# Thread: Proof of algebraic number

1. ## Proof of algebraic number

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!

2. If a is algebraic of degree 5 over $\displaystyle \mathbb{Q}$ then by definition there is a minimal irreducible polynomial of degree n $\displaystyle p(x)\in \mathbb{Q}[x]$ such that a is a root ie p(a)=0.

But then consider the polynomial q(x)=p(x+5).
q(a-5)=p((a-5)+5)=p(a)=0.

It clearly also has degree n, and by the hint it seems you have a theorem to show you why q(x)=p(x+5) is also irreducible.

If you cannot find it, a proof by contradiction is very easy, suppose q(x) were reducible, then you would have a factorization of p(x) which would make p(x) not irreducible which is a contradiction. Thus you have a minimal irreducible polynomial of degree n for which a-5 is a root, thus a-5 is algebraic of degree n.

3. I forgot to thank you for this the other day. This helped me a lot.