Answer
Proof by mathematical induction,
Base Case: Show that P(0) is true:
P(0) = 03 – 7*0 + 3 = 3 and 3 is divisible by 3. So, P(0) is true.
Inductive Case: Show that for all integer k ≥ 0, if P(k) is true then P(k+1) is true:
Let k be any integer such that k ≥ 0 and suppose P(k) is true.
This means k3 - 7k + 3 is divisible by 3. Now we must show that P(k+1) is true.
P(k+1) = (k+1)3 - 7(k+1) + 3
= k3 + 3k2 + 3k +1 -7k -7 + 3
= k3 + 3k2 -4k -3
= k3 – 7k + 3 +3k2 +3k – 6 [expressing above line in the form of P(k)]
Now we see that k3 – 7k + 3 is equal to P(k) and P(k) is divisible by 3.
Let (k3 – 7k + 3) / 3 = x. Where x is some integer
Then, (k3 – 7k + 3) = 3x
So, P(k+1) = 3x + 3k2 + 3k – 6
= 3(x+ k2 + k -2) which is divisible by 3.
So, by definition of divisibility, n3 – 7n + 3 is divisible by 3
Work Step by Step
Steps:
1. Take any number k and show that the statement holds true for that value of k.
2. Show that if the statement is true for k, the statement holds true for the value k+1. This can be done by taking the statement where the value is k+1 and reducing it into the statement where the value is k.