This is what i thought the definition of a constructible number is :
A real number is constructible if and only if, given a line segment of unit length, one can construct a line segment of length | r | with compass and straightedge.
So then a line segment would be constructible ( by...
Hi,
I was trying to prove the following theorem:
if x is a constructible number <=> it can be obtained from Q by taking a the square root a finite number of times ( or applying a finite # of field operations).
I managed to get the proof for <= this way, but I am not really sure on...
let F be a field with char p. Let a, b be in the field, with a not equal b^p .
show that f(x) = x^p - a is irreducible
i was thinking to start by contradiction
assume f(x) is not irreducible...than f(x) = (x-a1)(x-a2)...(x-an)
where no a can be equal to a p th power of b.
in order for...
10 sounds right
put the problem in mathematical terms:
juniors = j , seniors = s
j+s=15.....(1)
3 more juniors and 7 more seniors join.. so we'll have j+3 juniors and s+7 seniors
now the ratio of new j to new s is 2:3
that means :
j+3/s+7 = 2/3......(2)
so now u have 2 equations (1)...
ok this is how the theorem goes:
let y = p(x) and g(x,y) = 0 be 2 curves. Assume y = p(x) contains 0= (0,0) and that (y-p(x)) does not divide g(x). Then the intersection multiplicity at 0( i assume I_0 .. I sub zero...means that) of y- p(x) and g(x,y) is the smallest degree of any non zero...
shmoe.. i didnt actually find the solutions.
i let t^3=y from the frist equation
and then expressed everything in terms of t in the second equation, and then by
a theorem ( which i don't know the name of).. the lowest power of the non-zero
terms is the number of time the curves intersect at 0.
i have to show how many times the curves intersect at the origin
y^4 = x^ 3 and x^2y^3 - y^2+ 2x^7= 0
i don't really know how to start solving this :rolleyes:
yeah except one thing:
when you multiply the faction by 10, it only goes on the top, not on the bottom. So then when you invert it, its going to be on the botton
(x^2)(y^2) / (10 z^6)
the rest is fine now u just need to cancel the y^2 and the z
but by doing what u guys said..that would give me the solutions to the congruence. I don't really need to find them. Is there maybe another way to do it, just to show that it has solutions, without actually finding them?
I have to show that this congruence has solutions:
x^2 == -23 ( mod 4*59)
i don't think i can use the legendre symbol for that bc 4* 59 is even.
can i use the jacobi symbol ? ( -23 /4*59) or does it have to be odd too ?
i'm not exactely sure what u mean.. but i think i got it more or less right..i'm going to write it like that.. maybe explain more what i did..
thx again for the hint
ok let me see if i got the first part now :
let p = 5 or 1 or 9 ( mod 20)
then p can be written as follows p = ax^2 + bxy + cy^2 ( for all p : you can let
a=p and x =1, y = 0 )
then 4ap == 1,4,9,16 ( mod 20)
then ap == 1,4,9 or 16 (mod 20) or ap=5
but we already said p =5 or p...
yes i did mean cy^2
for the frist part i guess i just assumed that p could be written as a binary quadratic form... i'll have to think about that
and for the second part i just took 4p = 0 ( mod 20), 4p = 1 ( mod 20 ) .. etc.. and solved for p
ok well if u put it that way it's not that hard:)
ok related to that question i have to show that if p = x^2+5y^2
than (p/5 ) ( the legendre symbol) = 1 or 0 and that p is represented like that iff
p =5 or p == 1 or 9 mod (20)
ok this is what i did:
discriminant = -20
i showed...