Question

Prove that n²-n is divisible by 2 for every positive integer n

Answer 1

To Prove that n²-n is divisible by 2 for every positive integer n.

Proof: (By Mathematical Induction)

P(n): n²-n

P(1) : 1² - 1

⇒ P(1) = 0

since, 0/2 is 0 ; P(1) is true  

Assume that P(k) is divisible by 2

∴ P(k): k² -k is divisible by 2

⇒ P(k): k²- k = 2m  [where m is some integer] ...... equ(1)

To prove p(k+1) is ture

⇒ P(k+1) is 2m [where m is some integer]

P(k+1): (k+1)² - (k+1)

or, P(k+1): k² + 2k +1-k-1

⇒ P(k+1): k²+k  ......equ(2)

Now, equ(1) = equ(2)

⇒ P(k+1) = 2m

⇒ P(k+1) is divisible by 2

Therefore, by principle of Mathematical Induction P(n) is true for every positive integer n.

Answer 2

To ProvE :

Prove that n²-n is divisible by 2 for every positive integer n.

SolutioN :

• Any positive integer be n.

By Euclid division lemma.

→ Dividend = Divisior * Quotient + Remainder.

☛ Condition : Remainder [ 0 ≤ r < 2 ]

$\rule{90}2$

✎ Case 1.

• r = 0.
• a = 2q.

→ n² - n = ( 2q )² - 2q

→ n² - n = 4q² - 2q

→ n² - n = 2q( 2q - 1 )

• Clearly n² - n is divisible by 2.

$\rule{90}2$

✎ Case 2.

• r = 1.
• a = ( 2q + 1 )

→ n² - n = ( 2q + 1 )² - ( 2q + 1 )

→ n² - n = 4q² + 1 + 4q - 2q - 1

→ n² - n = 4q² + 4q - 2q

→ n² - n = 4q² + 2q

→ n² - n = 2q( q + 1 )

• Clearly n² - n is divisible by 2.

Hence Proved