Question

use euclid's division lemma to show taht the squre of any positive integer is either of the form 3m or 3m+1 for some integer m​

Answer 1

Step-by-step explanation:

HEY MATE HERE IS YOUR ANSWER

LET A BE ANY POSITIVE INTEGER AND B 3

A= 3Q + R

A = 3Q3 = 27 q3= 9 ( 3 q) 3= 9m

where m is an integer such that m= 3q3

a= 3q+1

à3= 27q3+ 27 q2+ 9q + 1,

à3= 9 ( 3 q3+3q2+q) +1

a3= 9m+1

à=3q+2

à3= 27q3+ 54q2+36q+8

à3= 9(3q3+ 6 q2+ 4 q) +8

a3= 9m +8

THEREFORE THE CUBE OF ANY POSITIVE INTEGER IS OF THE FORM 9 M+ 9 M+1. OR 9 M+8 .

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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.