[quote=R.D. Silverman;218562](4) Prove the rational roots theorem:

If alpha = a/b is a rational root of f(x) then b divides a_n
and a divides a_0.

.[/quote]

The rational roots theorem:
If A=a/b (where (a,b)=1) a rational root of f(x) then b divides a_n
and a divides a_0.

Proof:
B={a0,...,an} the coefficients of f(x).
f(a/b)=an(a/b)[SUP]n[/SUP]+...+a0=0.
We will multiplying each side by b[SUP]n[/SUP].
an(a[SUP]n[/SUP])+...+a0(b[SUP]n[/SUP])=0.
And by simple algebra:
a(an(a[sup]n-1[/sup])+a(n-1)(b*a[sup]n-2[/sup])+...+a1(b[sup]n-1[/sup])=-a0(b[sup]n[/sup]).
b(a(n-1)(a[sup]n-1[/sup])+a(n-2)(b*a[sup]n-2[/sup])+...+a0(b[sup]n-1[/sup])=-an(a[sup]n[/sup]).
These two phrases, each gives: a|a0b[sup]n[/sup], b|an(a[sup]n[/sup]) respectively.
Becuase (a,b)=1 we have: a|a0, b|an.

[QUOTE=blob100;220965]Here is a shorter proof.

Proof:
We define f(x) as a polynomial of degree n with,
A={a0,...,an} the set of its coefficients.
r is a constant.
1) We try to prove: f(x)=f(r)(mod x-r).
This next is equivallent to (1): anx^n+...+a0=anr^n+...+a0(mod x-r),
And is proved easily by the next theorem:
The next: z-y|z^b-y^b
Takes place for all z=/=y, b natural numbers.[/QUOTE]

Forget about working with ideals. A direct proof is much simpler.

[quote=R.D. Silverman;220938]This works, but you are "hitting a thumbtack with a sledgehammer".

The "proof from the book", ala P. Erdos takes only two lines.[/quote]Tomer: what Bob says is true, but don't get discouraged. A valid proof you find for yourself is praiseworthy, and much better than one you merely copy out of a book, even though other proofs may be shorter and/or more elegant.

For some theorems, there appears to be a sort of unofficial competition to find the largest number of different proofs. The proof that there are an infinite number of primes is a famous example.

Paul

[quote=xilman;220991]Tomer: what Bob says is true, but don't get discouraged. A valid proof you find for yourself is praiseworthy, and much better than one you merely copy out of a book, even though other proofs may be shorter and/or more elegant.

For some theorems, there appears to be a sort of unofficial competition to find the largest number of different proofs. The proof that there are an infinite number of primes is a famous example.

Paul[/quote]

I'm not discouraged.
But I know I must learn how to make the shortest proof and make valid proofs.

[quote=R.D. Silverman;220989]Forget about working with ideals. A direct proof is much simpler.[/quote]

To try proving it with another direction or to continue to the next problem?

Hint: show that there is a polynomial [I]g[/I]([I]x[/I]) such that [I]x[/I][SUP][I]n[/I][/SUP] = ([I]x[/I]-[I]r[/I]) [I]g[/I]([I]x[/I]) + [I]r[/I][SUP][I]n[/I][/SUP].

[QUOTE=akruppa;221054]Hint: show that there is a polynomial [I]g[/I]([I]x[/I]) such that [I]x[/I][SUP][I]n[/I][/SUP] = ([I]x[/I]-[I]r[/I]) [I]g[/I]([I]x[/I]) + [I]r[/I][SUP][I]n[/I][/SUP].[/QUOTE]

This is not correct.

Instead, put:

f(x) = (x-r)g(x) + h(x). It is also true that h(x) is constant, but this info is
not needed.

i.e. simply factor (x-r) from f(x).

I think it works with [TEX]g(x) = \sum_{i=0}^{n-1} r^{n-1-i} x^i.[/TEX]

Then

[TEX](x-r)g(x) + r^n=[/TEX]
[TEX]x \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) - r \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) + r^n=[/TEX]
[TEX]\left(\sum_{i=0}^{n-1} r^{n-1-i} x^{i+1}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/TEX]
[TEX]\left(\sum_{i=1}^{n} r^{n-i} x^{i}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/TEX]
[TEX]x^n + \left(\sum_{i=1}^{n-1} r^{n-i} x^{i}\right) - \left(\sum_{i=1}^{n-1} r^{n-i} x^i\right) - r^n + r^n=x^n.[/TEX]

If not, where did I make the error?

[QUOTE=akruppa;221065]I think it works with [TEX]g(x) = \sum_{i=0}^{n-1} r^{n-1-i} x^i.[/TEX]

Then

[TEX](x-r)g(x) + r^n=[/TEX]
[TEX]x \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) - r \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) + r^n=[/TEX]
[TEX]\left(\sum_{i=0}^{n-1} r^{n-1-i} x^{i+1}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/TEX]
[TEX]\left(\sum_{i=1}^{n} r^{n-i} x^{i}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/TEX]
[TEX]x^n + \left(\sum_{i=1}^{n-1} r^{n-i} x^{i}\right) - \left(\sum_{i=1}^{n-1} r^{n-i} x^i\right) - r^n + r^n=x^n.[/TEX]

If not, where did I make the error?[/QUOTE]

Look at the degree of the LHS and the degree of the remainder term.

[quote=akruppa;221065]I think it works with [tex]g(x) = \sum_{i=0}^{n-1} r^{n-1-i} x^i.[/tex]

Then

[tex](x-r)g(x) + r^n=[/tex]
[tex]x \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) - r \left(\sum_{i=0}^{n-1} r^{n-1-i} x^i\right) + r^n=[/tex]
[tex]\left(\sum_{i=0}^{n-1} r^{n-1-i} x^{i+1}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/tex]
[tex]\left(\sum_{i=1}^{n} r^{n-i} x^{i}\right) - \left(\sum_{i=0}^{n-1} r^{n-i} x^i\right) + r^n=[/tex]
[tex]x^n + \left(\sum_{i=1}^{n-1} r^{n-i} x^{i}\right) - \left(\sum_{i=1}^{n-1} r^{n-i} x^i\right) - r^n + r^n=x^n.[/tex]

If not, where did I make the error?[/quote]

Is this the easier proof? I think I gave a much simpler way to prove this theorem.

[QUOTE=blob100;221142]Is this the easier proof? I think I gave a much simpler way to prove this theorem.[/QUOTE]

The proof takes two steps.

(1) Put

f(x) = (x-r) g(x) + r(x) where r(x) is the remainder after dividing by (x-r).
Since (x-r) is linear and the degree of the remainder is less than the divisor,
we have r(x) = constant, but this is not needed.

(2) Put x = r yielding f(r) = r(r) = constant. QED.