JasonM
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Definition of irreducible element of a ring
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1 votes

Being in $U(R)$ is the natural analogy of being $1$. In $\mathbb{N}$, $1$ is the only unit. In $\mathbb{Z}$, there are two; $\pm 1$. In $\mathbb{Z}[i]$ there are four; $\pm 1$ and $\pm i$. In $F[t]...

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Can $b+c$ in the pythagorean triplets $(a, b, c)$ be a prime number?
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7 votes

If the Pythagorean Triple $(a, b, c)$ is not primitive, then $b+c$ is trivially composite. Thus, let it be primitive. Then $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$ where $b$ and $a$ are interchangeable....

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find the the greatest value of $m$ such that $\text{lcm}(1,2,3,..,n)=\text{lcm}(m,m+1,..,n).$
0 votes

Let $L_n$ be the left-hand side, and let $R_{m,n}$ be the right-hand side. Note that $L_n$ is divisible by all of the greatest prime powers $p^r \leq n$. In fact, $$L_n= \prod_{p \leq n} p^r$$ where ...

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What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?
17 votes

The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$

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Why does simplifying a function give it another limit
5 votes

It doesn't get a new limit, it actually just didn't have a value before (because it was $0/0$), but it still had that same limit. The original function and the new function are actually different ...

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When is $1-(1-p)^n \sim pn$
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1 votes

Before his edit Pretty sure it only works for functions $p(n)=o(1)$ (using little-$o$ notation). Otherwise, $p(n)=\Omega(1)$, so $1-(1-p(n))^n=\Omega(p(n)^n)$. Assuming $p \rightarrow \infty$, we ...

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What is the meaning of $e^{int}$ notation
2 votes

I believe from context $\varphi_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}=\frac{1}{\sqrt{2\pi}}(\cos(nt)+i\sin(nt))$ $i^2=-1$ $n \in \mathbb Z$ with argument $t$.

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A problem related to sequence and series
2 votes

$0<\frac{a_n}{n}<\frac{a_1}{n} \rightarrow 0$, so Squeeze Theorem, and $\sum \frac{a_n}{n^2} < \sum \frac{a_1}{n^2}=a_1\pi^2/6$

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Prove that $\sin x=[1+\sin x]+[1-\cos x]$ has no solution in $x\in \Bbb R$
2 votes

Note that $\sin(x)$ is an integer. But then $\lfloor \sin(x) \rfloor=\sin(x)$, so your equation reduces to $$\lceil \cos(x) \rceil=2$$ which is not possible.

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Use of greatest common divisor to calculate unknown
1 votes

Yes of course. Let $y=k\frac{z}{\gcd(x,z)}$ and $x=x'\gcd(x,z)$. Then $$\frac{xy}{z}=\frac{x'\gcd(x,z)\cdot k\frac{z}{\gcd(x,z)}}{z}=x'k$$ Furthermore, if $xy \equiv 0 \mod z$, then $$\frac{xy}{z}...

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Prove that $4$ divides $3^{2m+1} - 3$
22 votes

$3^{2m+1}-3=3(3^m-1)(3^m+1)$, and both of the factors $3^m \pm 1$ are even, so their product is divisible by 4.

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Solutions to $\{x^3\}+\lfloor x^4\rfloor=1$
0 votes

$\{x^3\}=0$ and $\lfloor x^4 \rfloor=1$ implies $$x \in \{\sqrt[3]{n} | n \in \mathbb{Z} \} \cap \left([1, \sqrt[4]{2}) \cup (-\sqrt[4]{2}, -1]\right)=\{\pm 1\}$$, so $x=\pm 1$

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Why do some rational functions approach the asymptote from the other side?
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2 votes

The reason is because the horizontal asymptote is identically the $x$-axis for your function. Also, an asymptote only describes end behavior, so it's perfectly fine for the function to cross it ...

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Solve $z^5=-32$ and draw its solutions in complex space, then describe their characteristic geometrical property.
1 votes

The solutions should just be $2e^{\frac{2\pi ki}{10}}$, for odd integers $0 \leq k <10$. They are the divisions of the circle of radius 2 centered at the origin, divided into 5 pieces, rotated ...

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Choosing Functions for the Squeeze Theorem
5 votes

Limits don't necessarily preserve strict inequalities. For example, $1-\frac{1}{n}<1+\frac{1}{n}$, yet they have the same limit as $n$ goes to $\infty$.

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Can someone tell me how to prove $\cap_{i\in I}(B \setminus A_i) = B \setminus (\cup_{i\in I} A_i)$
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0 votes

By set inclusion: Consider $x \in \cap_{i\in I}(B \setminus A_i)$. Then $x \in B \setminus A_i$ for all $i \in I \implies x \notin A_i$ for all $i \in I \implies x \notin \cup_{i\in I} A_i \implies ...

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minimal polynomial $\zeta_n$ and $\zeta_n^p$ is the same for any prime $p$ not dividing $n$
0 votes

We know the minimal polynomial of $\zeta_n$ is the $n$th cyclotomic polynomial $\Phi_n (x)$, which is defined as $$\Phi_n (x)=\prod_{k \in \mathbb{Z}_n^{\times}}(x-\zeta_n^k)$$ where $\mathbb{Z}_n^{\...

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A question about the existence of
1 votes

I assume the $p_i$ are meant to be distinct. Let $y_i$ satisfy $$-x_1p_1 \equiv y_1 \mod p_1$$ $$-x_1p_1 \equiv y_2 \mod p_2$$ $$-x_1p_1 \equiv y_3 \mod p_3 \\$$Clearly $y_1 \equiv 0 \mod p_1$ ...

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Divergence of $\sum\limits_{n=0}^{\infty} \frac{n^n}{n!e^n}$: fast proof?
3 votes

Stirling's approximation for $n!$ says $n!$ can be estimated by $\sqrt{2\pi n}(\frac{n}{e})^n$ for large values. The terms of your series are therefore asymptotic to terms of the form $\frac{1}{\sqrt{...

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