Reputation
29,533
Next tag badge:
score
answers
Badges
4 46 94
Newest
 Good Answer
Impact
~489k people reached

May
16
comment Example of an odd prime p that remains prime in $\mathbb{Z}[\sqrt{2}]$
Because there are many $\alpha$ and $\beta$ such that $5|\alpha\beta$ - your argument isn't exactly wrong per se, but for instance you could take $\alpha=375\sqrt{2}$ and $\beta=11$. It's much easier to try saying $5=\alpha\beta$; the rest of your argument goes through intact, but then you do in fact have $N(\alpha)=25$ or $1$.
May
15
comment Example of an odd prime p that remains prime in $\mathbb{Z}[\sqrt{2}]$
I think you want $=$ rather than $\mid$ in your proof (i.e., $5=\alpha\beta$, etc.)
May
15
comment Trying to find a function such that $\lim_{x\to\infty} f(x)=0$ but $\lim_{x\to\infty} f'(x) \ \text{does not exist}$.
@marco11 What do you think your Alpha link shows?
May
15
comment Trying to find a function such that $\lim_{x\to\infty} f(x)=0$ but $\lim_{x\to\infty} f'(x) \ \text{does not exist}$.
Consider a scaled sine function: $f(x)=x^{-a}\sin(g(x))$, for some $a$ (we can even try taking $a=1$). You know the oscillations will be shrinking in height; what would you have to do to keep them from 'flattening out' completely?
May
15
comment Question on von Neumann integers and power set
They're certainly isomorphic as sets because they have the same number of elements (and that's all you need to be isomorphic as sets), but they might not be isomorphic with any additional structure layered on top of them. For instance, if $f:3\mapsto 2$ is a function from $3$ to $2$ and $g$ is another such function, then any meaning that you might assign to $f+g$ is highly unlikely to correspond to ordinal addition on the integers.
May
15
answered Twelve identical circles touching one another on the surface of a sphere
May
15
awarded  abstract-algebra
May
13
comment A seemingly easy combinatorics brain teaser
Such a clean final formula begs for a combinatorial explanation!
May
13
answered A seemingly easy combinatorics brain teaser
May
12
answered Does this summation (involving binomial) have a closed form? If so, what is it?
May
12
comment Let $f(x) =7x^{32}+5x^{22}+3x^{12}+x^2$. Then find its remainder in the following cases.
A broad hint: what is $x^2$ equivalent to $\pmod {x^2+1}$?
May
12
comment Let $f(x) =7x^{32}+5x^{22}+3x^{12}+x^2$. Then find its remainder in the following cases.
@salmonkiller it means that OP will have to do 16 'steps' of long division, which is long for a problem.
May
12
comment Covering a rectangle of size $n\times1$ with dominos
@user21820 While it's 'obvious' once you know combinatorics, I think if you asked 'how many ways are there to tile a $2\times 1$ square? How many are there to tile a $1\times1$ square? Now how many are there to tile a $0\times 1$ square?', most people by far (including students) would say 'zero' to the last.
May
12
comment Non-Commutativity Implies Non-Associativity?
To disprove an implication, all you need is one counterexample - your concatenation case is already enough to prove that non-commutativity doesn't imply non-associativity, because you've found a case that's non-commutative but not non-associative. (And canonically, any group is associative, but many many groups are non-commutative.)
May
11
comment If $d \equiv 1 \pmod 4$, is $\mathbb Q[\sqrt d]$ the field of fractions of $\mathbb Z\left[\frac{1+\sqrt d}{2}\right]$?
The key is the difference between $\mathbb{Z}[\tau]$ and $\mathbb{Q}[\tau]$; all fractions are equivalent under the latter, so it makes no difference whether we speak of adding the element $\sqrt{d}$ or $(1+\sqrt{d})/2$ or even $\frac13(5+7\sqrt{d})$ - the extension of $\mathbb{Q}$ by any of these will wind up being equivalent to the others. It's only when looking at the integer ring that we need to be fussy.
May
10
revised If $d \equiv 1 \pmod 4$, is $\mathbb Q[\sqrt d]$ the field of fractions of $\mathbb Z\left[\frac{1+\sqrt d}{2}\right]$?
rolled back to a previous revision
May
10
comment If $d \equiv 1 \pmod 4$, is $\mathbb Q[\sqrt d]$ the field of fractions of $\mathbb Z\left[\frac{1+\sqrt d}{2}\right]$?
What is the difference between the two? (Hint: can you exhibit $a,b\in\mathbb{Q}$ such that $\sqrt{d}=a\cdot\dfrac{1+\sqrt{d}}{2}+b$?)
May
9
comment Please clear up this misunderstanding I have involving limits.
Keep in mind that a function is completely different from an expression - e.g., you could have $f(x)=\frac{x^2-4}{x-2}$ if $x\neq 2$ and $f(2)=54301$; that's a totally legitimate function, but plugging in 2 to it gives you the value $54301$, which clearly isn't the same as the limit. Instead, you want to plug in a sequence of values steadily closer to 2 but not equal to it.
May
9
comment Could all iterates of $s(n)=2n+1$ be composite for some starting $n$?
@AndréNicolas My answer wouldn't have existed if I hadn't read yours; I encourage you to go ahead and undelete your corrected version so that I can upvote it. :-)
May
8
comment The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$
It should be obvious that $\sum_{i=0}^{20}{30\choose i}{30\choose i+10}$ can't be equal to $30\choose 10$; this value is the first member of the sum (=${30\choose 0}{30\choose 10}+{30\choose 1}{30\choose 11}+\ldots$) and all the other terms are positive.