I'm learning inductive proofs in one of my computer science classes. Our homework was graded and this was one of the solutions. I understand all of it except for one line. If anyone could break down how the marked line in the induction step was derived, that would be much appreciated. Thank you.
5 divides (6^n − 1)
Proof
Base case: When n = 0, (6^n − 1) becomes (6^(0) − 1) = 0. We have 5 divides 0, because 0=5×0. Thus, 5 divides(6^n−1) when n=0.
Induction step: Let n be an arbitrary natural number larger than or equal 0. Assume 5 divides (6^n − 1). We will prove that 5 divides (6^n+1 − 1). Since 5 divides (6^n − 1), there exists an integer x such that 6^n − 1 = 5x. Then we have
6^n+1 −1 = 6^n × 6 − 1
= 6^n × 5 + 6^n −1 <------------ // THIS LINE
= 6^n × 5 + 5x (by the inductive assumption)
= 5(6^n + x)
Since both 6n and x are integers, so is the sum (6^n +x). Thus 6^n+1 −1 equals 5 multiplied with an integer, which means 5 divides 6^n+1 − 1. By induction, we have 5 divides (6^n − 1) for all natural integers n ≥ 0.
Well, $$6n\times 6-1=6n\times (5+1)-1=6n\times 5+6n\times 1-1=6n\times 5+6n-1.$$ The second equality is just the distributive law: $A\times (B+C)=A\times B+A\times C$.
This might be easier to see if we first ignore the "$-1$":
$$6n\times 6=6n\times (5+1)=6n\times 5+6n\times 1=6n\times 5+6n.$$
Note the order of operations: "$\times$" binds more tightly than "$+$," so e.g. "$A\times B+C\times D$" means "$(A\times B)+(C\times D)$," not "$A\times (B+C)\times D$".