Question

show that any positive integer is of the form of 9 M and 9 M + 1 or 9 M + 8 use Euclid Lemma

Answer 1

Answer:

Using Euclid Lemma, we have

Step-by-step explanation:

Let the positive integer be a.

Now, a=3q+r

where, r=0,1,2 (since 0<=r<3)

When r=0,

a=3q

=》 a^3=(3q)^3

=27q^3

=9(3q^3)

=9m, where m= 3q^3

When r=1

a=3q+1

=》 a^3=(3q+1)^3

=(3q)^3+1^3+3×(3q)^2×1+3×3q×1 [Using the identity (a+b)^3 ]

=27q^3 +1+27q^2 +9q

=9(3q^3+3q^2+q)+1

=9m+1 , where m=3q^3+3q^2+q

When r=2,

a=3q+2

=》 a^3=(3q+2)^3 [cubing both sides]

=27q^3+8+54q^2+36q [ using the identity (a+b)^3 ]

=9(3q^3+6q^2+4q)+8

=9m+8, where m=3q^3+6q^2+4q

Therefore, the cube of any positive integer is in the form of 9m, 9m+1 or 9m+8.

Hence, proved.