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High school senior studying graph theory.


Jul
21
comment Prove that $l = k/\gcd(m,k)$.
$l | 2l$ as well, so $m/(m,k) = 2$ is a possibility. $l | x$ does not mean $x = l$; the last statement does not follow.
Jul
4
answered If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Jul
3
answered Showing $ x-y\in\mathbb{Q}$ is an equivalence relation?
Jul
3
answered Subgroup of $GL_{n+m}(\mathbb K)$
Jul
3
answered Proof of $A \subseteq B \Leftrightarrow A \cap B = A$ (Check chain of implications)
Jul
2
awarded  Curious
May
11
reviewed Leave Open $\Delta \subset \Phi$ is a base in a root system imples $\Delta^\vee \subset \Phi^\vee$ is a base in a root system
May
11
answered If A and B are two Matrices , than find $(A-B)^2$.
May
11
comment Graph Theory Question On Exam Involving colorability of certain planar graph
@royherma The difficulty with this sort of exhaustive proof is that you need to prove that these comprise all the cases. This would be quite difficult, and there would be too many cases to cover. It is almost always easier to look for general methods like the one above.
May
10
comment About the $\lim_{x\to a} \frac{f(x)}{1+|f(x)|}$
@MarterJs $f(x) < 1 + |f(x)|$ is stronger than $f(x) \le 1 + |f(x)|$. It implies it. If it's true (as it is), then the hypothesis of the implication is satisfied.
May
10
comment About the $\lim_{x\to a} \frac{f(x)}{1+|f(x)|}$
@MarterJs No it is not possible. $f(x) < 1 + |f(x)|$. However, it is possible for the limit to be $1$: take $f(x) = x$ and let $x \to +\infty$. And to answer your second question, while the statement is correct and the left hand limit will be $\le 1$, your original proof was incorrect for the reason you state.
May
10
answered About the $\lim_{x\to a} \frac{f(x)}{1+|f(x)|}$
May
10
revised Graph Theory Question On Exam Involving colorability of certain planar graph
deleted 198 characters in body
May
10
answered Graph Theory Question On Exam Involving colorability of certain planar graph
May
10
comment The limit of the derivative of an increasing and bounded function is always $0$?
How are you supposed to get a construction like Hagen von Eitzen's out of this?
May
10
comment The limit of the derivative of an increasing and bounded function is always $0$?
This alone does not seem that useful in answering the question (see Hagen von Eitzen's answer).
May
10
comment How can I solve: ${\left [{x+1}\over2\right]}={x-1\over 3}$?
$[x+n] = [x] + n$ if $n$ is an integer. Here, $\displaystyle \left[ \frac{3k+2}{2} \right] = \left[ k+1+\frac{k}{2} \right] = k+1+\left[ \frac{k}{2} \right]$
May
10
comment How can I solve: ${\left [{x+1}\over2\right]}={x-1\over 3}$?
Yes, but what is the integer part of -2.5? -2 or -3?
May
10
answered How can I solve: ${\left [{x+1}\over2\right]}={x-1\over 3}$?
May
10
comment How can I solve: ${\left [{x+1}\over2\right]}={x-1\over 3}$?
By integer part of a number, do you mean the least integer function, even for a negative number?