Question

show that any positive integer is of the form 3q, 3q+1, 3q+2, for some integer q.

Answer 1

let a and b any positive integer where b=3.                                                    by using this theorem a=bq+r where q>0.                                                     r=0,1,2 because 0 <r<3.therefore a=3q(or)3q+1(or)3q+2.                              this is correct because i'm in 10 i wrote this in my homework.so give as brainlist.

Answer 2

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

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

let a be any positive integer

then

b= 3

a= bq+r

0≤r<b

0≤r<3

r= 0,1,2

case 1.

r=0

a= bq+r

3q+0

3q

case 2.

r=1

a= 3q+1

3q+1

case3.

r=2

a=3q+2

hence from above it is proved that any positive integer is of the form 3q,3q+1 and 3q+2

$\huge \boxed{ \boxed{ \green{HOPE\: IT \: HELPS}}}$

$\huge{ \pink{thanks}}$