use Euclid division lemma to show that the cube of any positive integer is either of the form 9m+9m+1 or 9m+8 for some integer m
Answers
Answer:
Step-by-step explanation:
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
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 = (3q)3 = 27q3
9(3q3) = 9m
Case 2: When a = 3q + 1,
a 3 = (3q +1) 3
a 3 = 27q 3 + 27q 2 + 9q + 1
a 3 = 9(3q 3 + 3q 2 + q) + 1
a 3 = 9m + 1
Where m is an integer such that m = (3q 3 + 3q 2+ q)
Case 3: When a = 3q + 2,
a 3 = (3q +2) 3
a 3 = 27q 3 + 54q 2 + 36q + 8
a 3 = 9(3q 3 + 6q 2 + 4q) + 8
a 3 = 9m + 8
Where m is an integer such that m = (3q 3 + 6q 2+ 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
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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 .