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Examples – Proof by Mathematical Induction for Divisibility Statements

Lata Agarwal Mathematical Induction, Mathpedia

mathematical induction, prove inequalities using mathematical induction examples with solution, divisibility statements, summation identities

Prove Divisibility Statements using Mathematical Induction examples with solutions

Question 1:

Prove the following statement by the principle of mathematical induction:

${10^{2n - 1}} + 1$ is divisible by 11.

Solution:

We can write,

P(n): ${10^{2n - 1}} + 1$ 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 $n = 1$ .

If $n = 1$ , then,

${10^{2n - 1}} + 1 = {10^{2 \times 1 - 1}} + 1 = {10^1} + 1 = 10 + 1 = 11$ ,

which is divisible by 11.

Hence, P(1) is true, that is, P(n) is true for $n = 1$ .

Step 2:

Assume that P(k) is true for some natural number k, that is,

${10^{2k - 1}} + 1$ is divisible by 11, or

${10^{2k - 1}} + 1 = 11q$ , or ${10^{2k - 1}} = 11q - 1$ where q is a natural number.

We need to prove that P(k +1) is also true. That is, we need to prove that ${10^{2\left( {k + 1} \right) - 1}} + 1$ is also divisible by 11 if ${10^{2k - 1}} + 1$ is divisible by 11.

Now we have,

${10^{2\left( {k + 1} \right) - 1}} + 1$

$= {10^{2k + 2 - 1}} + 1$

$= {10^{2k - 1 + 2}} + 1$

$= {10^{2k - 1}} \cdot {10^2} + 1$

$= {10^{2k - 1}} \cdot 100 + 1$

$= \left( {11q - 1} \right) \cdot 100 + 1$ (Because P(k) is true, and so ${10^{2k - 1}} = 11q - 1$ )

$= 11q \cdot 100 - 100 + 1$

$= 1100q - 99$

$= 11\left( {100q - 9} \right)$ , (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:

${10^{2n - 1}} + 1$ is divisible by 11,

is true for all natural numbers n.

Question 2:

Prove the following statement by the principle of mathematical induction:

${7^n} - 6n - 1$ is divisible by 36.

Solution:

We can write,

P(n): ${7^n} - 6n - 1$ 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 $n = 1$ .

If $n = 1$ , then,

${7^n} - 6n - 1 = {7^1} - 6 \cdot 1 - 1 = 7 - 6 - 1 = 0$ ,

which is divisible by 36.

Hence, P(1) is true, that is, P(n) is true for $n = 1$ .

Step 2:

Assume that P(k) is true for some natural number k, that is,

${7^k} - 6k - 1$ is divisible by 36, or

${7^k} - 6k - 1 = 36q$ , or ${7^k} = 36q + 6k + 1$ , where q is a natural number.

We need to prove that P(k +1) is also true. That is, we need to prove that ${7^{k + 1}} - 6\left( {k + 1} \right) - 1$ is also divisible by 36 if ${7^k} - 6k - 1$ is divisible by 36.

Now we have,

${7^{k + 1}} - 6\left( {k + 1} \right) - 1$

$= {7^{k + 1}} - 6k - 6 - 1$

$= {7^{k + 1}} - 6k - 7$

$= {7^k} \cdot {7^1} - 6k - 7$

$= {7^k} \cdot 7 - 6k - 7$

$= \left( {36q + 6k + 1} \right)7 - 6k - 7$ (Because P(k) is true, and so ${7^k} = 36q + 6k + 1$ )

$= 36 \cdot 7q + 6 \cdot 7k + 7 - 6k - 7$

$= 36 \cdot 7q + 42k - 6k$

$= 36 \cdot 7q + 36k$

$= 36\left( {7q + k} \right)$ , (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:

${7^n} - 6n - 1$ 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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Mathematical Induction

About Lata Agarwal 268 Articles

M.Phil in Mathematics, skilled in MS Office, MathType, Ti-83, Internet, etc., and Teaching with strong education professional. Passionate teacher and loves math. Worked as a Assistant Professor for BBA, BCA, BSC(CS & IT), BE, etc. Also, experienced SME (Mathematics) with a demonstrated history of working in the internet industry. Provide the well explained detailed solutions in step-by-step format for different branches of US mathematics textbooks.

Examples – Proof by Mathematical Induction for Divisibility Statements

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