how do you prove that (a-1) divides ((a^n)-1) evenly using mathematical induction
To prove that (a-1) divides ((a^n)-1) evenly using mathematical induction, we need to follow a step-by-step approach. Here's how you can do it:
Step 1: Base case
Start by proving that the statement holds true for the base case, which is usually n = 1. Substitute n = 1 into the equation and verify that (a-1) divides (a^1 - 1) evenly, i.e., there is no remainder when dividing.
Step 2: Assume the statement is true for n = k
Next, assume that the statement holds true for n = k, where k is an arbitrary positive integer. This is called the induction hypothesis. So, assume that (a-1) divides (a^k - 1) evenly.
Step 3: Prove the statement is true for n = k+1
Now, we need to prove that the statement holds true for n = k+1. Substitute n = k+1 into the equation and manipulate the expression to show that (a-1) divides (a^(k+1) - 1) without any remainder.
One approach is to write a^(k+1) - 1 as a^k · a - 1 and factorize it further, so you have:
a^(k+1) - 1 = a^k · a - 1
To simplify this expression, we can use the assumption made in step 2, which states that (a - 1) divides (a^k - 1) evenly. Hence, we can rewrite a^k as (a - 1) · m for some integer m. Substituting this into the equation, we get:
(a - 1) · m · a - 1 = (a - 1) · (m · a - 1)
From this equation, we clearly see that (a - 1) divides (a^(k+1) - 1) without leaving any remainder.
Step 4: Use mathematical induction
By establishing the base case (step 1) and proving that if the statement holds for n = k, then it also holds for n = k+1 (step 3), we can conclude that the statement holds for all positive integers n by mathematical induction.
Hence, we have shown that (a-1) divides ((a^n)-1) evenly using mathematical induction.