[quote=R.D. Silverman;218562](3) Prove the remainder theorem:

The remainder left when f(x) is divided by (x-r) equals f(r).

.[/quote]
I guess I need to solve this problem.
I can't understand this theorem....

[quote=blob100;220729]I guess I need to solve this problem.
I can't understand this theorem....[/quote]Let's see if this helps you understand. First an explanation of the notation:

f(x) is a polynomial in x. That is f(x) is shorthand for a_0 +a_1 * x + a_2 * x^2 + ... where the a_i are all constants and there are finitely many non-zero a_i.

(x-r) is a polynomial in x where a_0 equals -r, a_1 equals 1 and all other a_i are zero. It's a linear and monic polynomial, though that's jargon you don't strictly need to solve the problem.

f(r) is the value of f(x) when x has the value r and r is a constant.


Now, just to ensure all this is nailed down, let's take a particular example where f(x) is 1+4x+3x^2 and r=2.

It's fairly easy to see that (1+4x+3x^2) = (x-2)*(3x+10) + 21, so the remainder when (1+4x+3x^2) is divided by (x-2) is 21.

Substituting the value 2 into the example polynomial gives 1 + 4*2 + 3*(2^2) = 1+8+3*4 = 21.

Your task is to prove that the stated result holds for all possible polynomials and all values of r.


Paul

[quote=xilman;220734]Let's see if this helps you understand. First an explanation of the notation:

f(x) is a polynomial in x. That is f(x) is shorthand for a_0 +a_1 * x + a_2 * x^2 + ... where the a_i are all constants and there are finitely many non-zero a_i.

(x-r) is a polynomial in x where a_0 equals -r, a_1 equals 1 and all other a_i are zero. It's a linear and monic polynomial, though that's jargon you don't strictly need to solve the problem.

f(r) is the value of f(x) when x has the value r and r is a constant.


Now, just to ensure all this is nailed down, let's take a particular example where f(x) is 1+4x+3x^2 and r=2.

It's fairly easy to see that (1+4x+3x^2) = (x-2)*(3x+10) + 21, so the remainder when (1+4x+3x^2) is divided by (x-2) is 21.

Substituting the value 2 into the example polynomial gives 1 + 4*2 + 3*(2^2) = 1+8+3*4 = 21.

Your task is to prove that the stated result holds for all possible polynomials and all values of r.


Paul[/quote]

So I'm needed to prove:
f(x)=f(r)(mod x-r)
Where f(x) is a given polynomial in x, f(r) is the result when x=r and x-r is a monic polynomial with a_0=-r and a_1=1.
Example:
f(x)=(1+4x+3x^2) = (x-2)*(3x+10) + 21 is equivallent to:
f(x)=21(mod x-2)
Where we know f(2)=21.

Here is the proof:
We try to show the next congruence as valid:
f(x)=f(r)(mod x-r).
Where we state f(x)=anx^n+...+a0.
A={a0,a1,...,an} the set of the coefficients of f(x) a polynomial in x of degree n.
f(r) is the result given by x=r, where r is a constant.
x-r is a monic polynomial a0=-r, a1=1.
We try to prove:
anx^n+...+a0=r^n+...+a0(mod x-r).
an(x^n-r^n)+a(n-1)(x^(n-1)+r^(n-1))+....+a1(x-r)=0(mod x-r).
This last phrase is trivially true by the next useful theorem:
Theorem:
If x=/=y and n>0 (natural number), then:
x-y/x^n-y^n.
We have shown that every such ai(x^i-r^i) in the sum is devided by
x-r, which gives the whole sum is devided by x-r.

[QUOTE=blob100;220893]So I'm needed to prove:
f(x)=f(r)(mod x-r)


Where f(x) is a given polynomial in x, f(r) is the result when x=r and x-r is a monic polynomial with a_0=-r and a_1=1.
Example:
f(x)=(1+4x+3x^2) = (x-2)*(3x+10) + 21 is equivallent to:
f(x)=21(mod x-2)
Where we know f(2)=21.[/QUOTE]

This is totally wrong. You keep failinbg to check your work.
if e.g. x=7, then f(x) = 176, x-2 = 5, so f(x) = 1 mod x-2, yet
you claim f(x) = 21 mod x-2.

This is a totally incorrect statement of the problem.

You need to prove that f(r) = the remainder when f(x) is divided by
(x-r).

[QUOTE=blob100;220894]Here is the proof:
We try to show the next congruence as valid:
f(x)=f(r)(mod x-r).
Where we state f(x)=anx^n+...+a0.
A={a0,a1,...,an} the set of the coefficients of f(x) a polynomial in x of degree n.
f(r) is the result given by x=r, where r is a constant.
x-r is a monic polynomial a0=-r, a1=1.
We try to prove:
anx^n+...+a0=r^n+...+a0(mod x-r).
an(x^n-r^n)+a(n-1)(x^(n-1)+r^(n-1))+....+a1(x-r)=0(mod x-r).
This last phrase is trivially true by the next useful theorem:
Theorem:
If x=/=y and n>0 (natural number), then:
x-y/x^n-y^n.
We have shown that every such ai(x^i-r^i) in the sum is devided by
x-r, which gives the whole sum is devided by x-r.[/QUOTE]

No! You are mis-stating the problem.

The proof of the correct proposition only takes 2 lines.

[quote=R.D. Silverman;220900]This is totally wrong. You keep failinbg to check your work.
if e.g. x=7, then f(x) = 176, x-2 = 5, so f(x) = 1 mod x-2, yet
you claim f(x) = 21 mod x-2.

This is a totally incorrect statement of the problem.

You need to prove that f(r) = the remainder when f(x) is divided by
(x-r).[/quote]

Do you mean:
f(x)/(x-r)=f(r)?
Or:
f(x)-f(r) is always devided by x-r?

[quote=blob100;220911]Do you say:
f(x)/(x-r)=f(r)?[/quote]No. The statement is that f(r) is the remainder ([B]not[/B] the quotient) when f(x) is divided by (x-r).
[quote=blob100;220911]
Or:
f(x)-f(r) is always devided by x-r?[/quote]Yes, except that a native English writer would use "f(x)-f(r) is a multiple of (x-r)" or, perhaps, "f(x)-f(r) is divisible by (x-r)". In the second of these, the proviso "without remainder" is generally understood but would be used if complete clarity were required

Paul

[quote=xilman;220913]No. The statement is that f(r) is the remainder ([B]not[/B] the quotient) when f(x) is divided by (x-r).
Yes, except that a native English writer would use "f(x)-f(r) is a multiple of (x-r)" or, perhaps, "f(x)-f(r) is divisible by (x-r)". In the second of these, the proviso "without remainder" is generally understood but would be used if complete clarity were required

Paul[/quote]

This is what I tought.
So, f(x)=f(r)(mod x-r).

Proof:
We would like to prove (x-r)|f(x)-f(r) where,
f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_0.
f(r)=a_nr^n+....+a_0.
A={a_0,....,a_n} the set of the coefficients of f(x).
Now, let's see the next equation:
f(x)-f(r)=a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r).
To prove that f(x)-f(r) is divisible by x-r we will show
a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r)=0(mod x-r).
I would like to state the next theorem:
The next: z-y|z^b-y^b
Takes place for all z=/=y, b natural numbers.

The next is true:
a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r)=0(mod x-r)
Becuase the last theorem.
And we have:
f(x)-f(r)=0(mod x-r).

[QUOTE=blob100;220914]This is what I tought.
So, f(x)=f(r)(mod x-r).

Proof:
We would like to prove (x-r)|f(x)-f(r) where,
f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_0.
f(r)=a_nr^n+....+a_0.
A={a_0,....,a_n} the set of the coefficients of f(x).
Now, let's see the next equation:
f(x)-f(r)=a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r).
To prove that f(x)-f(r) is divisible by x-r we will show
a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r)=0(mod x-r).
I would like to state the next theorem:
The next: z-y|z^b-y^b
Takes place for all z=/=y, b natural numbers.

The next is true:
a_n(x^n-r^n)+a_(n-1)(x^(n-1)-r^(n-1))+....+a_1(x-r)=0(mod x-r)
Becuase the last theorem.
And we have:
f(x)-f(r)=0(mod x-r).[/QUOTE]

This works, but you are "hitting a thumbtack with a sledgehammer".

The "proof from the book", ala P. Erdos takes only two lines.

[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]

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.