Question

It’s question 2.3.7 that needs to be answered but only do (iv) please explain with details and circle your answer

Theorem 2.3.6. The following statements hold for all a,b,c,d Z. (i) a | 0, 1 1 a, and a l a. (ii) a l 1 ifand only ifa = ±1. (ii) If a | b and c |d, then ac | bd. (iv) Ifa | b and b | c, then a | c. (v) Ifa | b and a,b >0, then a < b (vi) If a | b and a | c, then a divides every linear combination of b and c. Proof. We prove parts (i). (v), and (vi). The rest are left as Exercise 2.3.7. Note that a-0=0. I-a=a, and a-l=a. This proves part (i). For part (v), we have the additional hypothesis a,b >0, so we assume that there exists some E e N such that al = b. Since 1 > 1, we conclude thata < al = b. To prove part (vi), assume that there exist integers k and 1 such that ak = b and al -c. Then for arbitrary x,y e Z, we have bx + cy -(ak)x(al)y Since kx +ly Z, we see that a l bx cy Exercise23,7, write proofs for parts (ii), (iii), and (iv) of Theorem 2.3.6.

Answer 1

As you said to answer only (iv), the answer for (iv) follows - It is given that a/b i.e b = ax_1 for some x_1 belongs to z Also, given that b/c i.e c = bx_2 for some k_2 belongs to Z' Now, C = bx_2 = a k_1 . k_2 (therefore b = ak_1) = ak_1 k_2 = a_m (therefore let k_1 k_2 = m belongs to Z') = C = a_m i.e a divider C i.e a/e. (proved).