Question

prove that n3-1 is divisible by 3 for every positive integer.

Answer 1

prove that n³ - n is divisible by 3 for every positive integers

n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2 [ ∵ if P = ab + r , then 0 ≤ r < a by Euclid lemma ]

∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer

Case 1 :- when n = 3r

Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ]

Case2 :- when n = 3r + 1

e.g., n - 1 = 3r +1 - 1 = 3r

Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3

Case 3:- when n = 3r - 1

e.g., n + 1 = 3r - 1 + 1 = 3r

Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3

From above explanation we observed n³ - n is divisible by 3 , where n is any positive integers.

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

PRINCIPLE OF MATHEMATICAL INDUCTION

Let P(n) be a mathematical statement

Where n is a non-negative integers n and n₀ be a fixed non-negative integer.

BASIS STEP :

Suppose P(n₀) is true i.e. P(n) is true for n = n₀

INDUCTIVE HYPOTHESIS

Whenever k is an integer such that k ≥ n₀ and P(k) is true

INDUCTION STEP

P(k + 1) is true

Then P(n) is true for all integers n ≥ n₀

TO PROVE

$\sf{( {n}^{3} - n )\: \: is \: divisible \: by \: 3 \: for \: every \: positive \: integer \: n\: }$

PROOF BY PRINCIPLE OF MATHEMATICAL INDUCTION

Let P(n) be the Statement

BASIS STEP

Now for n = 1

$\sf{P(1) : ({1}^{3} - 1) = 0 \: which \: is \: divisible \: by \: 3}$

INDUCTIVE HYPOTHESIS

Let P(k) is true for some positive integer k

$\sf{( {k}^{3} - k )\: \: is \: divisible \: by \: 3 \: \: }$

Let

$\sf{( {k}^{3} - k )\: = 3m } \: \: \: for \: some \: integer \: m$

$\implies \: \sf{{k}^{3} = 3m + k } \:$

INDUCTIVE STEP

We show that P(k+1) holds

Now

$\sf{ \: {(k + 1)}^{3} - (k + 1) \: }$

$= \sf{ \: {k}^{3} + 3 {k}^{2} + 3k + 1 - k - 1 \: }$

$= \sf{ \: {k}^{3} + 3 {k}^{2} + 2k \: }$

$= \sf{ \: 3m + k + 3 {k}^{2} + 2k \: }$

$= \sf{ \: 3m + 3k + 3 {k}^{2} \: }$

Since each of terms is divisible by 3

So

$\sf{ {(k + 1)}^{3} - (k + 1)\: \: is \: divisible \: by \: 3 \: \: }$

Hence P(k+1) is true

So by the Principle of Mathematical Induction P(n ) is true for all Positive Integer n

Hence

$\sf{( {n}^{3} - n )\: \: is \: divisible \: by \: 3 \: for \: every \: positive \: integer \: n\: }$