Prove the following Let a epsilon Z Then hcfa a 1 1 Let a

Prove the following: Let a epsilon Z. Then hcf(a, a - 1) = 1. Let a, 6 epsilon Z. Then hcf (a, b) = hcf(a + 176, 6). Let a, 6, c epsilon Z. Show that if hcf(a, b) = 1 and hcf (b, c) = 1, then hcf (ac, b) = 1.

Solution

1>

a = a-1 +1

=> a -(a-1)=1

=> there exists integers u=1 and v=-1 such that au+(a-1)v=1

=> hcf (a,a-1)=1