Recent content by 83956

Yes, I see the correction in my error of the base case...got it now. However, I'm still not getting the n+1 case. So the induction hypothesis is 1/(1-x) = 1 + x + x^2 + ... x^n/(1-x). I want to show 1/(1-x) = 1 + x + x^2 + ... + x^n/(1-x) + x^(n+1)/(1-x). ... Basically for the n case...

Homework Statement Prove that 1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) for n>=2 Homework Equations The Attempt at a Solution I'm not really all that sure how to begin. The base case would be 1/(1-x) = x2/(1-x) and the induction hypothesis would be 1/(1-x) = 1 + x + x2 + x3 +...

Ok I am thoroughly confused and getting frustrated...

Well Xb=1 so then ca=1 has a=b

Xa-1 has a roots Xb-1 has b roots but since a divides b b=a*s for some positive integer s so set a=a*s => s=1 so the number of roots in each equation is the same ?

So because a and b are real numbers, the roots of unity are 1 and -1, right? So this implies Xa-1 divides Xb-1?

Homework Statement Let q=pm and let F be a finite field with qn elements. Let K={x in F: xq=x} (a) Show that K is a subfield of F with at most q elements. (b) Show that if a and b are positive integers, and a divides b, then Xa-1 divides Xb-1 i. Conclude that q-1 divides...

This is closed under addition because each ai & bi are coefficients (scalars) of the field, so their sum will be a coefficient of the field. Likewise, bi+ai is commutative since they are both coefficients of the field. Addition is associative: anxn+an-1xn-1+...+a1x+a0 +...

Ok, so take two polynomials: anxn+an-1xn-1+...+a1x+a0 bnxn+bn-1xn-1+...+b1x+b0 and sum: (an+bn)xn+(an-1+bn-1)xn-1+...+(a1+b1)x+(a0+b0) where each ai+bi is a unit take "s" to be an integer, so we can multiply: s(anxn+an-1xn-1+...+a1x+a0) =sanxn+san-1xn-1+...+sa1x+sa0

Isomorphism still is kind of a difficult topic for me to pick up, however, I won't pick it up if I don't use it. So let's go that route, as I assume that is what my instructor may be looking for.

The only definition in my book states: A linear map from a vector space V to a vector space W is said to be an isomorphism if it is one-to-one and onto. We say that V is isomorphic to W if there is an isomorphism from V to W. If v1, v2, ... vn is a basis for a vector space V over F, then the...

Homework Statement Let F be a field. Prove that the set of polynomials having coefficients from F and degree less than n is a vector space over F of dimension n. Homework Equations The Attempt at a Solution Since the coefficients are from the field F, the are nonzero. So, if...

b is prime because one of its factors is a unit

either uc or d is a unit?

well since a=ub we can plug in b=a/u so a/u=cd, a=ucd. I'm sorry - I am taking a class in which I teach myself completely, and needless to say it is very frustrating and stressful. This problem should be so easy, and yet it has me stumped ten times over.