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visits member for 2 years, 5 months
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May
31
awarded  Enthusiast
Apr
25
comment If G is a finite abelian group and $a_1,…,a_n$ are all its elements, show that $x=a_1a_2a_3…a_n$must satisfy $x^2=e$.
What you have defined is obviously not a group. E.g. you've already seen that 22=6=42, but then (22)3=(42)3 and -- if associativity would hold -- 2=4.
Apr
24
answered 5 digit number $a6a41$ divisible by 9
Apr
11
answered Equivalence relation: prove that $(X \cap Y) $\ $E $ $\subset (X$ \ $E) \cap (Y$ \ $E)$
Apr
9
comment Equivalence relation: prove that $(X \cap Y) $\ $E $ $\subset (X$ \ $E) \cap (Y$ \ $E)$
Are you sure that you have written the inclusion in the right direction?
Apr
9
comment How do I find arccos(-16.503)
So, the problem seems to lie in an earlier step. How did you arrive at the expression $\arccos(-16.503)$?
Apr
4
comment cartesian product $A^2 = A$, possible?
Am I missing something, or should it be $A_\omega\times A_\omega\subset A_\omega$?
Apr
3
awarded  Yearling
Apr
3
comment Limit definition by ordinal numbers
Principia Mathematica, *207: "A term x is said to be the "upper limit" of alpha in P if alpha has no maximum and x is the sequent of alpha. In this case, x immediately follows the class alpha, though there is no one member of alpha with x immediately follows."
Apr
3
answered How multiple of number is determined?
Apr
2
comment Can two function $f$ and $g$ have same values through out a given interval and different values outside that interval?
@user136561: This merely shows the difference (within a certain interval) is smaller than the resolution of the computer monitor.
Apr
1
answered Show that for triangle ABC, with complex numbers for the coordinates, that we have the following equation
Apr
1
comment Why does $f(x)=ax^2 + bx + c \ge 0\ \forall x \in \mathbb R$ imply $f$ has at most one real distinct root and discriminant $D \le 0$?
@Sabyasachi: Take $a=1$, $b=c=0$.
Mar
31
comment proving that $f:\mathbb N\to\mathbb N\times\mathbb N$ is countable using Cantor's diagonal method
Sorry for the terrible punctuation in my comment above. $f$ may be defined as $\{(n,f(n)): n\in\mathbb N\}$.
Mar
31
comment proving that $f:\mathbb N\to\mathbb N\times\mathbb N$ is countable using Cantor's diagonal method
This set of functions, i.e. $(\mathbb N\times\mathbb N)^\mathbb N$ certainly is not countable. You can use Cantor's diagonal method to prove that. Note that <i>the set of functions</i> and <i>the function</i> (which one) are different concepts. A function $\mathbb N\to\mathbb N$ can be defined as a set (of pairs $(n,f(n)$, and this would indeed be countable. But this is almost certainly not what you want to know. Could you edit the question to incorporate answers you gave in the comments?
Mar
28
comment Show from the axioms: Addition in a quasifield is abelian
@azimut: Thanks, I've added the answer accordingly, fixing an incorrect eq reference along the way.
Mar
28
revised Show from the axioms: Addition in a quasifield is abelian
Fix an incorrect reference; incorporate suggestions from comments
Mar
28
answered Show from the axioms: Addition in a quasifield is abelian
Mar
25
comment Is there any number $n$ such that $nm=0$, $n\neq 0$, and $m\neq 0$?
@YiyuanLee: It is not a group but merely a monoid (there are no zero divisors in a group since they do not have an inverse). The group $(ℤ/6ℤ)^*$ exists, but it does not contain the elements 2 and 3.
Feb
24
answered Why is an algebra not a $\sigma$-algebra by induction?