Hello,
I have two questions to ask regarding uniform convergence for sequences of functions.
So I know that if a sequence of continuous functions f_n : [a,b] -> R converge uniformly to function f, then f is continuous.
Is this true if "continous" is replaced with "piecewise continuous"...
Hey,
There's one thought experiment that I can't seem to wrap my head around.
If we had an extremely long rod (say > 1 lightsecond) and rotate it about one end rapidly enough (say 1 rad/second), would the far end of the rod be moving faster than the speed of light (using v=omega*r)...
Sorry...i don't know what you mean.
I can come up with counter examples if it said "isomorphic as field extensions", because then i can just use the result that F(a) and F(b) are F-isomorphic iff a and b have the same irreducible poly.
But I'm stuck in this case, with general field isomorphisms...
Hey all,
I need to prove (or disprove) the following statement:
F1 and F2 are two finite field extensions of a field K. Assume [F1:K]=[F2:K]. Then F1 and F2 are isomorphic as fields.
Some help would be much appreciated.
I know the statement is false if i replace "isomorphic as fields"...
Hey,
Does anyone know how to show that these fields are equal:
Q(\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_k})=Q(\sqrt{p_1}+\sqrt{p_2}+...+\sqrt{p_k}),
where p_1,...,p_k are distinct primes in Z.
One inclusion is clear to me, but I'm having problems showing they're equal. Thanks!
Yes, great, thanks for the tip. After a little tedious work, it turns out that it is irreducible over F3, so my original polynomial was irreducible over Q. Yay, thanks a lot :biggrin:
Hmm..okay. I was hoping I wouldn't have to resort to brute force hehe
OK, so I have shown that the polynomial x^6+x^3+2 has no linear or quadratic factors over F3, but how did you show it can't factor into cubic terms? I don't really know the irreducible cubic polynomials over F3, and it...
Hi,
I need to figure out whether or not the polynomial
x^6-2x^3-1
is irreducible (over Q).
I don't think Eisenstein works in this case, and performing modulo 2 on this i get x^6-1
which is reducible over F2.
Any ideas? Incidently, if i let y=x^3, then i get
y^2-2y -1
which is...
Great, thanks a lot guys!
Following Zurtex's method, i calculated:
x^6-9x^4-8x^3+27x^2-72x-11=0 for x=\sqrt{3}+\sqrt[3]{4}.
(There was a little error with zurtex's calculation i think)
I managed to prove that this was irreducible tediously...but (this might be a dumb question) how did you...
Does anyone know how to find the degree over Q of this number:
\sqrt{3} + \sqrt[3]{4}
In fact I'm having trouble finding any generic polynomial that this satisfies! Please help :biggrin:
By rank i mean the maximum number of linearly independent rows/columns of the matrix.
I guess you have to be careful about terminology when you're dealing with modules instead of vector spaces, I just find it a little confusing...I understand your point though, and i'll try to follow your line...
So if they are rank preserving, then the diagonal matrix A'' will have the same rank as A over Z, which is m (since the v's are linearly independent over Z), and since A'' clearly has rank m over R also, applying the inverses of the same integer elementary operations will give me back A, and I...