SOLUTION: 1. Whenever we encounter a new proposition, it is a good idea to explore the proposition by looking at specific examples. For example, let a =20, b = 12, and t = 4. In this case,

Question 500817: 1. Whenever we encounter a new proposition, it is a good idea to explore the
proposition by looking at specific examples. For example, let
a =20, b = 12, and t = 4. In this case, t given a and t given b. In each of
the following cases, determine the value of (ax + by) and determine if t
divides (ax + by).
(a) x = 1; y = 1 a. yes
(b) x = 1; y = -1 b. yes
(c) x = 2; y = 2 c. yes
(d) x = 2; y= -3 d. yes
(e) x = -2; y = 3 e. yes
(f) x = -2; y = -5 f. yes
2. Repeat Part (1) with a = 21, b = -6, and t =3.
a. yes d. yes
b. yes e. yes
c. yes f. yes
3. We started the forward-backward process for the proof of Proposition 4.15
following the discussion of this proposition. Complete the following proof
of Proposition 4.15.
Proposition 4.15. Let a, b, and t be integers with t ≠ 0. If t divides a and t
divides b, then for all integers x and y, t divides ax + by.
Proof. Let a, b, and t be integers with t ≠ 0, and assume that t divides a and
t divides b. We will prove that for all integers x and y, t divides (ax + by).
So let x is an element of Z and let y is an element of Z. Since t divides a, there exists an integer m such that �.

Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Pretty straightforward. If t divides a and t divides b, then a ≡ 0 (mod t) and b ≡ 0 (mod t). Then we can multiply a and b by integers and the residue is still 0 mod t (i.e. ax ≡ 0, by ≡ 0). Then we can add them and we obtain our desired result, ax + by ≡ 0 (mod t) <--> t divides ax + by.
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