edit: Likewise, the recursion $$a\to\frac{a-u}{a+v},$$ is periodic with period $n$ for every starting point if and only if there exists some integer $k$ such that $$\cos^2\left(\frac{k\pi}n\right)=\frac14\frac{(1+v)^2}{u+v}.$$ If $u=v=1$, $n=4$ and $k=1$ solve this.

This is a simple attempt using basic algebra.


Cauchy-Riemann Equations and differentiability at a single point I am trying to figure out when can we conclude a complex function $f$ is differentiable at a point (and only that point). For example the function $|z|^2$ is differentiable only at $z=0$. However, I am not sure if I can use CR. The version we went over in class states that $f$ is differentiable on an open set $O$ if for each $z \in O$ we know C-R are satisfied and that the derivatives are continuous on $O$. However, a single point is clearly not an open set, so I do not know if C-R are satisfied at $z_0$, and $u,v$ are continuously differentiable at $z_0$, then f is differentiable at $z_0$.

$f(z)=|z|^2=x^2+y^2$ doesn't satisfy the C-R equations in a nbhd of $0$.
Thus you need a different way to conclude that $f$ is differentiable there.

See "Complex Differentiable -- from Wolfram MathWorld" http://mathworld.wolfram.com/ComplexDifferentiable.html


How to round 0.4999... ? Is it 0 or 1? If you want to round the repeating decimal 0.4999... to a whole number what is the right answer? Is it 0 or 1?

On the one hand you only look at the first digit when you round numbers which in this case is 4 so the answer should be 0.

On the other hand 0.4999... is only another representation for 0.5 which makes the result 1.

My question
Which of these rules wins? (My gut feeling is that the most consistent result should be 1)

Edit
Just for clarification: What is meant by rounding here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method.

It's $1$, because $0.49\ldots$ is the same as $0.5$. If rounding is to be well-defined, it can't map one real number to two integers, so whatever it maps $0.49\ldots$ to, it better maps it to the same integer as $0.5$. You could round both to $0$, of course, but that wouldn't then be the way we usually round.

What this shows you is that rounding doesn't commute with limits, i.e. there's a difference between find the limit of a sequence and then rounding, and rounding first and then finding the limit. As you correctly observed, all values $a_n = 0.4\underbrace{9\ldots 9}_{n\text{ times}}$ are rounded down to zero. Thus, $$
  \lim_{n\to\infty} \text{round }(a_n) = 0
$$
On the other hand, $\lim_{n\to\infty} a_n = 0.5$, and thus $$
  \text{round }\left(\lim_{n\to\infty} a_n\right) = 1
$$

There's another word for functions which don't commute with limit - they're called non-continuous. So what you have discovered is simply that rounding is not a continuous function.

If we round c to d, then c approximately equals d.  If d also consists of an integer, then c lies closer to d than any other integer i.  In other words for all integers i such that i does not equal d, ABS(ci-)>ABS(cd-), where ABS(ci-) indicates the number by which c and i differ by (the absolute value of the difference of c an i).  This implies that we can use the distance metric between real numbers to see how rounding works as follows.

Consider the set S=={[l, lu+2/), (lu+2/, u]} where ul-=1.  In other words, l and u differ by 1, and we consider the set of all numbers between x and y except for the midpoint m==lu+2/ between them.  We can use the metric function D(x,y)=|xy-|=ABS(xy-), where x belongs to S and y belongs to {l, u}, to determine how to round all members of S by following rule:

If D(x, l)>D(x, u), then round to u.
If D(x, u)>D(x, l), then round to l.

Since .5=.49999..., it follows that D(.49999..., 1)=D(.49999...., 0)=.4999...=.5.  Thus, the distance function D(x, y)=ABS(xy-) does not give us any guide as to how to round here.

Consequently, we have several different choices that we can make which will not go against our intuition of rounding as follows:

Choose to round .49999... to 1 (maybe you value larger numbers in general),
Choose to round .49999... to 0 (maybe you value smaller numbers in general),
Choose to allow .49999... to get rounded to both as one pleases or sees fit at particular points in time, or
Choose to round .49999... to neither 0 nor 1 (maybe you don't want to make a choice here).

None of those possibilities win or lose until we have a standard to measure a winner or loser for the rules.  There also doesn't exist a "right answer" here unless we have a means to determine what properties a "right answer" has in this context.  I suspect that many people would resist 3. with the possibility that one could round .4999... to 1 at one time, and then round .4999... to 0 at another as one pleases, thus indicating arbitrariness.  But, it does not follow that such comes as an improper procedure unless one first rates such "arbitrariness" as bad, and something so utterly awful and terrible that we or others must never, ever engage in such "arbitrariness".  Perhaps many people would see 4. as "evasive", but unless there exists a logical basis to picking either 0 or 1, anyone accused of "evasiveness" by picking 4. could very well respond to such a charge, that anyone who sees choosing 4. as indicating "evasiveness", insists dogmatically that all such questions absolutely must have answers, and/or that the people claiming "evasiveness" have copped out of making choices on the basis of logic as much as possible.

So, what sort of choice do you want to make?


How to work out the angle of a line passing through a plane I have a triangular plane composed of three points. From this it it easy to deduce that the plane is in fact composed of two vectors which must touch at some point. because all of this is relative, one of the points can be taken as being at reference (0,0,0). How do I calculate the equation for a line that passes through the plane which is perpendicular to it? Its actual location is of no great importance as the angle is more use - it will be used to show how 'bright' the plane should be due to facing light.

My maths isn't great so an answer in plain English would be appreciated!

Hint: you have two vectors (in the plane), call them $\vec a$ and $\vec b$. Then their cross product $\vec p$ will be a vector perpendicular to both of them, which is also therefore perpendicular to the plane:
$$\vec p = \vec a \times \vec b$$

Why not find the x direction, y direction and z direction of the plane? Then, rotate it so that the plane is, say, the plane at y=0. You're then going to find the 'y-angle' of the vector hitting the plane. Do the same for x and z. This way probably isn't particularly elegant, but it's simple.


Possibility of representing the Möbius strip by ways other than specifying a parameterization? I have looked up some texts, they all represent the Möbius strip by ways of specifying a parameterization. This is the only way how the Möbius strip is represented founded in the Wikipedia currently.

Question 1:

Can the Möbius strip be represented as the graph of a map?

For example, a map $\phi$ from a subset $K$ of $\mathbb{R}^2$ to $\mathbb{R}$ is specified, and the Möbius strip is exactly of the form $\{(V,\phi(V)) : V$ $\in$ $K \subseteq \mathbb{R}^2 \}$, which is called the graph of $\phi$.

Question 2:

Can the Möbius strip be represented as the locus defined by constraint equations? (With restrictions on allowable range of the variables.)

For example, an equation $f(X)=0$ is specified, where $X$ is in $\mathbb{R}^3$, $f$ map $\mathbb{R}^3$ to $\mathbb{R}$, and the Möbius strip is exactly of the form $\{X \in \mathbb{R}^3 : f(X)=0 \}$, which is called the locus defined by the constraint equation $f(X)=0$.

Neither kind of representation is possible, because the Möbius strip is not orientable.

Suppose $M$ is a regular level set of a smooth function $F: U \to \mathbb R$, where $U$ is an open subset of $\mathbb R^3$.  (A regular level set in this context is a level set on which the gradient of $F$ does not vanish.) Then $\operatorname{grad} f$ is a nowhere-vanishing normal vector field along $M$, so $M$ must be orientable.

Similarly, suppose $M$ is the graph of a smooth function $f: V\to \mathbb R$, where $V$ is an open subset of $\mathbb R^2$.  Then $M$ is a regular level set of the function $F(x,y,z)= z-f(x,y)$, so again $M$ must be orientable.

Since the Möbius strip does not live in $\mathbb{R}^2$ the first is not possible.
To have it as a function $f(x,y,z)=0$see:

$$-R^2y+x^2y+y^3-2Rxz-2x^2z-2y^2z+yz^2=0.$$

I copied it from http://mathworld.wolfram.com/MoebiusStrip.html
trula


$\gcd(a,b,c)=\gcd(\gcd(a,b),c)\,$ [GCD Associative Law] Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.

Hint $ $ By $\,\color{#c00}{\rm U}= $ gcd Universal Property (see below) we have
$$d\mid(a,b,c)\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid a,b,c\!\!\color{#c00}{\overset{\rm\ U}\iff} d\,\mid (a,b),c\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid ((a,b),c)\qquad$$
So $\,(a,b,c)\,$ & $\,((a,b),c)\,$ divide each other (let $\,d\,$ equal them above) so they're equal, being $> 0$.
This is the associative property of the GCD.  In the same way, by induction, we can erase (or normalize) brackets in $n$-argument gcds,  showing general associativity of the gcd, e.g. see here.
Remark $\ $ For completeness, below is a proof of $\,\color{#c00}{\rm U}= $ gcd Universal Property.
Lemma $\ \ d\mid a_1,\ldots,a_n\!\iff d\mid (a_1,\ldots,a_n)\ \ \ $ [GCD Universal Property]
${\bf Proof}\quad\ d\mid a_1,\ldots,a_n\Rightarrow\, d\mid (a_1,\ldots,a_n) = j_1 a_1\!+\ldots+j_n a_n\,$ for some $\, j_i\in\Bbb Z,\,$ by Bezout.
$\qquad\qquad\, d\mid (a_1,\ldots,a_n)\mid a_1,\ldots,a_n\,\Rightarrow\, d\mid a_1,\ldots,a_n\,$ by transitivity of $ $ "divides",
The second divisibility in the prior line is due to the gcd being a common divisor of its arguments.
Remark $ $ Dually we have the universal property of LCM
Lemma $\ \ a_1,\ldots,a_n\mid m\iff {\rm lcm}(a_1,\ldots,a_n)\mid m\ \ \ $ [LCM Universal Property]
These universal properties are the definitions of GCD & LCM in more general rings  - where the Bezout identity need not hold true,  e.g. in $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where the gcds $\,(x,2) = 1 = (x,y)\,$ cannot be written as linear combinations. Follow the links for further details.

Proof: Let $d=gcd(a,b,c)$ for positive integers $a$,$b$, and $c$. Then $d\mid a$, $d\mid b$, and $d\mid c$. Hence, $d\mid gcd(a,b)$ and $d\mid c$ and so $d\mid gcd(gcd(a,b),c)$. Thus, $d\mid f$ where $f=gcd(e,c)$ and $e=gcd(a,b)$. Since $f=gcd(e,c)$, then $f\mid e$ and $f\mid c$. Hence, $f\mid a$, $f\mid b$, and $f\mid c$, so $f\mid gcd(a,b,c)$ which means $f\mid d$. Since $d\mid f$ and $f\mid d$, $d=f$.


Strategies to prove inequalities with interval notation How to prove a inequalities with interval notation, for example:

  Find minimum of $a^3+b^3+c^3$ with $a,b,c \in [-1;\infty), a^2+b^2+c^2=9$

Clearly $a, b, c$ cannot be all negative.

If $a, b, c$ are all non-negative, then by Power mean inequality $(\frac{a^3+b^3+c^3}{3})^2 \geq (\frac{a^2+b^2+c^2}{3})^3=27$, so $a^3+b^3+c^3 \geq 9\sqrt{3}$ with equality when $a=b=c=\sqrt{3}$.

If exactly 1 of $a, b, c$ is negative, then WLOG assume that $c$ is negative, so $a, b$ are non-negative. Then $c \geq -1$, so $c^2 \leq 1$ so $a^2+b^2 \geq 8$. By Power mean inequality $(\frac{a^3+b^3}{2})^2 \geq (\frac{a^2+b^2}{2})^3 \geq (\frac{8}{2})^3=64$, so $a^3+b^3+c^3 \geq 2\sqrt{64}+c^3 \geq 16-1=15$, with equality when $a=b=2, c=-1$. (and permutations)

If exactly 2 of $a, b, c$ are negative, then WLOG assume that $b, c$ are negative and $a$ is non-negative. We have $b, c \geq -1$, so $b^2, c^2 \leq 1$, so $a^2 \geq 7$, so $a^3+b^3+c^3 \geq (\sqrt{7})^3-1-1=7\sqrt{7}-2$, with equality when $a=\sqrt{7}, b=c=-1$. (and permutations)

Thus $a^3+b^3+c^3 \geq \min(9\sqrt{3}, 15, 7\sqrt{7}-2)=15$, with equality when $a=b=2, c=-1$ and permutations.


Induction principle for $n!<n^n$ How can I prove that $n!<n^n$ for every $n>1$ using the induction principle?

If I put $n=2$ I get $2<4$ so I know that $n!<n^n$ is true.

Now I don't know how to prove that $(n+1)!<(n+1)^{n+1}$. I can write $(n+1)!=n!(n+1)$ but then I don't know how to go on.

HINT

$$(n+1)!=(n+1)\cdot n! \stackrel{\color{red}{n!<n^n}} <(n+1)\cdot n^n\stackrel{\color{red}{n^n<(n+1)^n}} <(n+1)(n+1)^n$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\pars{n + 1}! & = {\pars{n + 1}! \over \pars{n + 1}^{n + 1}}
\,\color{blue}{\pars{n + 1}^{n + 1}} =
\color{blue}{\pars{n + 1}^{n + 1}}\,{n! \over \pars{n + 1}^{n}} =
\color{blue}{\pars{n + 1}^{n + 1}}\,{n\pars{n - 1}\cdots 2 \times 1 \over n^{n}\pars{1 + 1/n}^{n}}
\\[5mm] & =
\color{blue}{\pars{n + 1}^{n + 1}}\,\ \overbrace{\underbrace{{\pars{1 - 1/n}\pars{1 - 2/n}\cdots
\bracks{1 - \pars{n - 2}/n}\bracks{1 - \pars{n - 1}/n} \over \pars{1 + 1/n}^{n}}}_{\ds{{\mbox{numerator}\ < 1} \atop {\mbox{denominator} > 1}}}}^{\ds{\Large\color{red}{< 1}}}
\\[5mm] & <
\bbx{\large \pars{n + 1}^{n + 1}}
\end{align}