Solution to $7^{2x-2} \equiv 4 \mod13$ I need to solve the congruence

$$ 7^{2x-2} \equiv 4 \mod13 $$

I think I need to use a primitive root to transform it into $2x-2 \equiv ??? \mod (\phi(13)=12)$ but I'm stumped on how to actually do it.

$$7^{2x-2}\equiv 4 \pmod{13}$$
$\iff$
$$49^{x}\equiv 4(49) \pmod{13}$$
$\iff$
$$10^{x}\equiv 4(-3) \pmod{13}$$
$\iff$
$$10^{x}\equiv 1 \pmod{13}$$

You can now check the powers of ${10} \pmod{13}$.

Alternatively, for a slight shortcut:

Integers mod ${13}$ form a group under multiplication of order $12$. So now you need to find the order of $10$, and you can do this by checking through the factors of $12$.

Then $x\equiv0\pmod{\textrm{order}(10)}$.


Diophantine equation $x^2 + y^2 = z^3$ I have found all solutions to the Diophantine equation $x^2 + y^2 = z^3$ when
$z$ is odd. I am having some difficulty finding the solutions when $z$ is even. I am asking for a proof that provides the solutions where $z$ is even. I want the proof to be elementary and use only Number theory and perhaps Calculus or basic ideas about groups and rings.

Unfortunately, there isn't (apparently) one complete polynomial parameterization to

$$x^2+y^2 = z^k\tag1$$

when $k>2$. For $k=2$, the complete solution is,

$$x,\,y,\,z = (a^2-b^2)s,\; (2ab)s,\; (a^2+b^2)s$$

where $s$ is a scaling factor. Using complex numbers $a+b i$, one can generalize the method. For $k=3$, it is

$$x,\,y,\,z = (a^3 - 3a b^2)s^3,\; (3a^2 b - b^3)s^3,\; (a^2+b^2)s^2\tag2$$

but you can no longer find rational $a,b,s$ for certain solutions. For example,

$\hskip2.7in$ $9^2+46^2 = 13^3\quad$ Yes

$\hskip2.7in$ $58^2+145^2=29^3\quad$ No

(You can click on the Yes/No links for Walpha output.) A related discussion can be found in this post while an alternative method is described here. For the case $k=3$, if $a^2+b^2=c^3$, then an infinite more can be found as,

$$(a u^3 + 3 b u^2 v - 3 a u v^2 - b v^3)^2 + (b u^3 - 3 a u^2 v - 3 b u v^2 +
        a v^3)^2 = c^3(u^2+v^2)^3\tag3$$

which should provide some solutions not covered by $(2)$.

Usually a view.  http://www.artofproblemsolving.com/community/c3046h1124891_almost_pythagoras
http://www.artofproblemsolving.com/community/c3046h1049554__

You can write the solution in this form.  http://www.artofproblemsolving.com/community/c3046h1054060_cubes_with_squares

http://www.artofproblemsolving.com/community/c3046h1048815___
http://www.artofproblemsolving.com/community/c3046h1047876____

But usually use the standard simple approach. In the equation.

$$X^2+Y^2=Z^3$$

$$X=ab+cd$$

$$Y=cb-ad$$

And receive such record.

$$(a^2+c^2)(b^2+d^2)=Z*Z^2$$

$$b^2+d^2=Z^2$$

$$Z=a^2+c^2$$

So.

$$d=a^2-c^2$$

$$b=2ac$$

Then the decision on the record.

$$X=3ca^2-c^3$$

$$Y=3ac^2-a^3$$

$$Z=a^2+c^2$$


Traveling salesman problem (TSP): what is the Relation with number of vertices and length of the found route? I know that there are many algorithms (exact or approximate) which implement the traveling salesman problem. I would like to know the relationship between the number of the vertices (i.e. the places to visit) and the length of the route found by these algorithms.

Intuitively, the fewer the vertices the shorter the route is. But can any one give me the mathematical relationship between the number of the vertices and the length of the route found by at least one of the existing traveling salesman algorithms?

Pick the hexagons that correspond to locations you want to visit.

If these hexagons constitute a connected area (there is hexagon-path between any two), then the shortest solution is of length $c_1 \cdot n$ and the solution found by a good approximate algorithm is $c_2 \cdot n$ where $c_1$ and $c_2$ are some constants (i.e. they depend on the size of the hexagon).

If these hexagons are not connected, then the length of the best solution can be arbitrarily big: imagine three hexagons separated by $k$ spaces each, then the best path has length of $3\cdot (k+1) \cdot d_\text{hex}$, despite having only $3$ places to visit.

I hope this helps $\ddot\smile$

Well, since you're required to visit each city, and to visit it only once, the number of edges in your route exactly equals the number of vertices (assuming that you return to your original location at the end). The total length of the road is the sum over all these edges multiplied by their weights. It totally depends on the weights and has little to do with the number of vertices, if I understood your question correctly.


Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that if $r \ne s$, then $u = o$, the zero vector in $V$ and If $u \ne o$, then $r = s$. Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that:

If $r \ne s$, then $u = o$, the zero vector in $V$.
If $u \ne o$, then $r = s$.

EDIT:
Attempt:
I know that this question is a little bit easy because it is a basic concept. But, although so,
I got a little confused about my answer below.
Let $r,s \in \Bbb F$, a field, and $u \in V$. Now,
\begin{align*}
r \cdot u &= s \cdot u \\
r \cdot u - s \cdot u &= o \\
(r-s) \cdot u &= o \\
\end{align*}
What next?
Assume that $V$ is over field, say $F$. Then, if $r \ne s$,
\begin{align*}
(r-s) \cdot (r-s)^{-1} \cdot u &= o \cdot (r-s)^{-1} \\
u &= o,
\end{align*}
and if $u \ne o$, then we must have $r-s = 0$ i.e. $r=s$.
Does it true?

EDIT: You are correct so far, but you must explain why $(r - s) \mathbf u = \mathbf 0$ implies that $r - s = 0$ or $\mathbf u = \mathbf 0.$ Equivalently, we can show that if $r - s \neq 0,$ then $\mathbf u = \mathbf 0.$ Given that $r - s \neq 0$ and $r$ and $s$ are elements of a field, what can you say about $r - s?$ Is there a way to "cancel" $r - s$ from the left-hand side? Once you study bases, you may use the below argument. (It can't hurt to practice.)

Consider a basis $\mathscr B$ of $V.$ Every vector of $V$ can be written uniquely as a finite $\mathbb F$-linear combination of basis vectors of $V.$ Particularly, there exist unique scalars $c_1, \dots, c_n$ and basis vectors $\mathbf v_1, \dots, \mathbf v_n$ such that $\mathbf u = c_1 \mathbf v_1 + \cdots + c_n \mathbf v_n.$ Given that $r \mathbf u = s \mathbf u,$ we have that $$(r c_1) \mathbf v_1 + \cdots + (r c_n) \mathbf v_n = (s c_1) \mathbf v_1 + \cdots + (s c_n) \mathbf v_n.$$ What can you say about the relationship between $r c_i$ and $s c_i$ for each integer $1 \leq i \leq n?$ From there, I believe that you can conclude the desired results.

I don't believe the question as posted is complete - unlike ordinary numbers, vector Dot Products do not have the cancellation property. From Wikipedia article for Dot Product
If $a ⋅ b = a ⋅ c$ and $a  ≠  0 $, then we can write: $ a  ⋅ ( b  −  c ) = 0$ by the Distributive Law; this just means that  $a$  is perpendicular to $( b  −  c )$, which still allows $( b  −  c ) ≠  0 $, and therefore allows $b  ≠  c $.
To put it another way, $r ⋅ u = s ⋅ u$ (with $r ≠ s$) can still be true if $u ⟂ (r - s)$ and $u ≠ 0$.
EDIT: Assuming a downvote means someone doesn't believe me, I'll work through a simple example with three perpendicular, non-zero unit vectors.
Let $r = (1,0,0)$, $s = (0,1,0)$, and $u = (0,0,1)$.
$$r ⋅ u = s ⋅ u$$
$$\left\lVert r \right\rVert \times  \left\lVert u \right\rVert \times \cos \theta_{ru} = \left\lVert s \right\rVert \times \left\lVert u \right\rVert \times \cos \theta_{su}$$
$$1 \times 1 \times \cos \theta_{ru} = 1 \times 1 \times \cos \theta_{su}$$
Obviously, being perpendicular unit vectors, $\theta_{ru} = \theta_{su} = 90^\circ$, so we get
$$1 \times 1 \times \cos 90 = 1 \times 1 \times \cos 90$$
$$1 \times 1 \times 0 = 1 \times 1 \times 0$$
$$0=0$$


Why is $1/i$ equal to $-i$? When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).

Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?


An explanation on variances I am having trouble understand what the following represent.

Let $\mathbb X^{p\times 1}$ denotes a column vector of random components $X_i,i=1(1)p.$ Then the expectation of $\mathbb x$ is defined as $$E(\mathbb X)=E\begin{pmatrix}X_1 \\ \vdots\\X_p\\\end{pmatrix}=\begin{pmatrix}EX_1 \\ \vdots\\EX_p\\\end{pmatrix}=\begin{pmatrix}\mu_1 \\ \vdots\\\mu_p\\\end{pmatrix}=\mathbb{\mu}
$$
Define $\text{Cov}(X_i,X_j)=E[(X_i-EX_i)(X_j-EX_j)^T]=\sigma_{ij}$. We extend the variance notation to the p-dimentional random vector $\mathbb X$ by the following matrix:
$$E[(\mathbb X-E \mathbb X)(\mathbb X-E \mathbb X)^T] \\=E[(\mathbb X- \mathbb \mu)(\mathbb X-\mathbb \mu)^T] \\=E\left[\begin{pmatrix}X_1-\mu_1 \\ \vdots\\X_p-\mu_p\\\end{pmatrix} \begin{pmatrix}X_1-\mu_1& \cdots &X_p-\mu_p\\\end{pmatrix}\right]\\=\begin{pmatrix}\sigma_{11} &\sigma_{12} &\cdots &\sigma_{1p}\\ \vdots & \vdots &\ddots &\vdots \\\sigma_{p1} &\sigma_{p2}&\cdots& \sigma_{pp}\\\end{pmatrix}=\Sigma$$ In similarly way co-variance between $X$ and $Y$ is defined.

That's all from me.


Special case of double integral where the upper bond of the inner integral is the integration variable of the out integral This is regarding a special type of double integral looking like the following:
$$\int_0^T  \exp\left(\int_0^\color{red}{t} f(s)\, ds\right) \,\color{red}{dt}. $$
Time being the integration variable here, one can think of $T$ as entire duration which sets the boundary of the outer integral, and the inner integral integrates from $0$ to each discrete time point $t$, and take the exponential of it. Here, the upper bond of the inner integral is the integration variable of the outer integral.
I suspect the expression can be simply to:
$$ \exp\left(\int_0^T f(t)\, dt\right). $$
but I'm not sure if this is correct. Can anyone enlighten me with the missing links in between and correct me if my suspicion is wrong.

You can look at it as
$$\int\limits_{0}^{T}e^{\int\limits_{0}^{t}f(s) \mathrm{d}s} \mathrm{d}t$$
for f$(s) = -\frac{s}{2}$, for example, you obtain well known
$$\int\limits_{0}^{T}e^{-\frac{t^2}{2}} \mathrm{d}t$$


Law of Probability question Question: 40% of Vancouverites are avid skiiers, whereas only 20% of people living in Canada are avid skiiers. Vancouverites make up about 2% of the population of Canada. You know that Johann lives in Canada, but you’re not sure where. Given that Johann is an avid skier, what is the probability that he lives in Vancouver?

I need help with this question and your help will be greatly appreciated if you can show me how to set it up and calculate the probability

so Far I set it up as:

A-> Vancouverites are avid skiers (4/10)

b-> People living in Canada are avid skiers (2/10)

c-> Vancouverites make up Canada's pop (0.2/10)

$(0.4)(0.02) = 0.008$

$\frac{0.008}{0.2} = 0.04$

Because 0.8% of the Canadian skiers are Vancouverites. If your a skier in Canada there's a 4% chance that your a vancouverite as 20% of Canadians are skiers.


Coin chosen is two headed coin in this probability question I have a probability question that reads:

Question:

  A box has three coins. One has two heads, another two tails and the last is a fair coin. A coin is chosen at random, and comes up head. What is the probability that the coin chosen is a two headed coin.

My attempt:

P(two heads coin| given head) = P(two heads coin * given head)/P(given head)
= 1/3/2/3 = 1/2

Not sure whether this is correct?

For such a small number of options its easy to count them

The possible outcomes are:

heads or heads using the double head coin
tails or tails using the double tail coin
heads or tails using the fair coin

All these outcomes are equally likely.  How many of these are heads and of those how many use the double headed coin?

      $$Answer =  \frac{2}{3}$$

1/2 is the only answer!! 50 50 chance, that's it period!!! The double tails is out for sure, so discard it. What is left besides the normal coin and 2 headed coin? Nothing! So, it's between two coins only. You can't rip the heads or tails from the coins and shake them up in the box. That would give you 3 heads and only 1 tail. That would be a 3 out of 4 chance! BUT THAT IS OUT! These are coins with no magical way of knowing if the coin chosen has a tail on the hidden side. So it's a 50 50 or 1/2 chance. Both the normal and 2 tail coins are out because it HAS to be a 2 headed coin chosen! So if you randomly draw out a coin with heads. You have a 50/50 or 1/2 chance. THAT'S IT, 1/2. No formula needed. I'm not great at math myself. This is common comprehension! Thanks Del


Proving equation is true Here is a question

Show that if $x>0$,$y>0$ than:

  $$\frac{1}{x}+\frac{1}{y}\geq\frac{4}{{x+y}}$$

How would you solve/start this?

Hint One way is simply to show that $$\frac1x+\frac1y-\frac{4}{x+y}\ge 0.$$

Then the rest is just simple algebra.


Mathematics of apportionment of representation in a legislative body It seems to be eclipsed by coronavirus, but today is the U.S. Census day for this coming decade.

If you are American, where you are living today, where your children are living today, is where they are enumerated in the Census. This will determine, first hand, how many U.S. Representatives your state will be apportioned. It will also affect how redistricting will occur in your state and, perhaps, in your city if it is divided politically into wards or districts.

Okay, this is about once the census populations of each state are finalized, about distributing (or "apportioning") a fixed number of representatives among the states.  I know about the Huntington-Hill method and this will eventually be about that.  But I want to get more fundamental about the problem.

This is what is given:

$N$: The number of U.S. states.  Currently 50.
$H$: The number of representatives in the House of Representatives.  Currently 435.
$P_n$: The census population of state $n$.

All of these parameters are positive integers $\in \mathbb{N}$.

What we need to determine:

$R_n$: The number of representatives apportioned to state $n$.

Also all of these $R_n \in \mathbb{N}$.

We know that the total population of the 50 states (this leaves out the District of Columbia, Puerto Rico, Guam, American Samoa, and other territories whose residents are American, but they get no voting representation in Congress) is the sum of the populations of all states:

$$ P_\mathrm{total} = \sum\limits_{n=1}^{N} P_n $$

And we know also that the total number of representatives is the sum of apportioned representatives of all states:

$$ H = \sum\limits_{n=1}^{N} R_n $$

Now, the simplest meaning of the concept of apportionment is that the number of representatives a state has is directly proportional to the population of that state.  That would say that there exists a constant (w.r.t. all of the states) of proportionality, $\alpha$, such that:

$$ R_n = \alpha \cdot P_n $$

Now, if there were no problems regarding fractional representatives, we know that:

$$\begin{align}
 \alpha \cdot P_\mathrm{total} &= \sum\limits_{n=1}^{N} \alpha \cdot P_n \\
\\
&= \sum\limits_{n=1}^{N} R_n \\
\\
&= H \\
\end{align} $$

So we can solve that for $\alpha$ and have an idea what it might be, but we cannot have a fraction of a representative, $R_n$ must be a positive integer, so then quantization or rounding to an adjacent integer is necessary.  We know that if rounding down or even rounding to nearest may cause the rounded value of $R_n$ for the least populous states to possibly be zero, and the U.S. Constitution does not allow for that.  It seems to me that the only consistent simple rule of rounding would be to always round up:

$$ R_n = \big\lceil \alpha \, P_n \big\rceil $$

where $\lceil x \rceil$ is the ceiling function which is the smallest integer $m$ such that $m-1 < x \le m$, or

$$ \lceil x \rceil \in \mathbb{Z} \\
\\
  x \le \lceil x \rceil < x+1 $$

So, it seems to me that the simplest consistent rule to guarantee that each state gets a whole number of representatives and at least one representative is to find the constant of proportionality, $\alpha$ such that:

$$\begin{align}
H &= \sum\limits_{n=1}^{N} R_n \\
\\
&= \sum\limits_{n=1}^{N} \big\lceil \alpha \, P_n \big\rceil \\
\end{align} $$

Now couldn't we simply define an increasing function $h(\alpha)$ as

$$ h(\alpha) \triangleq \sum\limits_{n=1}^{N} \big\lceil \alpha \, P_n \big\rceil $$

and, starting at $\alpha=0$ (and we know that $h(0)=0$), then increase that value $\alpha$ until $h(\alpha)=H$?  Then we know the number of representatives for all of the states $R_n = \big\lceil \alpha \, P_n \big\rceil$ for all $n$.

Is this consistent with the Huntington-Hill method?

If needed, I will explain the Huntington-Hill method here, but I need to figure out a good set of symbols that is consistent with the symbols I use above.  Give me a couple hours to do that.

Your method of apportionment is the one that John Quincy Adams proposed in 1832. It has never been used to apportion seats in the U.S. House of Representatives.

Most methods of apportionment in widespread use can be described in one of two equivalent ways.

One way is the one you use. A common divisor is defined for all states (or parties, when apportioning seats after an election), and each state’s population (or each party’s vote tally, in case of an election) is divided by this divisor and then rounded in some prescribed manner. The divisor is selected such that the desired total of seats results. The three obvious choices of rounding down, up or to the nearest integer are typically referred to by different names in Europe and the United States:

\begin{array}{c|c}
\text{rounding}&\text{European name}&\text{United States name}&\text{divisor offset}\\\hline
\text{up}&\text{Adams}&\text{Adams}&0\\
\text{nearest}&\text{Sainte-Laguë}&\text{Webster}&\frac12\\
\text{down}&\text{D'Hondt}&\text{Jefferson}&1\\
\end{array}

Equivalently, the seats can be apportioned one by one, with each state or party being assigned a divisor of its own that is determined by the seats already allocated to it. An offset $\Delta$ is added to this seat count, that is, the quotients

$$
\frac{\text{population or #votes}}{\text{#seats}+\Delta}
$$

are calculated, and the state or party with the largest current quotient gets the next seat. The divisor offsets corresponding to the three methods above are shown in the table.

The equivalence of the two ways of describing these methods can be shown by noting that in the first approach, when the global divisor is adjusted to achieve the desired total, a state gets an additional seat whenever the divisor crosses its rounding threshold, and this is precisely when its population is $\text{#seats}+\Delta$ times the global divisor.

The Huntington–Hill method currently in use in the U.S. House of Representatives is easier to describe in the second way. Here the divisor is the geometric mean of $\text{#seats}$ and $\text{#seats}+1$, that is, of the values that correspond to rounding up and down, respectively. Thus, it is somewhat similar to the Sainte-Laguë / Webster method, which uses the arithmetic mean $\text{#seats}+\frac12$ instead of the geometric mean. For large seat counts, the two are very similar, and even when a single seat has been apportioned, the geometric mean of $1$ and $2$ is $\sqrt2\approx1.414$, not all that different from $\frac32=1.5$; but when zero seats have been apportioned, the divisor is $0$ rather than $\frac12$, thus ensuring that all states get at least one seat. So in a sense Huntington–Hill is Webster with a smooth transition to the guaranteed minimum seat provided by Adams.

Here’s Java code that implements all four methods of apportionment. Applying them to the $2010$ United States census data yields the following differences relative to the Huntington–Hill method (which I’m using as a reference since it’s the one actually being used):

\begin{array}{l|r|c}
\text{state}&\text{rank}&\text{Adams}&\text{Webster}&\text{Jefferson}\\\hline
\text{California}&1&-3&&+2\\
\text{Texas}&2&-2&&+1\\
\text{New York}&3&-1&&+1\\
\text{Florida}&4&-1&&+1\\
\text{Illinois}&5&&&+1\\
\text{Pennsylvania}&6&-1&&&\\
\text{Ohio}&7&&&+1\\
\text{Georgia}&9&-1&&&\\
\text{North Carolina}&10&&+1&+1\\
\text{New Jersey}&11&&&+1\\
\text{Missouri}&18&+1&&&\\
\text{Minnesota}&21&&&-1\\
\text{South Carolina}&24&&&-1\\
\text{Louisiana}&25&+1&&&\\
\text{Oregon}&27&+1&&&\\
\text{Oklahoma}&28&+1&&&\\
\text{Iowa}&30&+1&&&\\
\text{West Virginia}&37&&&-1\\
\text{Nebraska}&38&&&-1\\
\text{Idaho}&39&+1&&\\
\text{Maine}&41&&&-1\\
\text{New Hampshire}&42&&&-1\\
\text{Rhode Island}&43&&-1&-1\\
\text{Montana}&44&+1&&\\
\text{Delaware}&45&+1&&\\
\text{South Dakota}&46&+1&&\\
\text{Vermont}&49&&&-1\\
\text{Wyoming}&50&&&-1\\
\end{array}

The remaining states get the same number of seats under all four methods.

The similarity between Huntington–Hill and Webster is evident, with the expected very slight advantage for smaller states under Huntington–Hill. The two “extreme” methods that always round up or down instead of taking one of the two means show a clear relative advantage for smaller and larger states, respectively. There would be quite significant changes if the Adams–Bristow-Johnson method were used instead of Huntington–Hill; so the answer to your question is “no”.

An interesting statistic that illustrates a sense in which the two methods that take a mean are more fair than the two others is the variance in the weight of voters in the House. Ideally, each voter should have exactly the same weight there, so this variance is a measure of the unfairness of the apportionment. The mean weight is the same for all methods; it’s just the number of representatives divided by the total population. In 2010 this was about $1.41\cdot10^{-6}$. The standard deviations under Huntington–Hill and Webster are quite similar; they are $6.54\cdot10^{-8}$ and $6.49\cdot10^{-8}$, respectively. The standard deviations under Adams and Jefferson are about twice as large, $1.14\cdot10^{-7}$ and $1.32\cdot10^{-7}$, respectively. So under these methods, the weight of voters in the House varies by about $±10\%$; whereas under the methods that use a mean, the variation is only about $±5\%$.

Perhaps consult one of the hundred or so books written on the mathematics of the topic.  For example

El-Helaly, Sherif, The mathematics of voting and apportionment. An introduction. Compact Textbooks in Mathematics. Birkhäuser/Springer, 2019


Are those statements Tautology? a.$$\forall x\forall y \exists z (x\neq y)\rightarrow (x\neq z)$$
  b. $$\neg\exists x\forall y \forall z (x=y)\rightarrow (x=z)$$

To revoke a. we need to find a case of $(x\neq z)\land (x=y)$ and for some $x,y$ (x=y) is false, therefore $(x\neq y)\land (x=z)$ is false and it always true.

b. is $\forall x\exists y \exists z (x\neq z)\lor (x=y)$ and it is always true, because for all $x$ there $y$ such $x=y$

a) This seems a tautology. Just take $z=y$.

b) I think you have mistaken in negating the implication.

One has to be careful with the possible domains for formulae like this.  For (a): If the domain has two or more distinct elements x, y, then as already posted one can take z=y.  If the domain has only one element, then (a) is still true, but now it is true vacuously because the condition (x $\neq$ y) can never be satisfied.  If the domain is empty, however, then $\exists$z is always false regardless of the expression that follows.  Conclusion: NOT A TAUTOLOGY.  [I leave (b) for the student]


How to reconstruct a quadrilateral ABCD only using compass and straight edge? Reconstruct a quadrilateral ABCD given length of its sides and the length of the midline between the first and third sides (namely all the segments drawn in the given figure) using compass and straight edge.  The method is parallel translation but i don't know how to do it.

This is really an interesting problem, which appears to be trivial. However, the only solution I can think of is not that trivial:

Let $M, N$ be the midpoints of $AC, BD$ respectively. Then $EMFN$ is a parallelogram, with $EM=BC/2, MF= AD/2$. So $EMFN$ is constructable from the given data. Now let $M_1$ be the reflection of $E$ about $M$ and $G$ the common midpoint of $MN$ and $EF$. It is trivial that $M_1$ is contructable, and that $\vec{EM_1} = \vec{BC}$. Thus $EBCM_1$ is a parallelogram and so $CM_1=EB = AB/2$. Therefore, we can construct $C$ (note that there are two possible solution here). Once we have $C$,

$D$ is the reflection of $C$ about $F$
$B$ is the reflection of $D$ about $N$
$A$ is the reflection of $B$ about $E$

And we are done.

During the construction:

Either BC is fixed by compass and then AD is automatically determined,

or,

AD is fixed and then BC is automatically determined.

However BC and AD cannot both be fixed together.

No construction is possible.

To visualize consider EBCF as a 4-bar mechanism of given link lengths made in steel with EF as fixed link and BC fixed, link AD would break unless made of thin elastic rubber.


Improbable vs Impossible? I was wondering how mathematics in general or any of its sub fields e.g.statistics, probability, define the words Improbable and Impossible.

I get their English meaning, that something is impossible means it is never going to happen and improbable means something is unlikely to happen.

Could someone please provide a mathematical description for these two words as used (if they are used) in some field of mathematics? May be in terms of size of the probabilities.

Thanks.

improbable = something that can happen, but its probability is comparatively low, but not zero. The distinction between probable and improbable is not, as far as I know, exactly defined.

if you roll one hundred dice, it is very improbable that all of them will land as sixes.

almost impossible = something that can happen, but its probability is exactly zero

if you roll a frictionless die, it is almost impossible that it will end up on its edge.

if you keep rolling a die until a six lands, it is almost impossible that you will never stop rolling.

impossible = something that cannot happen.

if you roll one hundred (standard) dice, it is impossible that at least one of them will land as a seven.

probable

if you do get one hundred sixes in a row, it is probable that the casino will kick you out.

almost definite

if you attempt to balance a frictionless die on its edge, it will almost definitely topple to one of its faces

definite

I think you will definitely try to balance a die on its edge. If you do succeed - remember it's due to friction.

Mathematically an event E can be called impossible if and only if $Pr(E)=0$.  I don't believe we can define improbable statically (i.e. in a way that doesn't change based on context).  One attempt, however, might be: given a threshold probability of $p_T$, any event $X$ is considered improbable if $Pr(X) \lt p_T$.


Computing an Integral Using Complex Analysis Does anyone know how to compute

$$
\int^{2\pi}_{0} \frac{1}{3 + \cos\theta} \, d\theta
$$

using tools from complex analysis? I'm not sure how to get started, or if complex analysis is even needed to do this.

Recall that $\cos\theta = (e^{i\theta}+e^{-i\theta})/2$, and change variables to $z=e^{i\theta}$.  Then you are integrating around a circle, and can look for poles.

Let $\cos \theta = \dfrac{z^2+1}{2z}$, $\,d\theta=\dfrac{dz}{iz}$.


Minimum difference of angles between points on square lattice I have integer grid of size $N \times N$.

If I calculate angles between all point triples - is it possible analytically find minimal non-zero difference between those angles?

By brute force, for $N = 4, 5, 6$ the smallest non-zero angle is formed between the segments from $(N-1,N-2)$ to $(0,0)$ and from $(N-1,N-2)$ to $(1,1)$, if the coordinates have range $0$ to $N - 1$ inclusive.

The first segment makes an angle
$\theta_1 = \tan^{-1}\left(\frac{N-2}{N-1}\right)$
with the edge of the lattice, and the second segment makes an angle
$\theta_2 = \tan^{-1}\left(\frac{N-3}{N-2}\right)$
with that edge.
The magnitude of the resulting angle between these segments is
$$| \,\theta_1 - \theta_2| =
 \tan^{-1}\left(\frac{N-2}{N-1}\right) - \tan^{-1}\left(\frac{N-3}{N-2}\right).
$$
(Since $0 < \theta_2 < \theta_1 < \frac{\pi}{4}$, the right-hand side of this equation is positive.)

While this exercise suggests that this might be the formula for larger $N$ as well, it is hardly a proof. I too am curious about the general result now.

Isn't the minimum angle just the one who's arms are along the diagonal (1,1) to (N,N) the back down to (1,2)? This gives the minimum angle as pi/4 - arctan((N-1)/N).


Find a group $G$ and elements $x,y,z ∈ G$ so that $|x|= 5$, $|y| = |z| = 7$, and $|xy| = 35$ but $|xz| ≠ 35$. I am doing an introductory group theory course, however this is one of the more advanced questions I have faced. This is my first question on here, so apologies if it isn't asked as well as it should be!

Question:

Find a group $G$ and elements $x,y,z ∈ G$ so that $|x|= 5$, $|y| = |z| = 7$, and $|xy| = 35$ but $|xz| ≠ 35$.

My attempt:

If we take $G$ as finite, $|G|$ must be divided by both $5$ and $7$, by Lagrange's Theorem.
This would mean the order of $G$ must some multiple of $\operatorname{lcm}(5,7)$, which is $35$.
In this case I can't see how $|xz|$ could be anything other than $35$, as I think it can't be of order $1$, $5$, or $7$, the other divisors of $35$.

By this I feel I should be looking for an infinite Group, however I am very unfamiliar with these and haven't been able to come up with any examples.

If anybody could tell me is my above reasoning correct, and point me in the right direction I would really appreciate it!

Orders of products of group elements can be quite unintuitive. For example, a lot of simple groups can be generated by two elements with order $2$ and $3$.

If you are familiar with cycle notation in $S_n$, consider $x=(1,2,3,4,5)$, $y=(6,7,8,9,10,11,12)$ and $z=(1,2,3,4,5,6,7)$ (I had to add commas because you get numbers with two digits).

Then $xz=(1,3,5,6,7,2,4)$, which has still order $7$.

The fact that $7$ divides $35$ is... a coincidence. You can do "worse" things! For example if you take $z=(5,6,7,8,9,10,11)$ you get $xz=(1,2,3,4,5,6,7,8,9,10,11)$ which has order $11$.

Hint: If $x$ and $y$ commute and $\gcd(|x|, |y|)=1$, then $xy$ has order $|x||y|$.


Number of roots of a quadratic equation modulo 325 Any Help solving this question ?

a) Find ONE solution $\overline x\in\Bbb Z/325\Bbb Z$ such that $x^2\equiv-1\pmod{325}$.  (Hint: CRT and lifting.)
b) How many solutions $\overline x$ to the above equation are there, and why?

$325=5^2\cdot 13$ so let's solve $x^2\equiv -1\pmod{25}$, $x^2\equiv -1\pmod{13}$.

Case $1$: $\mod{25}$

If $x^2\equiv -1\pmod{25}$, then $x^2\equiv -1\pmod 5$ so $x\equiv\pm2\pmod 5$

$$(5k+2)^2=25k^2+20k+4\equiv 20k+4\equiv-1\pmod{25}\implies 4k\equiv -1\pmod 5\\\implies k\equiv 1\pmod 5$$
$$(5k-2)^2=25k^2-20k+4\equiv -20k+4\equiv-1\pmod{25}\implies 4k\equiv 1\pmod 5\\\implies k\equiv 4\pmod 5$$
So we have $x^2\equiv-1\pmod{25}\implies x\equiv\pm7\pmod{25}$

Case $2$: $\mod 13$

We see upon inspection that $x^2\equiv -1\pmod{13}\implies x=\pm 5$

See if you can use CRT to find all solutions to $x^2\equiv -1\pmod{325}$

We can prove by using  Discrete Logarithm and Linear Congruence Theorem that $x^2\equiv a\pmod m$ has zero or two solutions if $m$ has  a primitive root.

Now, $\displaystyle  x^2\equiv-1\pmod{325}\equiv-1\pmod{25}$

$\displaystyle x^2\equiv-1\equiv49\pmod{25}\equiv7^2\implies x\equiv\pm7\pmod{25}\ \ \ \  (1)$

Again, $x^2\equiv-1\pmod{325}\equiv-1\pmod{13}$

$\displaystyle x^2\equiv-1\pmod{13}\equiv25\equiv5^2\implies x\equiv\pm5\pmod{13}\ \ \ \  (2)$

Now apply CRT on $(1),(2)$ to find four in-congruent solutions


What is the differantial equation which has the particular solution $y_p=(x^2-1)e^{-x}+x$? What is the lowest ordered differantial equation which has the particular solution $y_p=(x^2-1)e^{-x}+x$?

For $e^{-x}$ we have $(D+1)$
and for $xe^{-x}\to(D+1)^2$,
For $x^2e^{-x}\to (D+1)^3$

I dont know how to combine them. answer is $(D+1)^3D^2y=0$ btw.

Given: $$y_p = (x^2-1)e^{-x}+x$$
Finding the total derivatives we obtain:
$$dy = [e^{-x}(1+2x-x^2) + 1]dx$$
$$\therefore \frac{dy}{dx} = 1 - e^{-x}(x^2-2x-1)$$
$$\qquad \qquad\qquad\qquad\qquad\qquad\qquad\text{is the required differential equation}  $$


Nash equilibrium in second price sealed-bid auction I'm trying to understand the basics of game theory and the topic of auctions has arisen. I understand the basic concepts of auctions but I'm struggling with second price sealed-bid auctions.

I understand that a weakly dominant strategy in a second price sealed-bid auction is to always bid the amount how much the item is worth to you. However I'm having trouble with some example questions that I've been given.

We have $n$ bidders, $n \ge 2$. There is only one object in the auction. Player $i, i = 1, . . . , n,$ evaluates the object by giving it a valuation $v_i$ , where:

$$v_1 > v_2 > v_3 > . . . > v_n > 0$$

Each player i submits a sealed bid $b_i \ , i = 1, . . . , n$. So, we can describe a bidding profile of all players as $(b_1, b_2, b_3, . . . , b_n)$.

Now what I what I want to understand is are both the listed bidding profiles below nash equilibrium?

A) bidding profile  $(v_1, 0, 0, . . . , 0)$

B) bidding profile $(v_2, v_1, 0, . . . , 0)$

Surely both are a nash equilibrium because every bidder will bid there valuation for the item thus no one has any incentive to change there bid? E.G a weakly dominant strategy

I just wanted some clarification, any help on the matter would be greatly appreciated.

In scenario $A$, only the highest bidder bids their true value and everyone else bids $0$. None of the other players can increase their payoff by changing their strategy alone, as no one values the item above $v_1$. This makes it a weakly dominant strategy.

In scenario $B$, the bidder valuing the item at $v_1$ bids $v_2$ and the bidder valuing the item at $v_2$ bids $v_1$, with $v_1 > v_2$. In this scenario, Player 2 is actually bidding $v_1 - v_2$ above their true value, making it beneficial for this player alone to switch to bidding lower than $v_2$ to obtain $0$ value instead of his current negative value. Thus this is not a weakly dominant strategy.


Show minimum distance to a convex set is a convex function. Show that

$$
g(x)=\inf_{z \in C}\|x-z\|
$$
where $g:\mathbb{R}^n \rightarrow \mathbb{R}$, $C$ is a convex set in $\mathbb{R}^n$ (nor close neither bounded), and $\|\cdot\|$ is a norm on $\mathbb{R}^n$.
Let $x,y$ be in $\mathbb{R}^n$. We need to show that

$$
g(\lambda x +(1-\lambda)y) \leq \lambda g(x)+ (1-\lambda)g(y) \tag{1}
$$

I tried the following:

$$
\|\lambda x +(1-\lambda)y-z\| \leq \lambda\| x -z\| + (1-\lambda)\| y-z\| \,\, \forall {z \in C}
$$
Since

$$
g(\lambda x +(1-\lambda)y)=\inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \leq \|\lambda x +(1-\lambda)y-z\| \,\, \forall {z \in C}
$$

So

$$
g(\lambda x +(1-\lambda)y)=\inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \leq \lambda\| x -z\| + (1-\lambda)\| y-z\| \,\, \forall {z \in C}
$$

I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general

$$
\inf_{z \in C}\|\lambda x +(1-\lambda)y-z\| \nleq \lambda \inf_{z \in C} \| x -z\| + (1-\lambda) \inf_{z \in C} \| y-z\|
$$

Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?

Suppose $C$ is also closed, there exists $z_1,z_2\in C$ such that $g(x)=||x-z_1||$, $g(y)=||y-z_2||$. Then $$\lambda g(x)+(1-\lambda)g(y)\ge ||\lambda x + (1-\lambda)y - \lambda z_1-(1-\lambda)z_2||.$$ Using the convexity of $C$, $z=\lambda z_1+ (1-\lambda)z_2\in C$. Hence, $$\lambda g(x)+(1-\lambda)g(y)\ge ||\lambda x + (1-\lambda)y - z||\ge g(\lambda x + (1-\lambda)y).$$ If $C$ is not closed, probably some limit argument may plus the above reasoning should work.

Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way


If a 3D-cake is cut by $n$ planes yielding the maximum number of pieces, then what is the number of pieces with the cake crust? It is known that a 3D-cake can be cut by $n$ plane cuts at most into $N$ pieces, defined by Cake Number $N=\frac {1}{6}(n^3+5n+6)$. However, some of the pieces would have a crust of the cake as one of their boundaries. How many pieces are with crust? (Entire surface of the cake is assumed to be crust).

To get the maximum number, the cutting planes are in general position.

Consider the 2-D surface (crust) of the cake. Let the intersection of the cutting planes and the crust be denoted as cutting circles. Remove 1 point that is not on any of the cutting lines. Project what remains onto the plane. We now have $n$ circles in general position cutting the plane. We know that the number of regions it divides it into is

$$n^2-n+2.$$

Number of pieces with crust $= ( n^2+n+2)/2$


Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not? Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not?
The question is asking me to prove that $\Bbb R[x]$ is a PID. I'm assuming you would go about this knowing that every field is a PID, but I'm stuck on the differences between $\Bbb R[x]$ and $\Bbb Z[x]$.

There are many ways to answer this question, at different levels, but I would say that I think the right reason is because $\mathbb{Z}$ is not a field.
I will give you 3 supporting evidences for that.

The proof that $\mathbb{R}[X]$ is a PID relies on long division of polynomials, and the fact that you can divide a polynomial by any non zero polynomial, and get a quotient and a remainder in $\mathbb{R}[X]$. This relies on the fact that any nonzero real number is invertible.

This does not work in $\mathbb{Z}[X]$: for example, if you divide $X$ by $2$, you will get a quotient which does not lie in $\mathbb{Z}[X]$. What goes wrong here is that the invertible elements of $\mathbb{Z}$ are $\pm 1$.
OK, this may be not satisfactory, because what this really proves is that $\mathbb{Z}[X]$ is not Euclidean for the degree function. It does not prove that $\mathbb{Z}[X]$ is not Euclidean for another function. Worse, there exists PID's which are not Euclidean.

Another reason (see Bernard's answer): in a PID, any non zero prime ideal is maximal.
This is not the case for $\mathbb{Z}[X]$, since $p\mathbb{Z}[X]$ is a prime ideal but not maximal, for any prime number $p$.

Note that this cannot happen in $\mathbb{R}[X]$ ( $p\mathbb{R}[X]=\mathbb{R}[X]$, since $p$ is invertible in $\mathbb{R}$).
Once again, the main difference is that you have a lot of non invertible elements in $\mathbb{Z}$. Ok, but this is still too vague. SO let's go for :

Thm. Let $A$ be a commutative ring with $1$. Then the following properties are equivalent:

i) $A$ is a field
ii) $A[X]$ is an Euclidean domain
iii) $A[X]$ is a PID.
Only iii)$\Rightarrow$ i) needs a proof: in a PID, any nonzero prime ideal is maximal.
Since $A[X]$ is a PID, it is an integral domain, and thus so is $A$ (this is a subring of $A[X]$)
Since evaluation at $0$ induces  a ring isomorphism $A[X]/(X)\simeq A$, $(X)$ is a nonzero prime ideal. Since $A[X]$ is a PID, $(X)$ is thus a maximal ideal. Consequently, $A[X]/(X)$ is a field, meaning that $A$ is a field.

Find an ideal that is not principal in $\mathbb Z[x]$. Hint: $\mathbb Z$ is a PID so your idea will need something with $x$ in it.

Show that every ideal in $\mathbb R[x]$ is principal. Hint: every number is related by multiplication.


What is the (mathematical) point of straightedge and compass constructions? The ancient discipline of construction by straightedge and compass is both fascinating and entertaining. But what is its significance in a mathematical sense? It is still taught in high school geometry classes even today.

What I'm getting at is this: Are the rules of construction just arbitrarily imposed restrictions, like a form of poetry, or is there a meaningful reason for prohibiting, say, the use of a protractor?

Okay, I seem to be ranting too much in comments, so let me try to put forth my points and opinion here (as a community wiki, since it is opinion).

First, to address the question: What is the significance of straightedge and compass constructions within mathematics? Historically, they played an important role and led to a number of interesting material (the three famous impossibilities lead very naturally to transcendental numbers, theories of equations, and the like). They are related to fascinating stuff (numbers constructible by origami, etc). But I would say that their significance parallels a bit the significance of Cayley's Theorem in Group Theory: although important historically, and relevant to understand the development of many areas of mathematics, they are not that particularly important today. As you can see from the responses, many find them "fascinating", many find them "boring", but nobody seems to have come forth with an important application.

Now, addressing the issue of teaching it at K-12. Let me preface this by saying that I am a "survivor" of the New Math, which came to Mexico (where I grew up) in the 70s. I would say I became a mathematician in large part despite having been taught with New Math, rather than because of it. Also, I did not attend K-12 education or undergraduate in the U.S.; I am a bit more familiar with undergraduate at the level of precalculus and above thanks to my job, but not very much with the details of curricula in sundry states in the U.S. So I may very well have a wrong impression of details in what follows.

Now, one problem, in my view, is that when we talk about "math education", we are really talking about two different things: numeracy and mathematics. This is the same phenomenon we see when we think about "English class". I suspect that English Ph.D.s are nonplussed at people who think they spend their time dealing with grammar, spelling, punctuation, etc, just as mathematicians are nonplussed that people think we spend our time multiplying really big numbers by hand. English education has two distinct components, which we might call "Literacy" and "Literature". Literacy is the component where we try to teach students to read and write effectively, spelling, punctuation, etc. Making themselves understood in written form, and understanding the written word. On the other hand, Literature is the component where they are introduced to Shakespeare, novelists, book reading, short stories, poetry, creative writing, historical and world literature, etc. We consider English education at our schools a failure when it fails in its mission with regards to literacy: we don't consider it a failure if students come out not being particularly enthused with reading classic novels or don't become professional poets, or even if they don't care or like poetry (or Shakespeare).  The professional English Ph.D. engages in literature, not in literacy. Literacy is the domain of the grammarian.

Mathematics education likewise has two components; numeracy (to borrow the term from John Allen Paulos) and mathematics. Numeracy is the parallel of literacy: we want children to be able to handle and understand numbers and basic algebra, percentages, etc., because they are necessary to function in the world. Mathematics is the stuff that mathematicians do, and which we all find so interesting and beautiful.

"Mathematics" also includes advanced topics that are necessary for someone who is going to go on to study areas that require mathematics: so the physicists and engineers need to know trigonometry; business and economists need to know advanced statistics; computer science needs to know discrete mathematics; etc. Much like someone going on to college likely needs more than simple command of grammar and spelling.

Part of the problem with math education is that it so often conflates numeracy with mathematics; another part of the problem is that mathematics was included in the "classical" curriculum for much the same reason as Latin and Greek were included: historical reasons, because every "gentleman" was expected to know some Latin, some Greek, and some mathematics (and by "mathematics", people meant Euclid). To some extent, we still teach geometry in K-12 because we've always taught geometry. But it is not part of the numeracy curriculum.

It would be wonderful if we had the time in K-12 to teach students both numeracy and mathematics; it is a fact that today we are failing at both. We don't perform better in teaching students numeracy by teaching them beautiful mathematics, even if they understand and appreciate them, just like making them really understand and care for Shakespeare's plays (through performance, say) will make them able to understand a set of written instructions, or write a coherent argument.

Trying to excite students about mathematics is all well and good; but numeracy should come first. Trying to excite students about the wonderful world of books, plays, and poetry is all well and good, but we need them to be able to read and write first, because that's part of what they will need to function in society.

Constructions by straightedge and compass can become an interesting part of mathematics education, just like geometry. I just don't think they have a place in numeracy education. But they are taking the time we need for that numeracy education. Trigonometry used to be part of the numeracy education that people needed; this is no longer the case today. Trigonometry, today, is a foundational science for more advanced studies, not part of numeracy. But we are still teaching trigonometry as a numeracy subject (hence the rules, recipes, mnemonics, and the like). We could try to turn trigonometry into a mathematical subject, sure; or we could postpone it until later and only teach it to those for whom it is an important foundation.

Added. To clarify: I don't mean to say that the "solution" is to do less math and more numeracy. I think the solution is likely to be complicated, but the first step is to identify exactly what parts of what is currently branded as "mathematics" are really numeracy, and which parts are mathematics. Trigonometry is taught as part of "advanced mathematics", but it is taught as rote and rules because it was really numeracy. We don't need to teach trigonometry as numeracy any more, so we shouldn't. If it is to be taught, it needs to be taught in the right context. Geometry is similar: geometry used to be taught as basic numeracy because "every educated person should know geometry"; (of course, "educated" at the time meant "rich and land owner, or with aspirations in that direction"). We don't need most of geometry as basic numeracy, we want it now as mathematics. So teaching geometry as numeracy is a waste of time, and it takes up the numeracy time that should be spent in other things. (It can still be taught during the mathematics time). I certainly don't say "drop all the math, concentrate on the numeracy". I say, "when dealing with numeracy, concentrate on the numeracy, not the math, and don't confuse the two." Nobody seems to confuse spelling rules with reading novels, because we separate literacy from literature. Too many people confuse arithmetic with mathematics, because we don't separate them.

The reason I talk about dropping trigonometry and doing some basic statistics is precisely that: trigonometry is being taught as part of the numeracy curriculum, when it shouldn't. Basic statistics, say at the level of the wonderful How to Lie With Statistics, is not taught as part of numeracy. But in today's world, there is a far better case for statistics being part of the basic numeracy education than trigonometry. Everyone coming out of High School should know that taking a 10% pay-cut and then getting a 5% raise does not mean you are now at 95% of your old salary (go do a spot check, see how many people think you are). They need to know the difference between average and median, so they are not misled by statements about "the average salary of the American worker". They should understand what "false positive" and "false negative" means. They should be able to interpret graphs (even the silly ones on the cover of every USA Today issue) and be able to spot the distortions created by chopping axes, etc. These are numeracy issues.

Likewise high school geometry: it is trying to be both numeracy and mathematics, and I think it generally fails at both. There are some components of geometry that are part of numeracy, they ought to be treated that way, but the parts that are mathematics should be separate.

One problem I have with Lockhart's lament is that he does not make clear the distinction between numeracy and mathematical education. The nightmare he paints for the musician and the artist is precisely that musical education is being turned into the equivalent of numeracy/literacy education, thus doing a disservice to music-as-an-art.

The science of mathematics that we all know and love has some intersection with numeracy and arithmetic, but we all know it is a limited intersection; just as the study of literature has some intersection with the study of grammar an spelling, but the intersection is limited. The main purpose of K-12 education (or at the very least, K-6 or K-8) should be numeracy, with some limited forays into mathematics (just as the main purpose will be literacy with some limited forays into literature). The way to teach numeracy is necessarily different from the way to teach mathematics. Numeracy requires that we memorize multiplication tables, for all the horror this will cause to modern education people; this of course is a far cry from teaching the mathematics of multiplication, which may very well be very interesting and awaken the child's curiosity and wonder at the world. That can be done within the context of mathematical education, but it shouldn't be done in the context, and at the expense of, numeracy.

Trigonometry used to be part of numeracy; it no longer is. It is now either mathematical or foundational for advanced studies, so it should be treated as such. Geometry used to be taught for reasons which no longer hold, and to some extent we continue teaching it as a historical legacy; we shouldn't. Those components which are numeracy should be taught as that, and we could move the rest (including constructions with compass and straightedge) to the more creative, mathematical education side of the equation.

Anyway, I've ranted long enough, and probably made myself a few detractors along the way...

There is a reason why Euclid's Elements are so revered.  It builds logic, critical thinking/reasoning, spatial skills, and some.  Also, it is beautiful. It has been as influential as any one book in the history of time.  I think the undertones of conversations like this usually imply one's (perhaps subconscious) preference for quality or for quantity.  Are you a Journey person or a Destination person?    So many modern ideas are so obviously sprung from this foundation work.  The thinking of so many that shaped our society was very much shaped by Elements (Government, Law, Architecture, Building Construction, City Planning, Science). It deserves more than a cursory glance for ourselves and our kids.  Euclid's Elements was such a QUALITY work that it was taught in more or less its original form for two thousand years.  Getting to your question, the axiomatic structure used in Euclid's straight edge and compass constructions has helped move the idea of proof along. Without the definitions, common notions, postulates we are talking about a bastardized version that is less mathematically significant yet still important in the development of spatial skills. Remember, math is not reality and is all about imposing arbitrary restrictions to find the truths they bring about (and often beauty).  As far as not using a protractor, this just creates an extra set of constraints to make the subject more challenging and make you work/think a little bit for the answer. Why not use a computer or a geometer for that matter? Why are you not allowed to phone a friend on Jeopardy? Constraints often require you to use more creativity to solve a problem and exist in the real world in spades.


Is any uncountable, scattered subset of [0,1] dense? Just out of curiosity, I am wondering if any uncountable, scattered subset $U\subset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=\{\frac{1}{n} : n\in\mathbb{N}\}$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.

But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).

There are no uncountable scattered subsets of $[0,1]$.  This follows from the theory of Cantor-Bendixson rank, for instance.  Given any scattered $A\subset[0,1]$, there must be some ordinal $\alpha$ such that the $\alpha$th Cantor-Bendixson derivative $A^\alpha$ of $A$ is empty.  The least such $\alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable.  Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.

An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,\frac{1}{2}]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set


What's the difference in the interpretation of $y = x$ and $y(x) = x$? Both $y = x$ and $y(x) = x$ describe a line, but what are some of the differences between their uses, and can they ever be used interchangeably?

$y=x$ simply represents a line passing through origin, while $y(x)=x$ here $y$ is behaving as a function of $x$, in this case your function is identity function , you have to specify a domain and a codomain. Other possible functions are like $y(x)=x^2$ and so on for example.

The only reason y(x) = x describes a line is because, for any value of x, y(x) will be the same, as per the definition of y(x). As Kushal Bhuyan said, you can define y(x) to be any function you want. Also, when you graph y = x, there will be an x - axis and a y - axis, while if you graph y(x) = x, there will be an x - axis and a y(x) axis (or any other function that you choose).


Is there more simple way to get a doughnut property? The terms and my question appear from the Halbeisen's book Combinatorial set theory with a gentle introduction to forcing.

For subsets $a$, $b$ of $\omega$ such that $b-a$ is infinite, define a doughnut as
$$[a,b]^\omega:= \{x\in [\omega]^\omega : a\subseteq x\subseteq b\}.$$
(Imagine the Venn diagram of $a$, $x$ and $b$ then you can guess why this set is called a doughnut.)

A collection $\mathcal{A}\subset [\omega]^\omega$ has a doughnut property if there are $a$ and $b$ such that either $[a,b]^\omega\subseteq \mathcal{A}$ or $[a,b]^\omega\cap \mathcal{A} = \varnothing$ hold. If previous conditions hold for $a=\varnothing$, we call $\mathcal{A}$ has the Ramsey property.

Under the axiom of choice we can find a collection $\mathcal{A}$ that has a doughnut property but not Ramsey property. Here is my attempt:

  Define $\sim$ as
  $$a\sim b \iff a\triangle b\text{ finite}.$$
  Then $\sim$ is an equivalence relation.
  For each $x\in [\omega]^\omega$ choose a representative $r_x$. (that is, choose $r_x$ such that $[x]=[r_x]$ for each equivalence class.) Take a maximal antichain $X\subseteq[\omega]^\omega$ which is not contain $\omega$.

  Let define $\mathcal{A}\subseteq[\omega]^\omega$ as follow: for given $y\in[\omega]^\omega$,

  if there is some $a\in X$ such that $a\subseteq y$, then $y\in\mathcal{A}$.
  if not, $y\in\mathcal{A}$ if and only if $|y\triangle r_y|$ is even.

  then for $a\in X$, $[a,\omega]^\omega\subseteq\mathcal{A}$. But for any $y\in [\omega]^\omega$, we can find some $a\in X$ such that $a$ and $y$ are comparable and

  if $y\subseteq a$ then we can find some subsets of $y$ contained in $\mathcal{A}$ and not in $\mathcal{A}$.
  if $a\subseteq y$, we can consider a set $b\subseteq a$ such that $a-b$ is infinite. (since $a$ is infinite) then $b$ has subsets lies on $\mathcal{A}$ and not on $\mathcal{A}$.

  thus $[y]^\omega$ is neither a subset of $\mathcal{A}$ nor disjoint from $\mathcal{A}$.

My questions are: Is my construction valid? and is there more simple construction?

I think your construction works fine, adding the additional condition on the maximal antichain $X$: it should contain an infinite set $a$, whose complement is also infinite (that such $X$ exists can be quickly established by Zorn's lemma). This condition is needed for the doughnut $[a, \omega]^{\omega} \subseteq \mathcal{A}$ (the definition of the doughnut requires that $\omega \setminus a$ is infinite).


Quantifying "weight" or "control" of a variable to the value of a function? Say I have an equation

$$x = f(x)$$

I know that here, the independent variable $x$ controls 100% of the variability of the value of the function. It is the only "knob" that I need to turn to manipulate the value of the function. That is, compared to another function:

$$xy = f(x,y)$$

In this function, $x$ does not control the entire equation. The value of $y$ also controls the equation. I have two "knobs" to control, and each has its independent effect on the function.

My question is, how do I measure how much effect a variable has over a function?

To clarify: say my equation was

$$x + \frac{y}{1000} = f(x,y)$$

Then increasing the value of $x$ by 1 will increase the value of the function by 1. But to get to the same effect using $y$, I have to increase its value by 1000. It seems easy enough to say that $x$ has a weight 1000 times that of $y$.

Similarly, in the first equation $x = f(x)$, $x$ has 100% of the control. In the second equation, each variable has 50% of the control.

But these values I'm saying - about how much "control" a variable has on the function's value - are only intuitive. I cannot prove their truth.

How do I know how much control a variable has over the value of the function, especially in the general case when an equation becomes complex, like when the variable cannot be factored out:

$$x + \frac{y}{x} = f(x,y)$$

Or when the value of the variables are "polluted" by constants or itself:

$$\frac{\sqrt{x + 1000}}{99} + \frac{y}{y+1} = f(x,y)$$

Or when there are more than two variables:

$$x + yz = f(x,y,z)$$

Sounds like the weight is just the partial derivatives with respect to the variable you wish to measure.

$$x+y1000=f(x,y)$$

The derivative with respect to $y$ is $1000=dxf(x,y)/dy$ using the power rule.

The derivative with respect to $x$ is $1=dyf(x,y)/dx$

If the equation were:

$$x^2+1000y^3$$

then respectively

$$3000y^2=dxf(x,y)/dy$$

and

$$2x = dyf(x,y)/dx$$

and the "weight" depends very much on the point at which $x$ is being evaluated, as it can change.

for $$x+y/x=f(x,y)$$, I am hazy on my rules for differentiation, but I looked it up:

for dx/dy we can rewrite:

$$x+y*(x^-1)$$

Y is a constant here.  $$dx/dy = 1 + -1*(x^-2)*y$$

Always look how you can rewrite an equation, and also try wolframalpha.com if you dont already know about it.


Number of permutations such that $a-b+c-d+e-f+g-h=0$ All possible permutations $\left\{a,b,c,d,e,f,g,h\right\}$
 of the set $A=\left\{1,2,3,4,5,6,7,8\right\}$ are formed.  How many of those permutations satisfy

$$a-b+c-d+e-f+g-h=0$$

My Try:

we have for example

$$(2-1)+(4-3)+(5-6)+(8-7)=0$$ and each of the number in brackets if we treat them as four letters, they can be arranged in $4!=24$ ways.Now in all these possible permutations if we multiply with negative sign we get a different permutation.

So total is $48$.

similarly for $$(2-3)+(4-1)+(5-6)+(8-7)=0$$ we get $48$ permutations.

but i feel this is an informal approach. Any clue for better approach?

The first thing to notice is that the sum of all the numbers from $1$ through $8$ is $36$ so the numbers that are added must sum to $18$ as must the numbers that are subtracted.  We need to find the number of ways to divide the numbers into two sets of four such that each set adds to $18$.  The $8$ has to go in one set, so we look for ways to have three numbers sum to $10$.  They are $721,631,541,532$ so there are $4$ partitions of the set.  For each partition we have two ways to choose which set is added, $4!$ ways to choose the order of the added set, and $4!$ ways to choose the order of the subtracted set.  This gives a total of $4\cdot 2 \cdot 4! \cdot 4!=4608$

The total of the numbers is $36$.  Divide by $2  = 18$.

So you may have split the numbers in the following groups $= ({1,3,6,8}$ and ${2,4,5,7})$ or (${1,4,6,7}$ and ${2,3,5,8}$) or (${1,4,5,8}$  and ${2,3,6,7})$.

Each group  could be permuted in $4!$ ways such that $a-b+c-d+e-f+g-h=0$ is true and each could pair could be swapped for add ad subtract and thus there are three pairs and hence 8 to give a total of $ 8\times24\times24 = 4608$


20 books 5 different shelves So I'm trying to answer this question and am not sure if my answer is correct. In the text book I'm using, this question asked before combinations are even introduced (only permutations) so I'm not entirely sure if I'm in the right direction. Is it possible to answer these using permutations?

Consider a bookcase with 6 shelves and suppose that we have 15 different books to place on the shelves. How many different ways are there to place the books on the shelves if the left to right order on each shelf is unimportant?

The order being unimportant implies it's a combination question, and having different "containers" lead me to use the stars and bars approach.

$C(n + k -1, k) = C(5 + 15, 15) = C(20,15)$ ways to put the puts on the shelf when order is unimportant.

The next part of the question says the order is important, assuming the above answer is right, would the number of ways with order simply be $15! * C(20, 15$)? Furthermore if each shelf is to get at least one book, is C(14, 9) correct when order is unimportant?

Some confirmation or correction would be appreciated, as would alternate methods for solving. Thanks!

If the left-to-right order on each shelf is unimportant, there are $6^{15}$ ways to place the books (as each book should be assigned with its shelf number).

If the left-to-right order is important, you have first order all books in the left-to-right and top-to-bottom order, and then place 5 "containers separators" between them (consider the similar task of writing up 15 different printable characters in a text file containing no more than 6 lines - i.e. no more than 5 newline characters). The answer would be $15! \times \binom{20}{5}$.

combination

$nCr = \dfrac{n!}{(n-r)r!}$

Here $n=$ factorial of no of units/items

$r=$ chance no

example $20$ similar books be placed on $5$ different selves...

$\dfrac{20!}{(20-5)5!}$


Proof that degree $n$ polynomial has at most $n$ roots (counting multiplicity) I am trying to prove the following statement:

  If the polynomial $f(x)$ of degree $n$ has roots $a_1,a_2,...,a_k$ with multiplicities $\alpha_1,\alpha_2,...,\alpha_k$ in a field $F$ (with $\alpha_i\geq 1$), then $f(x)$ has $(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}$ as a factor. In particular, a polynomial of degree $n$ in one variable over a field $F$ has at most $n$ roots in $F$, even counted with multiplicity.

This is the proof that I wrote

  For the first part, we proceed by induction on $k$. If $k=1$, then by definition, we may write $f(x)=(x-a_1)^{\alpha_1}q(x)$. Now, suppose that the statement is true for up to $k$ roots, and consider a collection of $k+1$ roots. Using the inductive hypothesis, we may write
  $$f(x)=(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}q(x)\quad\text{ and }\quad f(x)=(x-a_{k+1})^{\alpha_{k+1}}q'(x)$$
  Equating
  $$(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}q(x)=(x-a_{k+1})^{\alpha_{k+1}}q'(x)$$
  and using the fact that $(x-a_{k+1})$ is an irreducible element that is distinct from any on the left hand side gives us that $(x-a_{k+1})^{\alpha_{k+1}}\mid q(x)$, which proves the first statement.

  Now, suppose that a polynomial $f(x)$ of degree $n$ as more than $n$ roots in $F$, counted with multiplicity. From what we showed, we must have $(x-a_1)^{\alpha_1}\cdots (x-a_k)^{\alpha_k}\mid f(x)$. But $F[x]$ is an integral domain so this is impossible, as the divisor has a higher degree than $f(x)$.

I was first wondering if the proof was correct. Moreover, in proving the first half of the statement I used the fact that an irreducible element is prime, which leverages the fact that $F[x]$ is a U.F.D. Is there a more general method of proof that works on general integral domains?

I'd state the result as follows:

  If $F$ is a field and $f \in F[x]$ not zero, then $f(x)=(x-a_1)\cdots(x-a_m)g(x)$, where $m\ge 0$, $a_i \in F$, and $g$ has no roots in $F$. In particular, $m \le n$, the degree of $f$.

and prove it by induction on the degree of $f$:

If $f$ has no roots in $F$, then take $m=0$ and $g=f$. In particular, this holds for $f$ nonzero constant.

If $f$ has a root $a_1$ in $F$, then $f(x)=(x-a_1)q(x)$ by polynomial division. The result follows by induction applied to $q$, whose degree is less than that of $f$.

No need to bring multiplicities into this argument until now. But then it is easy to conclude a result about multiplicities by grouping equal factors $x-a_i$.

Note that unique factorization is not a part of the argument.


Using variable spearation to solve a linear ODE of first order Suppose I want to solve the ODE:

$$f'(x) + 2f(x) = 3$$

I want to use variable separation, so I get:

$$ f'(x) = 3 - 2f(x)$$

Now I want to divide by $3-2f(x)$ but I am unsure what I need to assume on $f$ in order to do that. That is, does that expression have to be non zero for all x, or only for a certain x? After that point I know what to do but I don't understand what I am supposed to do here.

thanks

The constant function $f(x)=\frac32$ is one solution. All other solutions must satisfy $f(x) \neq \frac32$ for all $x$ since solution curves can't cross (by the uniqueness theorem), so they are given by $f'(x)/(3-2 f(x))=1$ and so on (the usual procedure).

If you want to solve the ODE, you 've first to solve the homogenueos equation, i.e. $f'(x) + 2f(x) = 0$ and then fing the particular integral. You can easily rearrange the homogeneous equation as:
\begin{equation}
\frac{f'(x)}{f(x)}=-2
\end{equation}
So that
\begin{equation}
\frac{df}{f}=-2 dx
\end{equation}
When integrating you get $f(x)=c e^{-2 x}$, where c is the integration constant. Your particular integral is a simple constant, since the right hand side of your differential equation is a constant. The solution is $f(x)=c e^{-2 x}+ 3/2$


Is there a proof that is true for all cases except for exactly one case? I was curious if there were any such proofs which state that a thing is true always EXCEPT for exactly one instance. As in, for some reason, there is only one instance where the proof is false, but it is true for all other objects. I understand that if it is not true in that one case that it is not necessarily a proof, I was just wondering if there were any "proof-like things" of this form.

Here's a famous one: $\mathbb{R}^n$ has a single differentiable structure (up to diffeo) except for $n=4$, in which case it has uncountably many.

These posts may be of interest:

https://mathoverflow.net/questions/16035/a-reference-for-smooth-structures-on-rn

https://mathoverflow.net/questions/24930/differentiable-structures-on-r3

Each object you can touch in your body is not your tongue, but it fails when the object you touch is your tongue.
Each person you look in this planet is not your mother, but it fails when you look at your mother.
Each person in this question is trying to showoff with advanced math concepts to answer a trivial question like this and is not called Voyska, but it fails for me.


Why the life of a light bulb follow an exponential law? Why the life of a light bulb follow an exponential law ? My teacher always said that, but he can't explain why it's this. So why did we decided that a light bulb follow an exponential law ? Where does this fact comes from ?

The exponential distribution comes from the assumption that light bulbs do not age. They just randomly, for each moment in time, decide whether to fail or to keep working, with the same probability regardless of how old they are. This leads to the exponential distribution.

Is this the way light bulbs actually works? I have no idea, but since this is such a ubiquitous example I would at the very least find it reasonable that it's close to the truth. Someone probably checked it once a long time ago, and it's just been the established truth since.

This was a great question, one I suffered with for a while.  If you accept the premise that a lightbulb can be modeled by a memoryless exponential function, then you are forced to accept some rather embarrassing results.

For instance, let's say you have $2$ lightbulbs, each modeled by a memoryless exponential with a mean of $30$ minutes.  An experiment is done where the first bulb is run until it fails, the $2^{\mathrm{nd}}$ one is immediately put in and let run to it fails and you record only the time of the $2^{\mathrm{nd}}$ bulbs failure, let's say $60$ minutes.  Using Bayesian Inference, you want to use the time the  $2^{\mathrm{nd}}$ bulb failed to update you knowledge about when the $1^{\mathrm{st}}$ bulb failed. With Baye's Rule it is straight forward to show that you've learned nothing, that the failure of the $1^{\mathrm{st}}$ bulb is just as likely to have been at one second in, $59$ minutes in, or its mean, $30$ minutes in.  This is preposterous, and if you apply this to bulbs or even, worse bridges, you are making a big mistake.

A coin flip is memoryless, the arrival of a single photon in a very low light environment is memoryless, not a lightbulb, because a lightbulb failure is not a single event, but rather a series of events where material is slowly lost from the filament finally leading to failure, and it cannot be modeled with a memoryless distribution.  I wish they would stop using lightbulbs as the cannonical example of exponential distributions and Bayesian Inference.


Reformulating Real Analysis using any basis of $\mathbb{R}^p$ To illustrate the question that I'm asking, consider the theorem:

  If the partial derivatives of the $f$ exists in a neighbourhood of $c$
  and are continuous at $c$, then $f$ is differentiable at $c$.

However, when we talk about partial derivatives, we talk about the directional derivative in the direction of the standard basis vectors (0,...,1,...0)of $\mathbb{R}^p$.

First of all, why do we choose and formulate the theorems with respect to these basis ? I mean would any orthonormal basis of $\mathbb{R}^p$ work in the same way ?

I mean, for example, if we think geometrically, for $\mathbb{R}^3 $the choice of $x,y,z$ coordinates is arbitrary, hence any orthonormal basis of $\mathbb{R}^3$ should work, but how about any basis of $\mathbb{R}^3$
?

Secondly, if we were to formulate, for example, the above theorem wrt any basis of $\mathbb{R}^p$, how could we do that ?

Edit:

Please provide an argument to your statements.

Edit 2:

In the book that I'm using (The elements of Real Analysis by Bartle), for example, the above theorem is given before stating that the derivative of $f$ can be written in terms of the gradient, so to answer to this question, I cannot use this fact.

First, your quoted theorem

  If the partial derivatives of the $f$ exists in a neighbourhood of $c$
  and are continuous at $c$, then $f$ is differentiable at $c$.

May or may not actually be true depending on how you define differentiable. If your are referring to Gâteaux differentiability then you are correct, but surprisingly not if you mean Fréchet differentiable. See this post for a counterexample.

Your question actually touches on something very subtle. Consider the Jacobian of the function $f$--that is, the $p \times p$ matrix $J$ (in the standard basis), defined by

$$
J_{ij} = \frac{\partial f_i}{\partial x_j}.
$$

Where $f_i$ is the $i$th component of $f$ and $x_j$ the $j$th coordinate in the standard basis. Suppose now we transform from the standard basis to a new basis by the transformation $T : (x_j) \mapsto (x_j')$. We now have a new function

$$
\tilde{f} = T\circ f\circ T^{-1}.
$$

(That is $\tilde{f}$ first takes the vector in the new coordinates and converts back to the old, applies $f$, and then converts back to the new coordinates.) Using the chain rule, it can readily be deduced that the new Jacobian $J'$ is related to the old by

$$
J' = PJP^{-1},
$$

where $P$ is the change of basis matrix of the transformation $T$. Thus, we are free to work in whichever coordinates we may like, and may readily convert to a different set of linear coordinates by applying the standard change of basis matrix from linear algebra. We are thus at liberty to take the standard basis because, as this argument shows, all the other partial derivatives in any other linear coordinate system exist and are given by a change of basis with respect to the standard coordinates.

Addendum: This reasoning also explains a certain puzzle: when we encounter a vector function $\mathbf{u} : \Bbb R^3 \to \Bbb R^3$ in physics, there are nine derivatives $\partial u_i/\partial x_j$ for $i,j = 1,2,3$. Why then do we only ever encounter two particular expressions $\nabla \cdot \mathbf{u}$ and $\nabla \times \mathbf{u}$ in our physics books?

It is because these combinations of partial derivatives are (in some sense) the only "derivatives" invariant under orthogonal transformations (and $\nabla \times \mathbf{u}$ is only "kind of" invariant under coordinate transformations, see this.)

We can see that $\nabla \cdot \mathbf{u}$ is invariant under coordinate transformations since it is the trace of the Jacobian, and the trace of a matrix is invariant with respect to change of basis.

Any basis would suffice for this theorem.
We just use the standard basis because its the easiest to work with.

The best way to look at this is given that the theorem holds for a basis $v_i$ of $\mathbb{R}^p$ and $e_j = \sum a_{i,j}v_i$ then $\frac{\partial f}{\partial x_j} = \sum a_{i,j}\frac{\partial f}{\partial v_i}$.


For which $a$, the DE $f'(x)=f(ax)$ has non- zero solution The question is

    For which $a$, the differential equation $$f'(x)=f(ax)$$ has non- zero solution(s)?

Of course, for $a=1$, we have $f'(x)=f(x)$ which has the solution $f(x)=ce^{x}$.

For $a=-1$, the DE is $f'(x)=f(-x)$ which gives
$$f''(x)=-f'(-x)=-f(x)$$
and this gives $$f(x)=c_1\cos x+c_2\sin x$$
where from $f'(x)=f(-x)$, we see $f'(0)=f(0)$ and thus for $a=-1$, the solution is
$$f(x)=c_1(\cos x+\sin  x).$$
For $a\ne\pm1$, I have no idea how to proceed.

Thanks for helps.

You may did as for $f'(x)=f(-x)$ for general case $f'(x)=f(ax)$.

Note that from $f'(x)=f(ax)$ we have
    $$f''(x)=af'(ax)=af(a^2x),\quad f'''(x)=a^3f'(a^2x)=a^3f(a^4x), \ldots$$
    and in general
    $$f^{(n)}(x)=a^{n(n-1)/2}f(a^nx)$$
    which gives $$f^{(n)}(0)=a^{n(n-1)/2}f(0).$$
    Therefore the $f(x)$ is given by
    $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n=f(0)(1+\sum_{n=1}^\infty \frac{a^{n(n-1)/2}}{n!}x^n)$$
    and by ratio test, we see that $$|a|\leq 1.$$

Hint. Let $y=f(x),\,w=ax,$ then we get $y'_x=y'_ww'_x=y(ax),$ which gives $$ay'=y,$$ where $y$ is now a function of $w.$ You should now be able to proceed.


Conditional probability that first card is spades given that second and third one are spades Attempt:
The original sample space is just the 52 cards. Now, Suppose the events that second and third card are spades have occured. We know now have a reduced sample space of $50$ cards which contains 2 spaces only, so
$$P( \text{first card spade} ) = \frac{2}{50} = \frac{1}{25} $$
but the correct answer should be 11/50 which I have no idea how they got it. What is the correct reasoning with this type of problems?

Added: Actually, thanks to the given hint I think I missunderstood.
So, basically we $\bf have$ 13 spades. Given the event that reduces
our sample space now we have $50$ cards and $13-2 = 11$ spades. Hence,
$$ P = \frac{11}{50} $$

You got the correct answer, though you didn't use conditional probability. Your method works here because, as you noted, knowing that the 2nd and 3rd cards are spades is basically equivalent to removing them from the deck before we picked the 1st card.

The way to do this with conditional probability would be to take the probability of both events -- i.e., the probability that the first three cards are all spades -- and divide by the probability of the observed event -- i.e., the probability that the second and third cards are both spades. We get

$$\dfrac{\frac{13}{52}\cdot \frac{12}{51}\cdot\frac{11}{50}}{\frac{13}{52}\cdot \frac{12}{51}} = \dfrac{11}{50}   $$

I think is goes like this:

$P(A):=$ 1st card a space; $P(B):=$ only 2nd & 3rd cards are spades, first might be a spade

Event E is the questions noted above.

$$P(E)=P(A \cap B)/P(B)$$

$$P(A \cap B)=\text{all three cards are spades}=13/52*12/51*11/50$$

$$P(B)=\text{all three cards are spades + 2nd & 3rd cards are spades, first might be a spade}$$

$$P(B)=\underbrace{13/52*12/51*11/50}_{\text{ all three are spades}} + \underbrace{39/52*13/51*/12/50}_{ \text{ first not spade, rest spades}}$$

Plug into $P(E)$ equation and you get 11/50!


Prove that $\lim \limits_{n\to \infty}\frac{n^{\alpha}}{(1+p)^n}=0$ Prove that if $p>0$ and $\alpha\in \mathbb{R}$ then $$\lim \limits_{n\to \infty}\dfrac{n^{\alpha}}{(1+p)^n}=0.$$

No ideas how to prove it. Can anyone help please?

Prove without using logarithm.

Consider $\sum_{i=1}^\infty \frac{n^\alpha}{(1+p)^n}$, and perform the Ratio Test for convergence of this series:

$$\lim_{n\to \infty} \left| \frac{(n+1)^\alpha}{(1+p)^{n+1}} \cdot\frac{(1+p)^n}{n^\alpha}\right| = \lim_{n\to \infty}\left|\left(1 + \frac{1}{n}\right)^\alpha \frac{1}{1+p} \right| = \frac{1}{1+p} < 1$$

As this limit is less than 1, the series is convergent, which is only possible if the sequence $\frac{n^\alpha}{(1+p)^n} \to 0$ as $n \to \infty$.

Take the logarithm and show that it goes to -infinity. If the logarithm goes to -infinity, then the sequence itself goes to 0.


Is there a standard category-theoretic way to express a loop or quasigroup? The standard way to encode a group as a category is as a "category with one object and all arrows invertible".  All of the arrows are group elements, and composition of arrows is the group operation.

A loop obeys similar axioms to a group, but does not impose associativity.  Inverses need not exist, but a "cancellation property" exists -- given $xy = z$, and any two of $x$, $y$, and $z$, the third is uniquely determined.

Quasigroups need not even have a neutral element.

Given the lack of associativity, arrows under composition do not work to encode loop elements.

Is there a natural way to do this?

As expressed by Qiaochi Yuan in this and this comment, the way that category theory applies to studying loops and quasigroups is in the form of a category whose objects are loops, resp. quasigroups.

Only a few structures can actually be described as categories having certain special properties (among which sets, groups, partially ordered sets). For a structure to have any chance of being a "special type of category", it is of course necessary that the defining properties for a category are somehow satisfied by the structure in question.

For loops and quasigroups, this is not obvious to say the least, due to the lack of associativity (which is all-important in category theory).

I know this is an older thread but I've been thinking about basically this question for a few months now
Something I've come up with is that a quasi-group requires certain automorphisms to exist on the categorical product of 3 elements.
An example of this, in the Category of sets the products AxB, CxA and BxC are all isomorphic to subsets of AxBxC which is isomorphic to CxAxB and BxCxA.
These isomorphisms are what allow us to distinguish between a binairy function and it's left and right inverses, and are generalizable withing the context of categorical products (I believe) as the proper morphisms should exist between categorical products of pairs and products of triples.
Once these automorphisms are established, I believe quasi-groups arise naturally from the existence of morphisms between categorical products of pairs and objects.
This is all really sketchy right now, I'm honestly just getting into category theory but I believe the above works?


Checking the existence of a solution for a set of linear equality and ineaulity equations I would like to know if there is a method to check the existence of the solution for a given set of linear equations composed with both equalities and inequalities?
I'm not interested in the solution, but whether there exist at least one solution or not.

Edit1: Mixed systems of linear equations and inequalities,

$a_{1,1} x_1 + \dots + a_{1,n} x_n = b_1$

$a_{2,1} x_1 + \dots + a_{2,n} x_n = b_2$

$\dots$

$a_{m-1,1} x_1 + \dots + a_{m-1,n} x_n \geq b_{m-1}$

$a_{m,1} x_1 + \dots + a_{m,n} x_n \geq b_m$

The number of equations $m$ might be less than, equal to or greater than $n$ (the number of unknowns).

Edit2: Example1,

$
\begin{cases}
 2x-y \geq -3 \\
 -4x-y \geq -5 \\
 -x+y=4
\end{cases}
$

There is no solution for above system. I'm looking for a systematic (algorithmic) way to determine if such systems of equations have any solution or not.

Yes.
You can easily find out the set of solutions given certain equations or inequations.
The various cases that you come across are :

1) You get unique solution.

2) You get infinitely many solutions.

3) You get no solution.


Let $T:\mathbb{P_2}\rightarrow\mathbb{P_2}$ be a linear transformation defined by $T(a+bt+ct^2)=3a+(5a-2b)t+(4b+c)t^2$ Let $T:\mathbb{P_2}\rightarrow\mathbb{P_2}$ be a linear transformation defined by $T(a+bt+ct^2)=3a+(5a-2b)t+(4b+c)t^2$

Find the basis C for $\mathbb{P_2}$ so that the matrix $[T]_c$ is a digonal matrix

My attmept: Suppose we take basis of C is ${1, t, t^2}$ i am little confusing can any   help me to solve

You have to find a basis $C=\{p_1,p_2, p_3\}$ of $\mathbb{P_2}$ and real numbers $t_1,t_2,t_3$ such that

$T(p_j)=t_jp_j$ for  $j=1,2,3$.

Then: $[T]_c=diag(t_1,t_2,t_3)$

FRED

Do The following operations on the T matrix w.r.t standard basis. $ C'_2\,=  C_2 \, - \, 4. C_3\,\, \\\\\, C'_1 = \, 2. C_1-\,\,5.C'_2 $. to get a diagonal matrix. Do the same operation on standard basis matrix to obtain $ [2,\, 5+t,\,-20-4t+ t^2] $ as basis. This basis will give a diagonal matrix.


$(C_{[0;1]}, \Vert . \Vert_1)$ is not a Banach space I'm going to prove that $C_{[0;1]}$ is not a Banach space w.r.s to the norm $\Vert x \Vert_1 = \int_{0}^{1} |x(t)| dt$ by consider the series $\sum_{n=1}^{\infty}x_n$ where $x_{n}(t)= t^{n} \cdot \sqrt{1-t}$.

By MCT we shall easy check that this series is absolutely convergence. Indeed $$ \begin{array}{ll}\sum_{1}^{\infty}\Vert x_n \Vert_1 & = \sum_{1}^{\infty}\int_{0}^{1}x_{n}(t)dt = \int_{0}^{1} \sum_{1}^{\infty} x_n(t)dt=\int_{0}^{1}\sum_{1}^{\infty}t^n\sqrt{1-t}\, dt \\ &= \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac{4}{3}  \end{array} $$
But I'm stuck in proving this series is not convergence in $(C_{[0;1]}, \Vert . \Vert_1)$. I think that if this series is convergence in $(C_{[0;1]}, \Vert . \Vert_1)$ to $x$, it may convergence to $x$ pointwise, but I can't prove this.

Thank you for your help.

Notice that, if $g(t)=\sum_1^\infty x_n(t)$, then, for $t\in (0,1)$
$$
g(t)= \frac{\sqrt{1-t}}{1-t}.
$$
You can prove this as you did, using the fact that we have a geometric series. On the other hand, since $x_n(0)=0$ for all $n$, we have $g(0)=0$, which is impossible since the sum representation above forces $g(0)=1$. Therefore the series doesn't converge in your space.

Your series are convergence in this norm indeed, and you have already proven it. For an example to show the noncompleteness, consider fn(x) to be a function such that it equal to 1 when x < 1/2, it equal to 0 when x > 1/2 + 1/n, and it is linear when 1/2 < x < 1/2 + 1/n. If such fn converges to a function f in the norm above, then view it as L1 we get a subsquence converges a.e. to f, thus f must be the function equal to 1 when x < 1/2 and equal to 0 when x > 1/2 a.e., and cannot be continuous.


Row swap changing sign of determinant I was wondering if someone could help me clarify something regarding the effect of swapping two rows on the sign of the determinant. I know that if $A$ is an $n\times n$ matrix and $B$ is an $n\times n$ matrix obtained from $A$ by swapping two rows, then
$$\det(B)=-\det(A)$$
but I don't know how to prove this.
I have been looking for proofs at internet, and read in both in textbooks and lectures notes that are available that this result is very hard to prove and most approaches rely on induction and so was wondering if there is something wrong with using that $\det(AB)=\det(A)\det(B)$ and then writing $B=EA$ where $E$ is an elementary matrix swapping two rows and using this result to get $\det(B)=\det(E)\det(A)=-\det(A)$ (since showing that $\det(E)=-1$ in this case is not that hard).

Yes, your method would work, and it is probably the most elegant possible.

We can without loss of generality assume that $E$ interchanges the first two rows.
This means that we can write $E$ in block-diagonal form:
$$
\left( \begin{array}[ccccc]
00 & 1 & 0 &\dots & 0 \\
1 & 0 & 0 &\dots & 0 \\
0 & 0 & 1 &\dots & 0 \\
... & ... & ... & 1 & ... \\
0 & 0 & 0 & ... & 1
 \end{array}\right)
$$

Now if you know how to calculate the determinant from the usual Laplace algorithm, starting at the bottom line, you see that the only nonzero terms are...

Also, why can we assume it interchanges the first two lines without loss of generality? (Think of what happens if we change a basis...)

If $~i~$ and $~j~$ are two rows of matrix $~A~$ that are interchanged to give matrix $~A^*~$.

Apply to $~A~$ and $~A^*~$ the general recursive formula twice (two-stage recursion) along the two interchanged rows  $~i~$ and $~j~$.

Using the fact that  $~A~$ and $~A^*~$  are identical when rows  $~i~$ and $~j~$ are deleted (which is the case for the determinants of the resulting matrices after the two-stage recursion) and rearranging the co-factors in the sums, you will find that $~\det A = -\det A^*$.


Let matrix $A$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric? Let matrix $A \in {M_n}(\Bbb R)$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric?

Here's a late answer.

First, a bit of notation. Since we will work over $\mathbb C,$ we will need the notion of "conjugate-transpose" of a complex matrix. Namely, if $M\in \mathbb C^{n\times n},$ then its "conjugate-transpose" is defined as
$$
M^H = (\overline M)^T,
$$
where $\overline \cdot$ denotes complex conjugation of the matrix entries and $\cdot^T$ denotes matrix transposition. This is a well known operation. Observe that $\overline \cdot$ and $\cdot^T$ commute. And, of course, for $N\in\mathbb R^{n\times n},$ we have $N^H = N^T.$ In the following, we will use only the operation $\cdot^H.$

On to the actual question. So we are given a matrix $A\in\mathbb R^{n\times n}$ which is normal, i.e. $A^HA = AA^H$ and such that $A^HA$ has $n$ distinct eigenvalues. We want to show that $A$ is symmetric, i.e. $A^H = A.$

Observe that, since $A$ is real, its characteristic polynomial has real coefficients. This means that the eigenvalues of $A,$ i.e. the roots of its characteristic polynomial, are either real or come in complex conjugate pairs. In other words, we can enumerate the eigenvalues of $A$ as a sequence
$$
\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s},
$$
with $r,s\in \mathbb N_0,$ $\lambda_j\in\mathbb R,$ and $\mu_k\in \mathbb C$ with $\Im\mu_k > 0.$ Repetitions are allowed, i.e. we can have $\lambda_j = \lambda_k$ or $\mu_j = \mu_k$ for $j\neq k.$

Now, since $A$ is normal, it can be diagonalized using a unitary transformation. This is a well known fact. In more detail, we have the following. Put
$$
D = {\rm diag}(\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s})\in \mathbb C^{n\times n},
$$
i.e. $D$ is the diagonal matrix with the eigenvalues of $A$ on the diagonal. Then, there is a unitary matrix $U\in \mathbb C^{n\times n}$ such that
$$
\tag{1}
A = U^HDU.
$$
Recall that "unitary" means
$$
\tag{2}
U^HU = UU^H = I,
$$
where $I \in \mathbb C^{n\times n}$ is the identity matrix. From $(1),$ we get
$$
\tag{3}
A^H = (U^HDU)^H = U^HD^HU.
$$
From $(1),$ $(2),$ and $(3),$ we conclude
$$
A^HA = \left(U^HD^HU\right)\left(U^HDU\right) = U^HD^HDU.
$$
This shows that the eigenvalues of $A^HA$ are the diagonal elements of the dialgonal matrix
$$
\begin{align}
D^HD & = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,\overline{\mu_1}\mu_1,\mu_1\overline{\mu_1},\ldots,\overline{\mu_s}\mu_s,\mu_s\overline{\mu_s}) \\
& = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,|\mu_1|^2,|\mu_1|^2,\ldots,|\mu_s|^2,|\mu_s|^2).
\end{align}
$$
Observe that every pair $\mu_j,\overline{\mu_j}$ of complex conjugate eigenvalues of $A$ produces twice the same eigenvalue for $A^HA.$ By assumption, this can't happen. So we must have that $s = 0,$ $A$ has only real eigenvalues, and in particular $D\in \mathbb R^{n\times n},$ i.e. $D$ is real. But then
$$
\tag{4}
D^H = D.
$$
Now we get from $(1),$ $(3),$ and $(4)$
$$
A^H = U^HD^HU = U^HDU = A,
$$
as desired.

$A$ is normal and $AA^T$ has $n$ distinct eigenvalues, so $AA^T=X \Lambda^2X^{-1}=A^TA $, where $\Lambda$ is diagonal and has $\lambda$'s on the diagonal.

Now $AA^T=(X \Lambda X^{-1})(X \Lambda X^{-1})=A^TA $ so that $A=A^T$.


How to find the limit $\lim_{n\to \infty}\frac{n}{2^n-1}$? I've been trying to find and prove the
$$\lim_{n\rightarrow\infty}\frac{n}{2^n-1}$$
but I haven't even figure out the limit, could anyone help?

You can use Stolz theorem:

$ \lim \frac{a_n}{b_n}  = \lim\frac{a_{n+1}- a_n}{b_{n+1}-b_n} $

provided that $ b_n \rightarrow \infty $ monotonously


Subgroups of Symmetric groups isomorphic to dihedral group Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n>2$?

Let $\left\{a_1,a_2,\ldots,a_n\right\}$ denote the vertices of an $n$-gon. The cyclic permutation $\alpha=(a_1,a_2,\ldots,a_n) \in S_n$. Choose any vertex, e.g. $a_1$ and consider it the fixed point of a reflexion of the plane (i.e the permutation $\beta=(a_2,a_n)(a_3,a_{n-1})\ldots$ of order $2$). This permutation also $\in S_n$. Now $D_n$ is generated by $\alpha, \beta \in S_n$ which shows that the $D_n$ thus defined is a subgroup of $S_n$

If you can build up a faithful action of $D_{2n}$ on a set of $n$ elements, then you are done. I'll adopt this definition of the dihedral group of order $2n$:
$$D_{2n}:=\langle r,s\mid r^n=s^2=(sr)^2=1\rangle \tag 1$$
Let's consider the action of $D_{2n}$ by left multiplication on the set of left cosets of some subgroup $H$ of index $n$, i.e. of order $2$; such an action, say $\varphi$, has kernel $\operatorname{ker}\varphi=\bigcap_{g\in D_{2n}}gHg^{-1}$, and hence if there is a $\tilde g\in D_{2n}$ such that $\tilde gH\tilde g^{-1}\cap H=\{1\}$, then the action is faithful. Now, take $H=\{1,s\}$ and $\tilde g=r$; then $\tilde gH\tilde g^{-1}\cap H=\{1,r^2s\}\cap\{1,s\}=\{1\}$ as soon as $n>2$. Therefore, indeed $D_{2n}\hookrightarrow S_n$ for $n>2$.


Product of two non-zero complex numbers equals zero Is it possible for the product of 2 non-zero complex numbers to be 0?

No, it is not possible.

For complex numbers $z_1,z_2 \in \mathbb{C}$ we have that
$$z_1\cdot z_2 = 0 \quad\implies\quad z_1 = 0\quad\text{ or }\quad z_2 = 0.$$

As a hint: non-zero complex numbers have multiplicative inverses.

It's not possible. the product of 2 complex number can equal 0 if and only if one of the numbers is zero.
   i.e a.b = 0 where a,b are complex numbers
     Then either a or b is 0


Expected value of a single variable Function $\displaystyle{ f_x(x) = 6x^5 \text{ for } 0 < x <1}$

A.) Find $\mathbb{E}\left[X\right]$

B.) Find $P\bigl[1/3 \le X \le 2/3\bigr]$


How do I calculate the number of unique permutations in a list with repeated elements? I know that I can get the number of permutations of items in a list without repetition using

(n!)

How would I calculate the number of unique permutations when a given number of elements in n are repeated.

For example

ABCABD

I want the number of unique permutations of those 6 letters (using all 6 letters).

There is a specific formula for such problems:

Permute all elements, and remove permutations of elements that are identical, viz.

$\dfrac{6!}{2!2!}$

Do you want the number of combinations of a fixed size? Or all sizes up to 6? If your problem is not too big, you can compute the number of combinations of each size separately and then add them up.

Also, do you care about order? For example, are AAB and ABA considered unique combinations? If these are not considered unique, consider trying stars-and-bars: How to use stars and bars (combinatorics)


how to memorize the sum and product of roots for an $n^{th}$ degree equation For my exams I need to know the following equations by heart:

for a polynomial equation: $a_nx+a_{n-1}x^{n-1}+...+a_1x+a_0,$ the sum and product of the roots are given by

$$\textrm{Sum}=-\frac{a_{n-1}}{a_n}$$
$$\textrm{Product}=(-1)^n\frac{a_0}{a_n}.$$

I have never been able to memorize these, and for some reason, they are not on the formula booklet. If anyone has any mnemonic or trick of some sort for memorizing them, it would be very useful to me.

Thank you very much in advance!

Write
$$a_n x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0 = a_n(x-r_1)...(x-r_n)$$

and develop. You see immediately what is the constant term and the term of degree $n-1$.

Hint:

Write the polynomial as $$(x-r_1)(x-r_2)...(x-r_n),$$ where the roots are $r_1, r_2, ... r_n.$


Determine amount multiplied in sum Suppose you have n , and r (result) = some value less than n (ltn) multiplied by 4, plus what's left of n after subtracting ltn (n - ltn).

Eg. n = 60; (r) = 25 (ltn) + (60 (n) - 25 (ltn) = 35) = 135.

Now, if I know the values of n (60) and r (135), is there a formula to determine the value of ltn?

I can "bruteforce" it by taking r (135) find the highest number that divides by 4 without fractions (132 / 4 = 33) and then deduct that from n (60 - 33 = 27) and add the left over (27) to 133 = 159. Well that's way higher than 135, so let's move down to 128 (next number divisible by 4) etc... but is there a simpler way? Thanks...

It gives a linear equation.

$r=4l+n-l $

$r=3l+n $

$l=\frac {r-n}{3}$


Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different? Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?

Probably a really dumb question but I just don't see it :(

Other answers (and comments) have already explained the potential pitfalls of multiplying both sides of an inequality by a quantity whose sign is sometimes negative. Here is an alternative approach that might help:
$$\begin{align}
{4\over x(4-x)}\ge1
&\iff{4\over x(4-x)}-1\ge0\\
&\iff{4-x(4-x)\over x(4-x)}\ge0\\
&\iff{4-4x+x^2\over x(4-x)}\ge0\\
&\iff{(2-x)^2\over x(4-x)}\ge0\\
&\iff x(4-x)\gt0\quad\text{(since }(2-x)^2\text{ is always non-negative)}\\
&\iff0\lt x\lt4
\end{align}$$
Remark: In general the quadratic in the numerator will not be a perfect square, in which case the final steps are a bit more complicated (you can wind up with more than one interval where the inequality holds). But the problem here was concocted to have a simple answer.

If $\dfrac4{x(4-x)}\ge 1$, $\:x(4-x)$ is positive, hence we can multiply both sides by $x(4-x)$ and therefore it implies $4\ge x(4-x)$.
However, the other way, $ 4\ge x(4-x)$ necessarily happens in particular if  $(4-x)<0$, and in this case it is equivalent to $\dfrac4{x(4-x)}\le 1$.


Are the laws of mathematics 'absolute' in this universe? We observe that almost all physical phenomena (which has been explained) can eventually be explained by the laws of mathematics. Mathematics seems ubiquitous- for example the form of the differential equation that governs the simple harmonic motion of a helical spring has the exact same form to explain the current in a RLC circuit. The parallels are endless virtually.
But it does not mean necessarily that the 'next to be discovered phenomenon' has to follow the laws of mathematics provided that there is a theorem out there which proves that every physical mechanism has to abide by the laws of maths.

Could anyone answer with a philosophical insight on this matter and has the Godel's Incompleteness theorem anything to say on this?

If you haven't already read Eugene Wigner's essay "On the unreasonable effectiveness of mathematics in the natural sciences". It addresses most of what you mentioned.

https://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html

I think they are not.  Consider, for any consistent first-order theory S in any language L, the set S* of consistent theories of greater provability strength.  The incompleteness theorems tell us that if S is incomplete, then S* can be arranged in an upside-down pyramid, where each S' in S* can be extended with at least con(S') and not con(S'), giving an upwards branching structure of incompatible extensions.

If mathematics was absolute, then there would be a subset T(L) of first-order theories in S* that is the "true" tower without branching structure that traces out this absolute mathematical reality for the language L.

Is there such an absolute mathematical reality for any meaningful language L?  I don't think we have a universally accepted methodology for getting T(L) for arithmetic or set theory.