Recent content by sihag

i don't have a copy of that book, so here's a conjecture (going by the looks of it): that's the ratio test. you determine the ratio of (n+1)th term to the n th term as n approaches infinity, and conclude depending on whether the value obtained is greater or less that 1, the series is cgt or...

bartle and sherbert !

this topic has so much of that feel good factor i needed ! : )

f(x) = x , if x is rational = 0 , if x is irrational on the interval [0,1] i just wanted to check if my reasoning is right. take the equipartition of n equal subintervals with choices of t_r's as r/n for each subinterval. calculating the integral as limit of this sum (and...

i'm stuck on this relatively simply integral for a differential equation problem ... | du / cos(pi/4 - u ) where | denotes the integral sign some help ?

oops, i missed that ! : )

something much neater! If a vector space V is finite dimensional , of dim n say, over a field F then, (assume, {v1,v2,...,vn} is a basis for V) V is isomorphic to F^n through the correspondence a1.v1 +a2v2 + a3v3 + ... + anvn -----> (a1,a2,a3,...,an) {the n component tuple of F^n)...

i did not understand the geometric bit. well i considered the principal ideals <2> and <3> their union includes 1 which is a unit in Z, so the ideal of the sum is nothing but Z itself right ? and that can't be prime by definition ? (since an ideal P is prime => P /= R (the ring in...

i was looking for a counter example. and, I've not been able to think of any.

let f(2) = 3n , f.s. n in Z f(2 + 2) = f(4) , and f(2.2) = f(4) f(2 + 2) = f(2) + f(2) = 6n f(2.2) = f(2)f(2) = 9n^2 => 6n = 9n^2 => n = 0 (contradiction as f is an isomorphism, so must be injective, so only 0 maps to 0) or, n = 2/3 (contradiction as n belongs to Z) Q.E.D...

I was proving 2Z is non-isomorphic to 3Z I tried it by contradiction, of course. If possible there exists a ring homomorphism f : 2Z ---> 3Z Then, f (x.y) = f(x).f(y) must hold x,y belong to 2Z So x = 2a, f.s. a belonging to Z y = 2b, f.s. b belonging to Z so x.y = 2z, f.s. z...

Suppose A : n x n A is invertible iff the columns (and rows) of A are linearly independent A is invertible iff det |A| is non-zero iff rank A is n iff column rank is n iff dim (column space is n) iff the n columns of A are linearly independent Well, this is a proof that I laid down...

the entire proof is there ! scroll up to my reply.

erm, where did my reply go ? this is strange.

Suppose ST = I To show T is injective: Let T(x)=T(y) , f.s. x,y in V => S(T(x)) = S(T(y)) [permitted since S is well defined from W to V, and T is of course from V to W] => (ST)(x) = (ST)(y) => I(x) = I(y) => x = y proving, T is injective. I think the converse is true only...