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Hopeful mathematics student with some competition level problem solving experience. Studying algebraic topology and algebraic geometry at the moment, but I feel like anyone who know anything know more than I do.


1h
comment Are these sets bounded or not?
@Quickbeam2k1 The first one is not a sphere (note the exponents). It is not limited, since $(n, -n, 1)$ is in the set for any $n$.
2h
revised partial fraction decomposition
edited body
3h
revised Which one is greater?
edited tags
3h
comment Question about group theory and order of elements
Note that the imlications go the other way as well.
6h
comment Irreducible Polynomial in $\mathbb{Q}[x]$
I would start with "If $x^4+2x^2+4=0$, then $x^2=\ldots$"
9h
comment How does look like an open set in one point compactification?
$C$ is a compact and closed set in $X$.
13h
comment prove or disprove $H$ is a subgroup
You prove it the same way you prove any other subset of a group is a subgroup: you find the identity element, you find inverses and you show that it is closed. You disprove it the same way you disprove any statement that says something is always true: you give a counterexample.
13h
comment Last 2 digits of $2345^{369}$
WolframAlpha agrees with you (and so do I). What the page says is true for numbers that end in $5$ and are one less than a multiple of $4$, though. So in half the cases it's true, and half the cases it's just $25$ all the way.
22h
comment Should this be rephrased into saying no common factors but 1?
The phrase "no common factors" usually does mean "apart from $1$", yes. To be more precise, it means "no common factor other than units", where unit is any factor of $1$, like $-1$ or $i=\sqrt{-1}$. This is done for the same reason that $1$ is not a prime: it saves three or four words every other sentence.
22h
comment If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.
Yes, that's exactly what I'm talking about.
1d
comment If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.
You're definitely on the right path. However, what you've proven is that from some point on, the sequence $b_n$ is bounded by $M+\epsilon$. There might be billions of terms before you get there. You have to point out that all the $b_n$ before that point has a bound as well.
1d
comment A game where starting with 3 boxes, with 10 balls in each, the goal is to remove as many balls as possible following the rules
Yes. At any point where an olympiad question asks "Is it possible to ...?" an answer containing a concrete algorithm for getting the result you want is considered a full justification. I remember hearing about someone at the IMO solving a question about why a knight-like piece on a rectangular chess board couldn't get from one corner to the other. He turned in a solution where he had drawn the chessboard and marked every single square the knight could possibly get to. He got full score (of course, had he missed a single square, it might've given $0$ points).
1d
comment A game where starting with 3 boxes, with 10 balls in each, the goal is to remove as many balls as possible following the rules
Yes, it's trial-and-error. The simplest and most straight-forward (although not always the easiest) way of proving that something is possible is to actually do it. Proving that something is impossible, however, is never justified by trial-and-error, although some trial-and-error could help you discover a deeper pattern (like how you could never end up with $2$ or $5$ balls). As to what would've happened if I started in the $7$-bin? I might've succeeded, and then that's what I wouldve written. I might've failed, and then I would've tried something different next time.
1d
answered A game where starting with 3 boxes, with 10 balls in each, the goal is to remove as many balls as possible following the rules
1d
comment A game where starting with 3 boxes, with 10 balls in each, the goal is to remove as many balls as possible following the rules
We can also eliminate the $0$ option, since no matter which move you do, there must be at least one ball left over afterwards. That means it's either $3$ or $6$.
1d
revised A game where starting with 3 boxes, with 10 balls in each, the goal is to remove as many balls as possible following the rules
Transcribed image.
1d
comment Generating a random binary matrix with fixed number of nonzeros
Depending on your language, there might be a shuffle method for arrays. If there is one, then just make a matrix with $c$ ones and shuffle it.
1d
comment What is so great about 7?
Which multiple of $7$ was the last (only) one you erased? Is that a coincidence?
1d
answered Solutions to $N=2^r-r-1$
1d
comment Irreducible polynomials over the reals
The intermediate value theorem solves this problem for odd degree polynomials. I don't even know where to start for even degree polynomials.