Question

using uclid devision lema prove that any positive integer is in the form of to 3m ,3m+1 .

Answer 1

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

 = 3 x ( 3q²)

 = 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

 =  9q² + 6q +1

 = 3 (3q² +2q ) + 1

 = 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

   = 9q² + 12q + 4

   = 9q² +12q + 3 + 1

 = 3 (3q² + 4q + 1 ) + 1

 = 3m + 1 where m = 3q2 + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.

Hence, it is solved .

THANKS

#BeBrainly .

Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{ \blue{here \: is \: answer}}}$

let a be any positive integer

and

b=3

$\bf{a = bm + r}$

$\bf{0 \leqslant r < b}$

$\sf{0 \leqslant r < 3}$
r= 0,1,2

case 1.

r=0

a=bq+r

a= 3q+0
3q^2
9q^2
3(3q^2)

let 3q^2 be m

a=3m

case 2.

r=1

a=bq+r

a=(3q+1)^2

(3q)^2+2*3q*1+1^2

9q^2+6q+1

3(3q^2+2q)+1

let 3q^2+2q be m

a= 3m+1

case 3.

r=2

a=bq+r

(3q+2)^2

(3q)^2+2*3q*2+(2)^2

9q^2+12q+4

9q^2+12q+3+1

3(3q^2+4q+1)+1

let 3q^2+4q+1 be m
a=3m+1

$\sf{hence \: from \: above \: we \: can \: \: see \: }$
$\sf{that \: any \: integer \: of \: the \: form \: 3m \: and \: 3m + 1}$

hence proved

$\huge \boxed{ \boxed{ \green{hope \: it \: helps}}}$
$\huge{ \mathfrak{ \pink{thanks}}}$