real analysis - Homeomorphisms between circles and rectangles

In our topology class we learn that in $\mathbb{R}^2,$ circles and rectangles are homeomorphic to each others.
I can understand the underline idea intuitively.
But can we find an explicit homeomorphic between them?
If so how?

Also our professor said that, "we can describe any point in the rectangle $[0,1]\times[0,1]$ using a single coordinate."
I wonder how such thing is possible.
As I think, for this we need a bijection between $[0,1]\times[0,1]$ and some (closed?) interval in $\mathbb{R}.$
Can some one explain this phenomena?

Answer

One very visual way of seeing such a homeomorphism is to place one inside the other, choose a point that's inside both (say point $P$), and let the image of a point $Q$ in the circle be where the line emanating from $P$ and through $Q$ intersects the rectangle.

EDIT: Added an example image:

abstract algebra - Two number fields with trivial intersection, not linearly disjoint but.... (continued)

Do there exist two finite extensions of $\mathbb{Q}$, $L=\mathbb{Q}(\alpha)$ and $K=\mathbb{Q}(\beta)$ such that $L\cap K=\mathbb{Q}$, the minimal polynomial $\mu_{\alpha}(X)\in \mathbb{Q}[X]$ of $\alpha$ over $\mathbb{Q}$ is not irreducible over $K$, but the factorization of $\mu_\alpha(X)$ over $K$ has no linear factors ?

(see also here)

Answer

I think this works. Let $f$ be an $A_4$-quartic, let $\alpha$ and $\gamma$ be two of its zeros in a splitting field $M$, let $\beta=\alpha+\gamma$. Note that $L$ has degree 4 over the rationals, and the Galois group of $M/L$ is cyclic of order 3, while $K$ has degree 6 over the rationals (as $\beta$ has 6 distinct conjugates over the rationals), and the group of $M/K$ is cyclic of order 2. If $L\cap K$ had any nonrational elements, they would have to be of degree 2 over the rationals, but $M$ has no subfield of degree 2 over the rationals, since $A_4$ has no subgroup of index 2. Also, $\mu_{\alpha}$ can't be irreducible over $K$, since it has degree 4, but $M/K(\alpha)$ only has degree 2. And $\mu_{\alpha}$ can't have any zeros in $K$, since $K$ would then be a field of degree 6 containing a subfield of degree 4.

I hope someone will check this.

EDIT: A simpler example. Let $\alpha=\root4\of2, \beta=(1+i)\root4\of2$. Then $\mu_{\alpha}(x)=x^4-2$, both $K$ and $L$ are subfields of the splitting field $M$ of $\mu_{\alpha}$, both have degree 4 over the rationals and index 2 in $M$. $L\cap K$ is the rationals, and $\mu_{\alpha}$ factors over $K$ as $$(x^2-(1+i)\root4\of2x+i\sqrt2)(x^2+(1+i)\root4\of2x+i\sqrt2)$$

elementary number theory - Divisibility by 7 rule, and Congruence Arithmetic Laws

I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let $n = (a_ka_{k-1}\ldots a_2a_1a_0)_{10} = \displaystyle{\sum_{j=0}^{k}}a_{k-j}10^{k-j}$. The expression $$ Q_{3}^{\prime}(n) = (a_2a_1a_0)_{10} - (a_5a_4a_3)_{10} + (a_8a_7a_6)_{10} -\ldots $$ are called alternating sum of the digits of third order of $n$. For example, $$ Q_{3}^{\prime}(123456789) = 789-456+123=456 $$ Proposition: $7 | n \ \Leftrightarrow \ 7 | Q_{3}^{\prime}(n)$.

proof. ??

Thanks for any help.

calculus - Are all limits solvable without L'Hôpital Rule or Series Expansion

Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?

For example,

$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$

$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$

$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$

$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$

$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$

$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

calculus - Function as a "constant of integration"

I'm reading a book Differential Equations with Applications and Historical Notes, 3rd edition, specifically section 8 about exact equations. The author is trying to prove that iff $\partial M/\partial y = \partial N/\partial x$ then equation
\begin{equation}
M(x,y)dx + N(x,y)dy = 0
\end{equation}
is exact differential equation.
At some point we integrate equation
\begin{equation}

\frac{\partial f(x,y)}{\partial x} = M(x,y)
\end{equation}
to get
\begin{equation}
f(x, y) = \int M(x,y)dx + g(y)
\end{equation}
The author states that function $g(y)$ appears as a constant of integration because if we take derivative of both sides with respect to $x$, $g(y)$ would disappear because it doesn't depend on $x$.
That's the part that I have trouble with, $y$ is a dependent variable and $x$ is independent variable so wouldn't derivative of $g(y)$ with respect to $x$ be
\begin{equation}
\frac{d\,g(y)}{dy} \frac{dy}{dx}
\end{equation}

and not $0$ ?

Answer

This is a common poorly written part in differential equations textbooks, because they don't want to spend time discussing differential forms.

At this point we forget that $y$ depends on $x$. Of course then the equation $M(x,y)dx+N(x,y)dy=0$ looks weird, and indeed it's wrong. What is meant there is that if we have a dependence of $x$ and $y$, a curve on $x$-$y$ plane, denoted $\gamma$, then the pullback of $M(x,y)dx+N(x,y)dy$ on $\gamma$ is $0$. For example, if we can parametrize $\gamma$ by $x$ (i.e. we can write $y$ as a function of $x$), then this condition says $\frac{dy}{dx} = -\frac{M(x,y)}{N(x,y)}$. That's why we want to find such $\gamma$.

The exactness condition means that $df=M(x,y)dx+N(x,y)dy$. Then the level sets of $f$, $\{(x,y)|f(x,y)=c\}$, give us such $\gamma$'s. Note that exactness follows from closeness on simply connected domains.

So, one can separate this problem into two stages, where $x$ and $y$ are independent, and then were we look for a required dependence.

Alternatively, instead of using differential forms, one can think of $(N,M)$ as a vector field on $x$-$y$ plane perpendicular to $\gamma$'s, the level sets of $f$, gradient of which is $(N,M)$.

calculus - Extended euclidean algorithm using table

I have to solve gcd$(133,99)=d$, $d = ax + by$ using the extended euclidean algorithm. I understand a part of it but this table we have to use/fill out confuses me a bit.

I get bit confused at step $4$. I sort of get the subtraction column I'm guessing $()$ set the precedence so $4$th step subtraction becomes $99 - ((133 - 99) \cdot 2)$

So first evaluate $(133\cdot 0 + 99\cdot 1)$, then evaluate $(133\cdot 1 - 99 \cdot (-1))$, then evaluate $99 - (34 \cdot 2)$ and finally $99 - 68 = 31$.

This is bit confusing but I guess understandable what confuses me the most is where I go wrong in the combined expression column on step $3$ seems I made the correct assumption however on step $4$ the result does not equal $31$.

Answer

Hopefully this layout for the algorithm will help:

summation - Infinite sum over the primes

I was recently trying to create my own unique proof that the harmonic series diverges, so what I did was realize that within the harmonic series is the reciprocal of powers of primes and for every prime, every reciprocal of the powers of that prime also are contained. By regrouping, you can show that for every prime, the powers of that prime add up to 1/(p-1) so we know that for every prime, the harmonic series at least

contains 1/(p-1) which means that the harmonic series is greater than the sum from 1 to infinity of 1/(p-1) where the sum is over the primes. How would I go about showing that this sum diverges?

Sorry about formatting I am currently doing this on a mobile device.