Question

Use euclid's division lemma to show that the square of positive integers is in the form 3m or 3m + 1 for some integer m

Answer 1

Answer:Let a be any positive integer and b = 3

Now, by Euclid division lemma, a = bq + r where r = 0 or 1 or 2 and q is some integer.

Now,

If r = 0

a =bq+r

= 3q + 0

=(3q)²

= 9q²

=3 * 3q²

=3m(where m = 3q²)

If r = 1,

a = (3q +1)²

=(3q)² + 2*3q*1 * (1)²

=9q² + 6q + 1

=3(3q²+2q) + 1

=3m + 1(where m = 3q² + 2q)

If r = 2,

a = (3q + 2)²

= (3q)² + 2*3q*2 + (2)²

= 9q² + 12q+ 4

=9q² + 12q+ 3 +1

=3(3q² + 4q +1) +1

=3m + 1(where m = 3q² + 4q +1)

Therefore, every positive integer if of the form 3m and 3m + 1 wherem is some integer.

Step-by-step explanation:

Answer 2

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.