Use euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 where m is an integer
Answers
Let ‘a’ is a positive integer and b = 4
Therefore, according Euclid’s division lemma,
a=3q+r where 0<_r<_3
so a=0,1,2
when a=0
a=3q
a^3=(3q)^3
a^3=27q^3
a^3=9(3q^3)
a^3=9m where m=3q^3.
when r=1
a=3q+1
a^3=(3q+1)^3
a^3=(3q)^3+1^3+3*3q*1(3q+1)
a^3=27q^3+1+27q^2+9q
a^3=9(3q^3+3q^2+q)+1
a^3=9m+1 where m=3q^3+3q^2+q
hence cube of any positive integer is of the form 9m or 9m+1.
Step-by-step explanation:
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ r = 0,1,2 .
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m =
Case 2: When a = 3q + 1,
a = (3q +1) ³
a = 27q ³+ 27q ² + 9q + 1
a = 9(3q ³ + 3q ² + q) + 1
a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .
Case 3: When a = 3q + 2,
a = (3q +2) ³
a = 27q³ + 54q² + 36q + 8
a = 9(3q³ + 6q² + 4q) + 8
a = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Hence, it is proved .