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visits member for 4 years, 4 months
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Software engineer and long-time dabbler in mathematics; 11th in the Putnams forever ago but I've long since atrophied.


11h
comment Arithmetic Progressions in an unusual sequence
@RossMillikan Agreed - fusible numbers are another and possibly even more pertinent example that springs to mind, where the hugeness comes about in the interplay between some König-style lemma guaranteeing solutions of all lengths and certain 'large countable ordinal' properties of the core problem itself. I wonder if Harvey Friedman would know anything...
12h
comment Set of integer p-adics-Proposition
It means almost exactly the opposite of that - it means that if $m=\sum_{i=0}^\infty m_ip^i$ is the $p$-adic expansion then $m_i$ is $0$ for $i\lt n$. Can you see why?
12h
comment Set of integer p-adics-Proposition
None of the components (in the particular version of the expansion that I'm talking about) is a multiple of $p^n$; they're all numbers $\lt p$. Think of it as the (infinitely-long) $p$-adic version of a decimal expansion, only going the 'other way'.
12h
comment Set of integer p-adics-Proposition
What do you mean by 'divisible by'? (For instance, is $\frac15$ divisible by $2$ in $\mathbb{Z}_2$? Is $\frac25$?) It's probably easier to think of elements $m\in \mathbb{Z}_p$ as infinitely long vectors $\langle m_i\rangle$ where $0\leq m_i\lt p$ for all $i$ - that is, as their '$p$-adic expansion'.
12h
comment Set of integer p-adics-Proposition
What do the members of $p^n\mathbb{Z}_p$ 'look like'? What would a number have to look like to be in all of them - that is, in $\bigcap_{n\in \mathbb{N}_0}p^n\mathbb{Z}_p$?
1d
answered Why is $\sum_{r=1}^{m-1} (2r+1)r=\sum_{r=1}^{m-1} 4\binom{r}{2} + 3\binom{r}{1}$?
1d
comment Where Does F' in Rubik's Cube Group Singmaster Notation Come From?
The presumable reason multiplication is used instead of addition, incidentally, is because additive notation is generally reserved for commutative groups, which the cube group clearly isn't.
1d
comment How to write $1-x-x^3+x^4+x^5+x^6-x^7 \cdots$ as a power series representation
It's not 'works for' - you need all those terms in the product.
2d
comment What are some interesting sole exceptions or counterexamples?
An example would go a long ways, too: 'take $a=3, b=2$. Then we have $3^2+2^2=13$ obviously divisible by $13$; $3^3+2^3=35=7\cdot 5$ is divisible by $3^1+2^1=5$, but it's also divisible by $7$; $3^4+2^4=97$ is prime; $3^5+2^5=275$ is divisible by $11$; $3^6+2^6=793$ is divisble by $13$; and so on'.
2d
comment What are some interesting sole exceptions or counterexamples?
That's arguably worse. I agree with the comments - this could as easily be expressed with words and would be much clearer into the bargain: 'for all $a, b\in \mathbb{Z}$ and all $n\geq 2$ there's a prime $p$ such that $p$ divides $a^n+b^n$ but doesn't divide $a^k+b^k$ for any $k\lt n$'.
2d
comment Can a piece of A4 paper be folded so that it's thick enough to reach the moon?
This is an even clearer way of making my point that doesn't rely on any specific properties of the fold in question - very well-put.
2d
comment Knapsack problem NP-complete
Your statement about showing that the problem is in NP is basically correct, though I prefer to use the 'verifier' model - we can verify a purported solution in polynomial time (and solutions are clearly smaller than the problem itself, so that's not a concern here).
2d
comment Is My Professor Wrong or I Am?
As it stands this question is very unclear as to precisely what's being asked, and even if it's clarified, its applicability to others is questionable. This may just not be a suitable question for this site.
Dec
18
comment Is chess Turing-complete?
It's relatively straightforward if you get 'infinite' pieces, which is what your mapping requires. The original question (infinite board, but only a finite number of pieces) is substantially more challenging.
Dec
18
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
You can certainly do a sort, but that doesn't solve the problem by itself - depending on the count of on bits in the original string, the 'dividing line' between 1 bits and 0 bits can be at an arbitrary place in the resulting string.
Dec
17
comment Product of numbers $\pm\sqrt{1}\pm\sqrt{2}\pm\cdots\pm\sqrt{n}$ is integer
Try finding an inductive proof. Suppose that $P_n$ is the given product; can you find a formula for $P_{n+1}$ in terms of $P_n$?
Dec
16
comment Is Cantor's diagonal argument dependent on the base used?
I rather disagree with the phrashing that 'Cantor's argument is not meant to be a machine that produces reals not in your list'. Diagonlaization is exactly that: it shows that no listing is exhaustive by providing an algorithm for, given any such list, producing a number not on that list.
Dec
16
comment Is indefinite integration non-linear?
@vadim123 Maybe it would be worth using a boldface or calligraphic zero (or even ${\bf 0}(x)$) to denote the zero function, for consistency with the usual ways of differentiating between vector and scalar zeroes? (And agreed with others that this is an excellent way of phrasing the core answer.)
Dec
16
comment Is indefinite integration non-linear?
@Rahul It's not even so much that it's defined that way, as that the vector space axioms force it to be so.
Dec
16
comment What did I do here? This can't be right… ($i = -1$)?
@AlphaMCubed Try plugging in other numbers there - there's nothing 'special' about $i$ in that regard. $-1(4^2)\neq (-4)\cdot(-4)$!