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Aug
18
answered Permutations of the elements of $\mathbb Z_p$
Aug
12
comment Galois group of a characteristic polynomial
As $0$ is a known root, it should be easy to find other roots by solving a quadratic and see that the splitting field is the rationals. And the Galois group of any field $F$ over itself is trivial.
Aug
12
answered Conditions for distinct real roots of cubic polynomials.
Aug
1
answered function: bending the y=x line
Jul
25
revised Is the following set open?
added 273 characters in body
Jul
25
answered Is the following set open?
Jul
23
revised Finding Linear Combination of Polynomials
corrected "denominator" to "divisor"
Jul
23
answered Generate an integer matrix such that all submatrices are non-singular
Jul
22
answered Which permutation am I? Or: what is a bijection $f:S_n \rightarrow \{1,2,\ldots,n!\}$ such that we can compute $f(\beta)$ easily?
Jul
22
answered Priority of the 3 axioms of groups
Jul
22
comment The definition of a vector space: closure under scalar multiplication
Take the set of all integers, positive. negative and zero. We can add them up and see it is closed for addition. But when we scalar multiply by, say 8.3 for many integers the result goes outside so it does not satify closure law. If it satisifies closure law it would have qualified to be declared a vector space. Thats all. These are all qualification conditions to be called a vector space. These conditions are distilled from the natural examples like the one I outlined in my answer.
Jul
22
revised The definition of a vector space: closure under scalar multiplication
added 5 characters in body
Jul
22
answered The definition of a vector space: closure under scalar multiplication
Jul
20
answered Find the smallest integer $n$
Jul
17
comment Nested… binomials coefficients?
It is already in closed form, enclosed by a so many parentheses. :D
Jul
17
answered Group acting on $X$ and element of normal subgroup $H$ fixes an element of $X$ implies $H$ fixes all of $X$
Jul
17
comment Subgroups of the multiplicative group of a Field
Your example is correct; so the exercise, in the form you have typed here, is wrong. However it is better to check if there are any assumptions before that exercise that will exclude this.
Jul
16
comment Locating the roots of a cubic polynomial.
I said $b,c,d>0$. You are bringing $-1$ as coefficient of $x^2$. Two conditions are needed.
Jul
16
answered Locating the roots of a cubic polynomial.
Jul
15
answered Find a polynomial with cubic values for consecutive integers.