Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8.
Answers
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 .
★To prove:-
- The cube of any positive integer is of the form 9m,9m+1 or 9m+8.
★Proof:-
Let us consider a and b where a be any positive number and b is equal to 3.
According to Euclid’s Division Lemma,
✦a = bq + r
where r is greater than or equal to zero and less than b (0 ≤ r < b)
➜a = 3q + r
so r is an integer greater than or equal to 0 and less than 3.
Hence r can be either 0, 1 or 2.
Case 1: When r = 0, the equation becomes
➜a = 3q
Cubing both the sides,
➜ a³ = (3q)³
= 27 q³
= 9 (3q³)
= 9m
where m = 3q³
Case 2: When r = 1, the equation becomes
➜a = 3q + 1
Cubing both the sides,
➜a³ = (3q + 1)³
= (3q)³ + 1³+ 3 × 3q × 1(3q + 1)
= 27q³+ 1 + 9q × (3q + 1)
= 27q³ + 1 + 27q² + 9q
= 27q³ + 27q² + 9q + 1
= 9 ( 3q³ + 3q² + q) + 1
= 9m + 1
Where m = ( 3q³ + 3q² + q)
Case 3: When r = 2, the equation becomes
➜a = 3q + 2
Cubing both the sides,
➜a³ = (3q + 2)³
= (3q)³ + 2³ + 3 × 3q × 2 (3q + 1)
= 27q³ + 8 + 54q²+ 36q
= 27q ³+ 54q² + 36q + 8
= 9 (3q³ + 6q² + 4q) + 8
= 9m + 8
Where m = (3q³+ 6q² + 4q)
Therefore,
The cube of any positive integer is of the form 9m or 9m + 1 or, 9m + 8.
Hence proved!
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