Let . Prove: If a is an algebraic number of degree n over Q, then a-5 is also an algebraic number of degree n over Q.
Hints: To show that a-5 is an algebraic number, and has the degree n over Q, we must find an irreducible polynomial of degree n such that q(a-5)=0.
Look over the unit for the connection between the irreducibility of polynomials p(x), p(x+r), and p(x-r) for a rational number r.
I suck at proofs. Royally. It's very rare that I get all the points for one. Here's what I have so far. (I think.)
Assume . Let . Assume that q(x) is reducible. Then there are nonconstant rational polynomials such that q(x)=f(x)g(x). Note that q(x-a)=f(x-a)g(x-a). Since f(x-a) and g(x-a) have the same degree as f(x) and g(x), the polynomials f(x-a) and g(x-a) are nonconstant. Thus, q(x-a) is reducible.
You know what, I just realized this does the opposite of what I need to do. I need to show that something is irreducible, not reducible. Help point me in the right direction please!