Coe Mod

Coe Mod
Coe Mod

help with number theory proof?

prove: Suppose f(x) = a(subn)x^n + a(subn−1) x^(n−1) + . . . + a(sub0) is a polynomial of degree n > 0
with integer coefficients. Let a, b and m be integers with m > 0. If a ≡ b (mod m), then f(a) ≡ f(b) (mod m).

Intuitively, it’s true because polynomials are composed entirely of multiplications, additions, and subtractions. These operations preserve equality in moduli. So, if a ≡ b (mod m), then:

a + c ≡ b + c
a – c ≡ b – c
ac ≡ bc

If we combine multiple additions, subtractions, and multiplications, we obtain the set of all polynomials, so for any polynomial:

a ≡ b ===> f(a) ≡ f(b)

More formally, we can prove this by induction on the degree of f. Suppose f is of degree 0. Then, for some constant c:

f(x) = c … for all x

Therefore, if a ≡ b, then f(a) = c = f(b) ===> f(a) ≡ f(b). Therefore, we have our base case. Suppose it works for all polynomials of degree k. Let f be a polynomial of degree k + 1, i.e.

f(x) = a_(k + 1)x^(k + 1) + a_k x^k + … + a_1 x + a_0
= x(a_(k + 1)x^k + a_k x^(k – 1) + … + a_1) + a_0

for some a_0, …, a_(k + 1), where a_(k + 1) is non-zero. Let g(x) = a_(k + 1)x^k + a_k x^(k – 1) + … + a_1. Then g(x) is of degree k, since a_(k + 1) does not equal 0, and:

f(x) = x g(x) + a_0

By the induction hypothesis, we know that:

b ≡ c ===> g(b) ≡ g(c)
===> b g(b) ≡ b g(c)
===> b g(b) ≡ c g(c)
===> b g(b) + a_0 ≡ c g(c) + a_0
===> f(b) ≡ f(c)

Therefore, by induction, b ≡ c ===> f(b) ≡ f(c) for all polynomials f.

CALL OF DUTY4 Mod COE

 




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