Reputation
13,911
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
1 34 66
Newest
 Enlightened
Impact
~346k people reached

Feb
4
revised How to prove that $k^3+3k^2+2k$ is always divisible by $3$?
include expression in title; misc. copyedits
Feb
4
revised Is it possible to solve the following equation without using the Rational Root Theorem?
add line breaks to keep the long equation from extending into the sidebar
Jan
22
revised Example of a relation that is reflexive but not symmetric
swap R2 and R3
Jan
20
comment How to prove that all odd powers of two add one are multiples of three
As hinted at in several answers below, this is exactly the kind of thing that modular arithmetic was invented for. While you can present a solution using "elementary" language, without using (or even being aware of) any terminology from modular arithmetic, that's a bit like driving a nail using a large stone instead of a hammer. Sure, either way works, but having the proper tool makes it so much less awkward.
Jan
14
comment Is $[p \land (p \to q)] \to q$ a tautology?
+1 for actually giving a clear and readable derivation using the rules in the OP's table.
Jan
1
comment Does this have a Euler circuit or a Euler path?
Possible duplicate of Is it possible to draw this picture without lifting the pen?
Dec
28
revised What's the smallest number that we can multiply with a given one to get the result only zeros and ones?
split up the long line of mathjax to make it wrap
Dec
28
answered Example of a relation that is reflexive but not symmetric
Dec
26
comment Subadditivity of square root function
This would be correct if you replaced $\implies$ with $\iff$ (and, perhaps, pointed out that the equivalence is true because $x \mapsto x^2$ is strictly monotone increasing, and thus order-preserving, for non-negative real $x$).
Dec
25
comment Why do we do mathematical induction only for positive whole numbers?
Basically, because you haven't yet studied enough math to run into other, more interesting forms of induction.
Dec
23
comment How to generate random points on a sphere
While this method is simple and clever, it can suffer from numerical accuracy issues near the poles (where ${\rm d}x/{\rm d}z \to \infty$). For example, let's assume that your random number generator internally generates a 32-bit integer $n$, which is then scaled to the desired interval as $z = 20n/2^{32}-10$. Thus, the minimum distance between $z$ values is $20/2^{32}\approx5\times10^{-9}$, while near the poles, the minimum distance between points along a line of longitude is $10\arccos(1-2/2^{32})\approx0.0003$. For a sphere the size of the Earth, that distance would be about 200 meters.
Dec
23
comment How to generate random points on a sphere
Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
Dec
19
comment Why do we need “span” in linear algebra?
Arguably, the fact that a span is closed under addition and scalar multiplication is implicit in the definition you give; the "set of all things you can make" using a certain set of operations, from any given starting points, will necessarily be closed under those operations -- otherwise you could make even more things with them.
Dec
19
comment On what sets can we define a group operation?
To be pedantic, this only works as stated for infinite sets: for finite $X$, $\Omega$ has strictly higher cardinality than $X$. Also, you might want to explicitly note that $\Delta$ denotes (as I assume) the symmetric set difference operator.
Dec
16
comment Is there a function whose antiderivative can be found but whose derivative cannot?
@Joshua: The reason why the Dirac delta "function" doesn't qualify is because, despite the widely used name, it's not really a function but a distribution.
Dec
13
comment If $n$ is composite, then $n$ divides $(n-1)!$.
Ps. See also Wilson's theorem, which says that $(n-1)! \equiv -1 \pmod n$ if and only if $n$ is prime.
Dec
12
revised After 6n roll of dice, what is the probability each face was rolled exactly n times?
simplify denominator; remove "edit" (posts already have an edit history; you don't need to mark edits within the post itself)
Dec
12
comment Some basic questions regarding rank-1 matrices
@User001: By the same argument given by Roland above, the product of two matrices cannot have a higher rank than either of the matrices alone. Since we know that $A^{-1}(AB) = (A^{-1}A)B = B$ has rank 1, this means that both $A^{-1}$ and $AB$ must have rank at least 1. (Of course, for $A^{-1}$, we actually know more; namely that, as invertible matrices, both $A$ and $A^{-1}$ in fact have full rank. In fact, we can prove this simply by replacing $B$ in the previous argument with, say, the identity matrix. But for the proof that $AB$ has rank 1, we don't really need this extra information.)
Dec
12
comment Some basic questions regarding rank-1 matrices
@User001: Yes, because then $A^{-1}(AB) = B$ has rank 1, and so $AB$ must have rank at least 1.
Dec
11
awarded  Enlightened