Question

prove that every equation of dimension n has n roots, and no more​

Answer 1

Answer:

Step-by-step explanation:

We have f(α1)=0. Then we can write f(x)=(x−α1)g(x) where g(x) is a polynomial with degree n−1.

So, g(x) must have a root α2. Then we can write f(x)=(x−α1)(x−α2)h(x) where h(x) is a polynomial with degree n−2 etc...

In general we have a proposition that a polynomial f(x) with degree n has n roots for n∈Z+.

So for n=1:

f(x)=a1x+a0=0⟹x=−a0a1 is the only root. So true for n=1.

Assume true for n. Now we must prove true n+1:

Let q(x) have degree n+1. So by the Fundamental Theorem of Algebra, there exists a root of q(x) which we will call α1. Then we can write q(x)=(x−α1)f(x) where f(x) is some polynomial with degree n. From our inductive assumption, we have that f(x) has n roots. Therefore q(x) has n+1 roots as required. Hence our statement holds ∀n∈Z+