7
votes
Accepted
The least positive integer ending in $7$ which quintuples if the $7$ is moved to the front
Your solution is correct, but the problem admits a significantly cleaner approach. Let $x$ be the desired number. Write $x=10t+7$, where $t$ is is the integer formed by the other digits and has $n$ ...
- 379
5
votes
Accepted
How to solve $ \lim\limits_{x\to+\infty} \!\!\left(\! \frac{x^{2}+3}{3x^{2}+1}\! \right)^{\!x^{2}}\!\!\!=0\;?$
$\color{brown}{\text{Trabajo Odoo}}$, your procedure is correct, but I think that you could calculate your limit in the following simpler way since $\color{blue}{\left(\dfrac13\right)^{\!\!+\infty}}\!$...
- 14.4k
4
votes
Proving that the set of the quotients of Fibonacci numbers is not dense in the positive reals
I don't find the argument you presented to be particularly rigorous or convincing, although you're very much on the right track.
One thing you could do to make this more rigorous is to find a specific ...
- 5,547
3
votes
Convergence of a Zeta-Zeta function
Let me expand my comment into a short answer; first since $|\rho_n|=|\sigma_n+i\gamma_n|\to \infty$ where we index the nontrivial roots by increasing $|\gamma_n|$, it is clear that we need $\Re s >...
- 33.2k
2
votes
Accepted
Proving that the set of the quotients of Fibonacci numbers is not dense in the positive reals
The fact that $\left|F_n - \frac{\tau^n}{\sqrt{5}}\right| \to 0$ as $n \to \infty$ does not imply
$\left|\dfrac{F_n}{F_m} - \tau^{n-m}\right| \to 0$ as $n \to \infty$ for fixed $m$. Indeed,
$$ \...
- 473k
2
votes
Accepted
Does $C \cong \left( B / A \right)$ imply the existence of a short exact sequence $0 \to A \to B \to C \to 0$
I believe your reasoning is correct. However, I have some comments.
for the inclusion $A \hookrightarrow B$ to be injective, it is unnecessary to assume that $A$ is a strict submodule of $B$ (hence ...
- 522
2
votes
How to solve $ \lim\limits_{x\to+\infty} \!\!\left(\! \frac{x^{2}+3}{3x^{2}+1}\! \right)^{\!x^{2}}\!\!\!=0\;?$
More simply
$$\left( \frac{x^{2}+3}{3x^{2}+1} \right)^{x^{2}}=\left(\frac{1}3 \right)^{x^{2}}\left( 1+\frac{8}{3x^{2}+1} \right)^{x^{2}}$$
with $\left(\frac{1}3 \right)^{x^{2}} \to 0$ and
$$0\le \left(...
- 164k
2
votes
How to solve $ \lim\limits_{x\to+\infty} \!\!\left(\! \frac{x^{2}+3}{3x^{2}+1}\! \right)^{\!x^{2}}\!\!\!=0\;?$
For $$ x \geq \sqrt 5$$
we have $$ \frac{ x^2 + 3}{3x^2 + 1} \leq \frac{1}{2} $$
so that
$$ \left( \frac{ x^2 + 3}{3x^2 + 1} \right)^{x^2} \leq \frac{1}{2^{x^2}}, $$
when $ x \geq \sqrt 5$
- 147k
1
vote
Accepted
Show that $w\in W$ if and only if $w\wedge w_1\wedge\cdots\wedge w_k = 0$ in $\wedge^{k+1}V$
Your $\impliedby$ proof looks correct to me.
However, I would caution that proofs like that are better written as proofs by contraposition, not contradiction. Specifically, if your proof looks like:
...
- 5,547
1
vote
The least positive integer ending in $7$ which quintuples if the $7$ is moved to the front
The number is $N=10X+7$ and $X$ must have the minimum possible number $n$ of digits. Further $$5N=50X+35=7\times10^n+X\iff49X+35=7\underbrace{00\cdots00}_{n\text{ times}}$$ This implies that $$35\...
- 33.7k
Only top scored, non community-wiki answers of a minimum length are eligible