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Aug
29
answered Infinite set always has a countably infinite subset
Aug
27
answered Prove by definition that $(x,2)\subset\mathbb Z[x]$ is a maximal ideal
Aug
27
revised Prove by definition that $(x,2)\subset\mathbb Z[x]$ is a maximal ideal
edited title
Aug
27
revised Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
correction
Aug
27
comment when ${\rm gcd} (a,b)=1$, what is ${\rm gcd} (a+b , a^2+b^2)$?
It is not the case that if $d\;\vert\;2ab$ then $d=1$ or $2$ or $d\vert a$ or $d\vert b$. For example, take $d=15,a=3,b=5$.
Aug
24
comment A Solution to the Classic “12 Marbles, find the one of different weight” Brainteaser
The 'clever' solution you have seen is a solution to the problem where you know that one marble has a different weight, but don't know whether it's lighter or heavier. If you know that the marble is heavier (or that it is lighter), then the problem becomes much easier and requires no clever trickery, as you have discovered for yourself.
Aug
24
answered Modulo question: $(\operatorname{rand}[0,n-1]+\operatorname{rand}[0,n-1]+\cdots+) \pmod n$?
Aug
14
comment What is the difference between operators, functions, sequences and vectors?
@FemaleTank Yeah, I understand; I just wanted to make you aware of how things stand in the field of rigorous mathematics, and to show you why the linear relationship you proposed does not in fact hold.
Aug
14
comment What is the difference between operators, functions, sequences and vectors?
$\mathbb R^n$ is defined to be the set of all $n$-tuples of real numbers, which we can endow with a vector space structure, so we are not really choosing a basis, implicitly or otherwise, when we write an element of $\mathbb R^n$ as a tuple. Of course, if we said 'let $V$ be an $n$-dimensional real vector space and then started writing elements of $V$ as tuples, we would be implicitly choosing a basis. One way of defining a basis of an ($n$-dimensional, real) vector space is: an isomorphism from that vector space to $\mathbb R^n$.
Aug
14
answered What is the difference between operators, functions, sequences and vectors?
Aug
12
answered Prove that $S=D$
Aug
12
comment Example of function that is differentiable, but the second derivative is not defined
@lalamer That's the problem with coming up with examples of functions that are continuous everywhere and differentiable nowhere. If you are only interested in a function that's not twice differentiable at one point, then $x^2\cos(1/x)$ is definitely an easier answer. This is just an alternative.
Aug
11
revised Example of function that is differentiable, but the second derivative is not defined
added 108 characters in body
Aug
11
answered Example of function that is differentiable, but the second derivative is not defined
Aug
9
revised Adelic topology on the group of ideles
small correction
Aug
9
comment Orders with no anti-symmetry requirement?
I view this as one reason why we mainly study the theory of partial orders, rather than the more general theory of preorders: the transition from preorder to partial order is fairly trivial, and difficulties only appear in the 'partial order part' of the preorder.
Aug
9
comment Orders with no anti-symmetry requirement?
@TnatsissaHCraeser That's a good way to think about it. For example, in the topological case, if $X$ is a topological space then we can put an equivalence relation on the points of $X$ by saying that $x\sim y$ if $x$ is topoogically indistinguishable from (i.e., is contained in exactly the same open sets as) $y$. Then the quotient space $X/\sim$ is $T_0$ (equivalently, its specialization preorder is a partial order). I imagine there's a similar way to go from preorders to partial orders in general.
Aug
9
awarded  Revival
Aug
7
comment How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$
No no no no no! The quadratic equation $(\lambda-3)^2 - (a+1)^2=0$ can be solved immediately by putting the $(a+1)^2$ on to the right, and then taking (plus or minus) square roots of both sides. Remember how the quadratic formula is derived: we complete the square and then take a square root. Your first equation is already in 'square completed' form, so what you've effectively done is multiplied it out and then put it back into the original form again. The quadratic formula is a last resort!
Aug
7
comment How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$
Oh yes. Nevertheless, it is perhaps easier just to solve the quadratic equation directly by observing that $(\lambda-3)^2=(a+1)^2$, so $\lambda-3=\pm(a+1)$. Equivalent, I know, but still easier.