Prove Divisibility Statements using Mathematical Induction examples with solutions
Question 1:
Prove the following statement by the principle of mathematical induction:
is divisible by 11.
Solution:
We can write,
P(n): is divisible by 11.
Step 1:
We first show that the basis for induction P(1) is true, that is, P(n) is true for .
If , then,
,
which is divisible by 11.
Hence, P(1) is true, that is, P(n) is true for .
Step 2:
Assume that P(k) is true for some natural number k, that is,
is divisible by 11, or
, or where q is a natural number.
We need to prove that P(k +1) is also true. That is, we need to prove that is also divisible by 11 if is divisible by 11.
Now we have,
(Because P(k) is true, and so )
, (where 100q−9∈N, because q∈N)
which is divisible by 11.
Thus, P(k +1) is true whenever P(k) is true.
Therefore, from the principle of mathematical induction, the statement:
is divisible by 11,
is true for all natural numbers n.
Question 2:
Prove the following statement by the principle of mathematical induction:
is divisible by 36.
Solution:
We can write,
P(n): is divisible by 36.
Step 1:
We first show that the basis for induction P(1) is true, that is, P(n) is true for .
If , then,
,
which is divisible by 36.
Hence, P(1) is true, that is, P(n) is true for .
Step 2:
Assume that P(k) is true for some natural number k, that is,
is divisible by 36, or
, or , where q is a natural number.
We need to prove that P(k +1) is also true. That is, we need to prove that is also divisible by 36 if is divisible by 36.
Now we have,
(Because P(k) is true, and so )
, (where 7q+k ∈ N, because q, k ∈ N)
which is divisible by 36.
Thus, P(k +1) is true whenever P(k) is true.
Therefore, from the principle of mathematical induction, the statement:
is divisible by 36,
is true for all natural numbers n.
Tags: mathematical induction examples with solutions, prove divisibility statements by induction
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