TITLE: Why do balloons deflate?
QUESTION [10 upvotes]: Title says it all. Every balloon that I have got, from parties, etc, all slowly deflated for no reason at all.
Why is that?

REPLY [10 votes]: Balloons are not perfectly air-tight. Air molecules can slowly diffuse through the surface, or leak through the knot. Of course, they can diffuse in just as well as out. The reason there is a net flux directed outward is that the pressure is higher on the inside, since the air there has the force of the wanting-to-shrink balloon to support in addition to atmospheric pressure.
You'll probably notice this effect proceeds faster with helium balloons. This is a result of helium existing as single, almost perfectly non-interacting atoms, rather than as molecules with two or three atoms. It simply has an easier time slipping between the cracks and diffusing through membranes.
One final note: The pressure imbalance is not particularly large. To get a feel for this, remember that atmospheric pressure is 14.7 pounds per square inch, so a balloon with a 5 inch radius has well over a thousand pounds of force being exerted on its surface by the exterior air. Since you can inflate a balloon with just your lungs, you know the total force induced by the stretched membrane is not terribly much. Thus you'd expect that the total number of air particles entering and leaving the balloon far exceeds the net outward flux.


TITLE: Functional Square Root of $1-x^3$
QUESTION [0 upvotes]: I'm completely stuck with the problem of finding such a continuous function $f$ that
$
f(f(x)) = 1 - x^3
$
or prove that it doesn't exist.
After some calculations I tend to the second case - continuous function with such property can not exist.
Does anyone have some ideas?

REPLY [2 votes]: Suppose $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(f(x)) = 1-x^3$, for all $x \in \mathbb{R}$.

Note that $f$ is injective, since
\begin{align*}
&f(x) = f(y)\\[4pt]
\implies\;&f(f(x))=f(f(y))\\[4pt]
\implies\;&1-x^3 = 1-y^3\\[4pt]
\implies\;&x^3=y^3\\[4pt]
\implies\;&x=y\\[4pt]
\end{align*}
Since $f$ is injective and continuous, it follows that $f$ is strictly monotonic.

If $f$ is strictly increasing, then
\begin{align*}
&x < y\\[4pt]
\implies\;&f(x) < f(y)\\[4pt]
\implies\;&f(f(x)) < f(f(y))\\[4pt]
\implies\;&1-x^3 < 1-y^3\\[4pt]
\implies\;&x^3 > y^3\\[4pt]
\implies\;&x > y\\[4pt]
\end{align*}
contradiction.

If $f$ is strictly decreasing, then
\begin{align*}
&x < y\\[4pt]
\implies\;&f(x) > f(y)\\[4pt]
\implies\;&f(f(x)) < f(f(y))\\[4pt]
\implies\;&1-x^3 < 1-y^3\\[4pt]
\implies\;&x^3 > y^3\\[4pt]
\implies\;&x > y\\[4pt]
\end{align*}
contradiction.

Hence, no such function $f$ exists.


TITLE: What if, for a body moving in a circle, the centripetal force is not equal to $mv^2/r$?
QUESTION [5 upvotes]: For example, I don't understand why the speed of a satellite moving in an orbit of radius 'a' around the Earth must be equal to $\sqrt{GM/a}$ If I release a particle in outer space with a velocity perpendicular to the line joining the particle and Earth but magnitude not equal to $\sqrt{GM/a}$ what will happen? For al I know, the acceleration will still be perpendicular to its velocity at all instants. And, an always perpendicular acceleration is only capable of changing the direction of velocity continuously. Then, wouldn't the particle still rotate even if its speed not equal to the one given by F=$mv^2/r$?

REPLY [2 votes]: For al I know, the acceleration will still be perpendicular to its
  velocity at all instants. And, an always perpendicular acceleration is
  only capable of changing the direction of velocity continuously.

Well, let's see.  For convenience, stipulate that at $t=0$, the particle is located at $r=R, \theta=\pi/2, \phi=0$ with velocity
$$\vec v(0)=v\;\hat{\boldsymbol{y}}$$
and acceleration
$$\vec a(0)=-\frac{GM}{R^2}\;\hat{\boldsymbol{x}}$$
The velocity at the next instant is then
$$\vec v(0 + dt)=-\frac{GM}{R^2}dt\;\hat{\boldsymbol{x}} + v\;\hat{\boldsymbol{y}}$$
while the acceleration is
$$\vec a(0 + dt)=-\frac{GM}{R^2}\left(\hat{\boldsymbol{x}} + \frac{v}{R}\,dt\;\hat{\boldsymbol{y}}\right)$$
and so the dot product of these velocity and acceleration vectors is
$$\vec v(0 + dt)\cdot \vec a(0 + dt)=-\frac{GM}{R^2}\left[-\frac{GM}{R^2}+\frac{v^2}{R}\right]dt$$
which is zero only if
$$\frac{v^2}{R}=\frac{GM}{R^2}$$
or
$$v=\sqrt{\frac{GM}{R}}$$
Thus, it isn't the case that the acceleration will still be perpendicular to the velocity at all instants for arbitrary $v$.


TITLE: Determining compactness of the set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$ in metric space $\mathbb R$
QUESTION [4 upvotes]: Given a set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$.
My attempt:
Claim $1:0$ is a limit point of set $S$.
Proof: Given any $\epsilon \gt 0, \exists N_\epsilon \in \mathbb N: \frac1{N_\epsilon}\lt \epsilon/2$. Choosing $m,n \ge N_\epsilon$,we have $\frac1{n}\le \frac1{N_\epsilon}\lt \epsilon/2$ and $\frac1{m}\le \frac1{N_\epsilon}\lt \epsilon/2 $. Thus $0\lt\frac{1}{m}+\frac{1}{n}\lt \epsilon$. Since $\epsilon \gt 0$ is arbitrary, we conclude that $0$ is a limit point of $S$.
Claim $2$: All elements of set $T=\{1/n: n\in \mathbb N\}$ are also limit points of $S$.
Proof: Given any $\epsilon \gt 0, \exists M_\epsilon \in \mathbb N: \frac1{M_\epsilon}\lt \epsilon\implies \frac1{m}\lt \epsilon \;\;\forall m\ge M_\epsilon\implies -\epsilon\lt \frac1{m}\lt \epsilon$. Hence for any $\frac1{n}\in T$, we have $\frac 1{n}-\epsilon\lt \frac 1{n}+ \frac1{m}\lt \frac1{n}+\epsilon$. Just as previosuly, it follows that all elements of $T$ are limit points of $S$.
Heine-Borel theorem states that a set $S$ in $\mathbb R$ is compact $\iff S$ is closed and bounded. Indeed, in the present case, $S$ is bounded as $2$ is one upper bound.
Since, $1/n \notin S$ for any $n\in \mathbb N$, $S$ is not closed and therefore, $S$ can't be compact.
Is my conclusion correct? Thanks.
Edit: How do I show that $S$ does not have limit points except $0$ and $\{1/n: n\in \mathbb N\}$?

REPLY [1 votes]: Consider this:
Let $x \not\in S$. We are going to show $x$ can't be a limit point of $S$.  As $S$ is bounded below by $0$ and above by $2$ we don't need so consider $x \le 0$ or $x \ge 2$.
So $x \ne \frac 1m = \frac 1{2m}+\frac 1{2m}$
Now $x > 0$ So there is a minimal $m \in \mathbb N$ so that $\frac 1m < x$.  If $m> 1$ then $\frac 1m < x < \frac 1{m-1}$ and if $m= 1$ then $\frac 1m = 1 < x$
Now $x -\frac 1m > 0$ so there is a minimal $n\in \mathbb N$ so that $\frac 1m + \frac 1n < x$.  And again if $n>1$ then $\frac 1m +\frac 1n < x < \frac 1m + \frac 1{n-1}$ and if $n=1$ then $\frac 1m + \frac 1n = \frac 1m + 1 < x$. (but it's not actually possible for $n = 1$)
It's easy to see that if $m =1$ then $\frac 1m + \frac 1n < x < 2$ so $n > 1=m$, and if $m> 1$ then $\frac 1m +\frac 1n < x < \frac 1{m-1}$ so $\frac 1n < \frac 1{m-1} + \frac 1m = \frac 1{(m-1)m} < \frac 1m$ and $n > m$.  SO $n > m\ge 1$.  So $\frac 1m +\frac 1n < x < \frac 1m +\frac 1{n-1}$
Now pick an $r$ so that $r < \min(x - (\frac 1m +\frac 1n),  (\frac 1m +\frac 1{n-1})- x)$
The $B_r(x)$ can't contain any points of $S$.
So $x$ is not a limit point of $S$ and then only limit points of $S$ (if any) are in $S$ so $S$ is closed.
(I say "if any" because we did nothing to find any limit points and we don't care.  If we actually cared we can see all $\frac 1m$ (including $1$) and $0$ are limit points [as there are infinitely many $\frac 1m + \frac 1n$ is any $B_r(\frac 1m)$] and for any $\frac 1m +\frac 1n\ne \frac 1t; t\in \mathbb N$ we can use a similar argument above is not t limit point.)
====
On the other hand a proof by finite subcover could be easier:
If $0 \in U$, an open set, then all but a finite number of $\frac 1t$ are in $U$.  For each of these finite number $\frac 1m \not \in U$ if $\frac 1m \in V_m$, then all but a finite number of $\frac 1m +\frac 1s \not \in V_m$. If each of these finite number of $\frac 1m + \frac 1{n_m}$ is in an open set $W_{m,n_m}$ then we have covered $S$.
For any open cover, $\mathscr U$, of $S$ we can find a $U\in \mathscr U$  so that $0 \in U$.  Then we can have a finite number of $\frac 1m \not \in U$ and we can find a finite number of $V_m \in \mathscr U$ so that $\frac 1m \in V_m$.  And for each of those we can find a finite number of $n_m$ so that $\frac 1m +\frac 1{n_m}\not \in V_m$ we can find a $W_{m,n_m}$ so that $\frac 1m +\frac 1{n_m} \in W_{m,n_m}$.
And $\{U\} \cup \{V_m\}\cup \{W_{m,n_m}\}$ is a finite subcover of $\mathscr U$.


TITLE: The largest equilateral triangle circumscribing a given triangle
QUESTION [10 upvotes]: Seven years ago, one of my many contributions to the March 2010 edition of Erich Friedman's Math Magic was a packing of eight circles of unit diameter and one equilateral triangle of unit side length into as small a circle as I could manage.

To minimise the bounding circle's radius $r$, I had to determine the largest equilateral triangle externally tangent to the three circles adjacent to it, then numerically adjust $r$ until said triangle had side lengths $1$. This is not a trivial problem, and back then I did trial and error in GeoGebra. ($2r=3.4133707107\dots$)
It is easy to see that the circles can be shrunk and the equilateral triangle enlarged about its centre while preserving tangencies, so the problem is equivalent to finding the largest equilateral triangle $\Delta^*$ where each edge is incident to one of $A,B,C$ where $A,B,C$ are the circle centres. The following construction doesn't count since one edge does not meet $\triangle ABC$:

Now, having reinvigorated my interest in triangles, I believe I have found a construction for $\Delta^*$ for arbitrary $A,B,C$, and am asking here for a more traditional proof of its optimality. My construction proceeds as follows:

Construct the first isogonic centre/Fermat point of the triangle, $X_{13}$ in the Encyclopedia of Triangle Centers (ETC). That is, construct outward equilateral triangles $A'BC, AB'C, ABC'$ on the sides of $\triangle ABC$, then $X_{13}$ is the concurrence of $AA',BB',CC'$.
$\Delta^*$ is (conjecturally) the antipedal triangle of $X_{13}$.$^\dagger$ That is, $\Delta^*$ is the triangle formed by perpendiculars at $A,B,C$ of $AX_{13},BX_{13},CX_{13}$ respectively; it is guaranteed to be equilateral because the lines meeting at $X_{13}$ are evenly spaced around it.

The centroid of the triangle thus constructed is $X_{5463}$ in the ETC, the reflection of $X_{13}$ in the centroid of $\triangle ABC$, and its area is $\frac{a^2+b^2+c^2}{2\sqrt3}+2\operatorname{area}(\triangle ABC)$ where $a,b,c$ are the side lengths. (Alternatively, the area is $2\left(1+\frac{\cot\omega}{\sqrt3}\right)\operatorname{area}(\triangle ABC)$ where $\omega$ is the Brocard angle.)

How can I prove that my construction actually produces $\Delta^*$, the largest circumscribing equilateral triangle?

$^\dagger$Without loss of generality, assume $AB$ is the largest side, and let $P,Q$ be the midpoints of $AC',BC'$ respectively. If $\overline{CP}$ or $\overline{CQ}$ is less than $\frac12\overline{AB}$, my construction will not produce an equilateral triangle actually incident to all of $A,B,C$. In that case, $\Delta^*$ has a side collinear with the triangle's shortest side $s$ and one vertex coincident with the original triangle's smaller angle incident to $s$. If $A,B,C$ are circle centres in a packing problem, this means that a larger equilateral triangle can be drawn by introducing a point contact.

REPLY [6 votes]: I'll use $\triangle PQR$ instead of $\triangle ABC$, to avoid some notational confusion.

First, a little prep work.
Given $\triangle PQR$, we erect on (directed) segment $\overline{PQ}$ an equilateral triangle $\triangle PQR'$ with a clockwise orientation. Similarly, we erect clockwise equilaterals $\triangle QRP'$ and $\triangle RPQ'$. As it happens, the lines $\overleftrightarrow{PP'}$, $\overleftrightarrow{QQ'}$, $\overleftrightarrow{RR'}$ meet at a common point, and do so symmetrically. When $\triangle PQR$ itself has a counter-clockwise orientation, the three equilaterals are external to it, and the three lines meet at the first isogonic center (Kimberling center $X_{13}$); when $\triangle PQR$ has a clockwise orientation, the equilaterals overlap its interior, and the common point is the second isogonic center ($X_{14}$).

In any case, we find that any triangle can be positioned such that each of its vertices lies on one of three concurrent, symmetrically-arranged lines. Taking the common point to be the origin, and one of the lines to be the $x$-axis, we can coordinatize $\triangle PQR$ thusly (abusing notation so that $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$) as
$$P := p \operatorname{cis} 0 \qquad Q := q \operatorname{cis}\tfrac23\pi\qquad R := r \operatorname{cis}(-\tfrac23\pi) \tag{1}$$
where we may assume $q$ and $r$ are non-negative (and not simultaneously zero). One can show that the $x$-intercept of $\overline{QR}$ is $-qr/(q+r)$; consequently, $\triangle PQR$'s orientation depends upon $p$'s relation to that value, and we can write
$$\text{The origin is}\;\triangle PQR\text{'s}\; \left\{\begin{array}{c}\text{first} \\ \text{second} \\ \text{(either)} \end{array}\right\}\; \text{isogonic center if} \;\; p q + q r +r p \left\{\begin{array}{c} > \\ < \\ = \end{array}\right\} 0 \tag{2}$$

Now to the topic at hand.
The sides of an equilateral triangle circumscribing $\triangle PQR$ are lines through $P$, $Q$, $R$ with symmetrically-arranged normal vectors, say, $$u := \operatorname{cis}\theta \qquad v := \operatorname{cis}\left(\theta+\tfrac23\pi\right)
\qquad w := \operatorname{cis}\left(\theta-\tfrac23\pi\right) \tag{3}$$
Let $T_\theta$ be the resulting equilateral triangle. I'll forego giving its vertices. The important thing is to compare the sizes of these triangles across all $\theta$, which we can do by computing the areas:
$$|T_\theta| = \frac{1}{\sqrt{3}}\;\left(\,p + q + r\,\right)^2\,\cos^2\theta \tag{4}$$
Clearly, $|T_\theta|$ is maximized with $\theta = 0$ (or $\pi$), giving a maximal area

$$|T_0| = \frac1{\sqrt{3}} (p+q+r)^2 \tag{$\star$}$$

when $u$, $v$, $w$ are parallel to $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$. When $O$ is $\triangle PQR$'s first isogonic center (see $(2)$), then $T_0$ matches OP's construction.
Now, while the $T_\theta$ triangles form an infinite family of equilaterals with side-lines containing the vertices of $\triangle PQR$, theirs is not the only such family. We get another by changing the signs in $(3)$; that is, by exchanging the roles of $v$ and $w$. The corresponding triangles, $T_\theta'$, have areas given by
$$|T^\prime_\theta| = \frac{1}{\sqrt{3}}\,\left(\,p \cos\theta + q \cos\left(\theta+\tfrac23\pi\right) + r \cos\left(\theta-\tfrac23\pi\right) \,\right)^2 \tag{5}$$
A quick derivative tells us that the critical values of $(4)$ occur for
$$\cot\theta = \frac{\sqrt{3}\,(q-r)}{2p-q-r} \quad\text{or}\quad
\tan\theta = \frac{\sqrt{3}\,(r-q)}{2p-q-r} \tag{6}$$
The former corresponds to an area of zero; the latter is maximizing, and we have

$$\phi := \tan^{-1}\frac{\sqrt{3}\,(r-q)}{2p-q-r} \quad\to\quad |T^\prime_\phi| = \frac1{\sqrt{3}} \left(p^2+q^2+r^2-p q-q r-r p\right) \tag{$\star\star$}$$

Comparing $(\star)$ to $(\star\star)$, we have
$$|T_0| - |T_\phi^\prime| = \sqrt{3} \left(p q + q r + r p\right) \tag{7}$$
which hearkens back to $(2)$. So, $T_0$ is maximal when $O$ is the first isogonic center of $\triangle PQR$; otherwise, $T^\prime_\phi$ is.
Importantly, the reader can verify that, at the key angle $\theta = \phi$ from $(\star\star)$, the lines through $P$, $Q$, $R$ with the direction vectors $u$, $v$, $w$ (the last two with their signs exchanged) meet at a point; specifically, they meet at $\triangle PQR$'s "other" isogonic center. This tells us that $T^\prime_\phi$ is actually the equilateral triangle obtained from OP's construction relative to that "other" center. Since, by $(7)$ and $(2)$, equilateral $T^\prime_\phi$ maximizes all triangles precisely when the "other" center is the first isogonic center, we have shown that

OP's construction relative to the first isogonic center is always maximal.

$\square$

Here are a couple of animations, showing differently-oriented $\triangle PQR$ (counter-clockwise vs clockwise); the origin (unmarked black dot) is either the first or second isogonic center, respectively. Triangles $T_\theta$ are green, while $T^\prime_\theta$ are light blue.
These images highlight that, while the various equilateral triangles have side-lines passing through the vertices of $\triangle PQR$, not all have $\triangle PQR$ in their interiors; hence, they are not all "circumscribing" in the traditional sense.


TITLE: How our eye see real images formed by pinhole camera?
QUESTION [0 upvotes]: How do we(eye) see real image form in pinhole camera?
1.Are rays that are forming image pass through screen and reach our eye help us too see it ?
OR

Light rays from hole from where we are looking at image , hit screen and form its image in our eye.

Here is a diagram I made

Which shows correct formation image of real image formed by pin hole camera in our eyes?
I have asked this question once but I haven't describe what actually doubt I am facing.
So now I done it, please don't downvote , I am super confused in this doubt. Please help. You all have better knowledge than me, tell me where I am wrong?

REPLY [1 votes]: Case 1 is correct. We see the image formed on the screen as the screen allows light to pass through it.


TITLE: If $\lim \limits_{x \to \infty}f(x) = L \neq 0$ must it be that $\lim_{x \to \infty} f(x) \sin x $ does not exist?
QUESTION [0 upvotes]: I got this question:
Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function that satisfies $\lim \limits_{x \to \infty}f(x) = L \neq 0$, Must it be that $\lim_{x \to \infty} f(x) sin x $ does not exist?
I tried to prove it and to find a counter example but I failed so far.

REPLY [4 votes]: If $\lim_{x\to\infty} f(x)=L$, with $L\ne 0$, then
$$\lim_{n\to\infty} f(2n\pi x)=L$$ and $$\lim_{n\to\infty} f(2n\pi x+\pi/2)=L.$$
But
$$
f(2n\pi x)\sin(2n\pi )=0,
$$
and hence $\lim_{n\to\infty} f(2n\pi x)\sin(2\pi n)=0$, while
$$
f(2n\pi x+\pi/2)\sin(2n\pi+\pi/2)=f(2n\pi x+\pi/2)\to L,
$$
as $n\to\infty$, and hence the limit $$\lim_{x\to\infty} f(x)\sin x$$ does not exist.


TITLE: What topological restrictions are there for a topological space to be a group?
QUESTION [8 upvotes]: I'm trying to provide a group structure for some Riemannian surfaces. I heard that the following result holds:

Let $X$ be a compact Riemannian surface. Then $X$ admits a group structure if, and only if it has genus $1$.

REPLY [3 votes]: Here's a proof for the result you provided (even if you need stronger assumptions):
Claim: If a closed $2$-dimensional smooth manifold admits a Lie group structure, then it is orientable and it has genus $g=1$ (i.e. it is the torus $T^2$).
Proof: If $G$ is a Lie group with $\dim G=n$, then $TG\cong G\times\mathbb{R}^n$, since you can use the action of the group structure by left multiplication to move around a basis you define at a point. This obviously implies that it must be orientable, by considering a global orientation on $TG$ induced by any choice of basis on $\mathbb{R}^n$. At the same time, it implies that you have $n$ never vanishing vector fields defining a basis at every point (corresponding to the standard basis $\partial_i$ of $G\times\mathbb{R}^n$). Thus the Poincaré-Hopf theorem implies that $\chi(G)=0$. In particular, if $G$ is as assumed in the claim, we must have $g=1$, by the formula $$\chi(G)=2(g-1)\stackrel{!}{=}0$$

We also have a partial converse to this result:
Claim: If a closed, orientable $2$-dimensional smooth manifold $M$ has genus $g=1$, then it admits the structure of a topological group.
Proof: By the classification of closed surfaces, $M$ is homeomorphic to the torus through some homeomorphism
$$h:M\to T^2$$
We define a multiplication $*$ on $M$ by
$$x*y=h^{-1}(h(x)\cdot h(y))$$
where $\cdot$ denotes the product on $T^2$. This makes $M$ into a topological group.


TITLE: consider a base-16 adder. explain how to modify the adder so that it can perform a base-10 addition
QUESTION [0 upvotes]: consider a base-16 adder. explain how to modify the adder so  that it can perform a base-10 addition
I found this when I searched in Google but  not understand

please guide me to understand this question so  that I can solve it

REPLY [1 votes]: Lets consider the case of a single digit addition:
you have two base-$16$ numbers $x$ and $y$  ($4$ bits each) and want to produce $z = x+y$.
Note that for $z$ we will allow $8$ bits (two base-$16$ digits) as we may have a carryover by the addition and we will need to store it somewhere.
Lets see how the base-$16$ adder works, through an example:
$$
\begin{array}{rlcccl}
 &x&      & 0011 &  (3)_{16}\\
+&y&      & 0100 &  (4)_{16}\\
=&z& 0000 & 0111 &  (7)_{16} & [0, 7]_{\text{BCD}}
\end{array}
$$
In this case, we added $3$ and $4$ and produced number $7$. There was no carryover and $z$ effectively can be stored in a single base-$16$ digit.
The important thing to realize is that the base-$16$ representation of $z$ is also good to go as a BCD representation: the left most digit is $0000$ ($=(0)_{10}$)and the right most is $0111 = (7)_{10}$.
So no modification is needed: all that changes is how  you read the output.
The problem is that each base-$16$ digit can store a value up to $15$ which is larger than $9$, which in turn is the largest value for a digit in the BCD representation. Consider the following example:
$$
\begin{array}{rlccc}
 &x&      & 0111 &  (7)_{16}\\
+&y&      & 0111 &  (7)_{16}\\
=&z& 0000 & 1110 &  (14)_{16}
\end{array}
$$
Now, the right-most base-$16$ digit of $z$ is $(14)_{16}$.
Instead, the BCD representation should be:
$$
\begin{array}{rlccc}
& 0001 & 0100 & & [1, 4]_{\text{BCD}}.
\end{array}
$$
So from the digit holding the value $(14)_{16}$, we would have to go to a digit holding the value $(4)_{16}$ and get a carryover to put in the left most digit.
This can be done by adding $(6)_{16}$ to $z$:
$$
\begin{array}{rlcccl}
 &z& 0000 & 1110 &  (14)_{16}\\
+&6&      & 0110 &  (6)_{16}\\
=&z^{'}& 0001 & 0100 &  (20)_{16} & [1, 4]_{\text{BCD}}
\end{array}
$$
The general rule is the following:
If the sum produced by the base-$16$ adder is greater than $9$ then you have to add $6$ to get the correct digit and propagate the carryover.
I will conclude with one more example to clarify something:
$$
\begin{array}{rlccc}
 &x&      & 1000&  (8)_{16}\\
+&y&      & 1000 &  (8)_{16}\\
=&z& 0001 & 0000 &  (16)_{16}
\end{array}
$$
The left-most digit of $z$ is $0000 = (0)_{16}$ which is smaller than $9$. But the sum is equal to $16$ (we have a carryover).
So the sum is greater than $9$. So we have to add $6$ to get the right BCD representation. Indeed:
$$
\begin{array}{rlcccl}
 &z& 0001 & 0000 &  (16)_{16}\\
+&6&      & 0110 &  (6)_{16}\\
=&z^{'}& 0001 & 0110 &  (22)_{16} & [1, 6]_{\text{BCD}}.
\end{array}
$$
Hope this helps.


TITLE: Existence of inscribed polygon in positive measure sets
QUESTION [0 upvotes]: Fix a convex $n$-gon $\Gamma$, is it true that for any subset $E$ of $\mathbb{R}^2$ with positive Lebesgue measure, there exist $n$ points in $E$, such that the associated $n$-gon of these points is similar to $\Gamma$ ? If this is true, could this be generalized to $\mathbb{R}^m$ with $m >= 3$?
I have already proved that the answer is affirmative when $\Gamma$ is regular, by considering rotations of $\mathbb{R}^2$ by $\frac{n}{2 \pi}$, and applying a lemma which asserts the existence of a square $S_{\epsilon}$ satisfying $\frac{m(S_{\epsilon} \cap E)}{m(S_{\epsilon})}>1-\epsilon$ for $\epsilon > 0$.
Any remark will be greatly appreciated.

REPLY [0 votes]: I see, direct application of (almost) the same method yields the desired result, namely for any ${p_{1},...,p_{n}}$ with prescribed similar type we consider still a order-$n$ subgroup of $Aut_{Top}(\mathbb{R^2})$ fixing ${p_{1},...,p_{n}}$, and let it act on a region containing ${p_{1},...,p_{n}}$.


TITLE: Is it necessary to have left and right limit at any limit point in a set A of R for any given continuous function on A?
QUESTION [1 upvotes]: Let $A$ be a subset of $\mathbb{R}$ and $c$ a limit point of $A$ but not in $A$.
Suppose $f$ is a continuous function on $A$.
Consider only in case that $c$ is not maximum point and minimum point.
Is it necessary that $f$ must have a left and a right limit at $c$?

REPLY [3 votes]: No. For example, $\sin \frac{1}{x}$ is continuous on $(-\infty, 0) \cup (0, +\infty)$, but has no left or right limits at $0$.


TITLE: Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification]
QUESTION [2 upvotes]: Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$.
My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\Rightarrow a^3\equiv b^3\pmod p\Rightarrow a^{3k}\equiv b^{3k}\pmod p.$
Next, observe that due to FLT we have $a^{3k+1}\equiv b^{3k+1}\pmod p.$ Now if $p\not|a$ and $p\not|b$, then $\gcd(a,p)=\gcd(b,p)=1.$ Therefore we can use can conclude that $$a^{\gcd(3k,3k+1)}\equiv b^{\gcd(3k,3k+1)}\pmod p\Rightarrow a\equiv b\pmod p.$$ Therefore $$a^2+ab+b^2\equiv 0 \pmod p\Rightarrow 3b^2\equiv 0\pmod p \text{ and } 3a^2\equiv 0\pmod p.$$  Which implies that  $p|3$ which is a contradiction. Hence Proved.
I would like to know whether this proof is correct or not. I am unsure about the use of $\gcd$ in the exponent. Moreover, I acknowledge that this question has been asked before, but I've not seen any answer using this fact explicitly. The fact being: Let $\gcd(a,m)=\gcd(b,m)=1$, then if $a^{x}\equiv b^x\pmod m$ and $a^y\equiv b^y\pmod m\Rightarrow a^{\gcd(x,y)}\equiv b^{\gcd(x,y)}\pmod m.$

REPLY [4 votes]: Here's a correct different solution. If $p=3k+2$ is an odd prime, $p\mid a^2+ab+b^2$, where $a,b\in\mathbb Z$, then $$a^2+ab+b^2\equiv 0\pmod{p}$$
$$\stackrel{\cdot 4}\iff (a+2b)^2\equiv -3a^2\pmod{p}$$
$$\iff (2a+b)^2\equiv -3b^2\pmod{p}$$
For contradiction, let either $p\nmid a$ or $p\nmid b$. Then either
$$\left((a+2b)a^{-1}\right)^2\equiv -3\pmod{p}$$
or $$\left((2a+b)b^{-1}\right)^2\equiv -3\pmod{p}$$
In both cases $-3$ is a quadratic residue mod the odd prime $p=3k+2$, which contradicts Quadratic Reciprocity.
If $p=2$ and $2\mid a^2+ab+b^2$ for some $a,b\in\mathbb Z$, then either $a\equiv 0\pmod{2}$ or $a\equiv 1\pmod{2}$; in the first case $2\mid b^2$, i.e. $b\equiv 0\pmod{2}$, and in the second case $$a^2+ab+b^2\equiv 1+b+b^2\equiv 1+b+b$$
$$\equiv 1+2b\equiv 1\not\equiv 0\pmod{2}$$

REPLY [1 votes]: Start with $a^3\equiv b^3\bmod p$.
If $p$ divides $b$, then $p$ divides $a$.
If $p$ does not divide $b$, there is $c$ such that $bc \equiv 1 \bmod p$. Then $(ac)^3 \equiv 1 \bmod p$.
If $ac \not\equiv 1 \bmod p$, then the order of $ac \bmod p$ is $3$ and so $3$ divides $p-1$, which contradicts $p=3k+2$.
If $ac \equiv 1 \bmod p$, then $a \equiv b$ and $a^2+ab+b^2 \equiv 3b^2$. If this is a multiple of $p$, then $p=3$ because $p$ does not divide $b$, which contradicts $p=3k+2$.