Question

GQ) Use Euclid division lemma
to show that any positive odd
integer is of the form 3p, 3p+1

Answer 1

Explanation :

Using euclid's division lemma

a = bq + r

Let a be any positive integer.

Here, b = 3 and r = 0, 1, 2 (0 ≤ r ≤ 3)

Case 1.

Now, take b = 3 and r = 0

→ a = 3q + 0

→ a = 3q

Squaring both sides

→ a² = 9q²

= 3(3q²)

Put 3q² = p

= 3p

Case 2.

Similarly, take b = 3 and r = 1

→ a = 3q + 1

Squaring both sides

→ a² = (3q + 1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

Put 3q² + 2q = p

= 3p + 1

Case 3.

Now, take b = 3 and r = 2

→ a = 3q + 2

Squaring both sides

→ a² = (3q + 2)²

= 9q² + 6q + 4

= 9q² + 6q + 3 + 1

= 3(3q² + 2q + 1) + 1

Put 3q² + 2q + 1 = p

= 3p + 1

•°• Squaring of positive odd integer is of the form 3p, 3p+1.

$\rule{200}2$

Answer 2

${\large\bf{\mid{\overline{\underline{Correct\:Question:-}}}\mid}}$

• Use Euclid's division lemma to show that the square of any positive integer is of the form 3p,3p+1.

$\Large\underline{\underline{\sf{Solution}:}}$

using Euclid division lemma,

let a = bq + r

where b = 3 and r = 0,1 and 2 , since r ≤ 0 ≤ 3 ...

______________________

Putting r = 0 we get,

→ a = 3q + 0

→ a² = 9q²

→ a² = 3(3q²)

putting 3q² = p now, we get,

→ a² = 3p -------------------(1)

_____________________

Putting r = 1 we get,,

→ a = 3q + 1

→ a² = (3q+1)²

[ (a+b)² = (a²+b²+ab) ]

→ a² = 9q² + 6q + 1

→ a² = 3(3q²+2q) + 1

Putting (3q²+2q) = p now,

→ a² = (3p+1) -------------------(2)

_____________________

putting r = 2 we get,

→ a = 3q + 2

→ a² = (3q+2)²

→ a² = 9q² + 12q + 4

→ a² = 9q² + 12q + 3 + 1

→ a² = 3(3q²+4q+1) + 1

putting (3q²+4q+1) = p we get,

→ a² = (3p+1) -------------------------(3)

_________________________

$\textbf{Hence, from all above cases}, \\ \: \textbf{it is clear that square of any positive integer } \\ \textbf{is either of the form 3p or 3p + 1.}$