Use euclids division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8
Answers
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as per euclids division lemma
a=b*q+r
let positive integer be a
and b=3
hence a=3q+r
hence r can be 0,1,2
now take r=0
a=3q+r
a=3q+0
a=3q
cubing both sides....
a³=(3q)³
a³=27q³
a³=9(3q³)
let 3q³ be m
so...
a³=9m
where m=3q³
now take r=1
a=3q+1
cubing both sides....
a³=(3q+1)³
a³=(3q)³+1³+3*3q*1(3q+1)
a³=27q³+1+9q*(3q+1)
a³=27q³+2q²+9q+1
a³=9(3q³+3q²+q)+1
a³=9m+1
where m=3q³+3q²+q
now at last we take r=2
a=3q+2
cubing both sides....
a³=(3q+2)³
a³=(3q)³+2³+3*3q*2(3q+2)
a³=27q³+8+18q*(3q+2)
a³=27q³+54q²+36q+8
a³=9(3q³+6q²+4q)+8
a³=9m+8
where m=3q³+6q²+4q......
thank uu...
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 .