show that any positive odd integer is of the from 6q +,6q +3 ,6q+5 ,where q is some integer
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Animesh Tripathi asked in Math
show that any positive odd integer is of the form 6q+1,6q+3 or 6q+5, where q is some integer. explain the solution
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Let a be a given integer.
On dividing a by 6 , we get q as the quotient and r as the remainder such that
a = 6q + r, r = 0,1,2,3,4,5
when r=0
a = 6q,even no
when r=1
a = 6q + 1, odd no
when r=2
a = 6q + 2, even no
when r = 3
a=6q + 3,odd no
when r=4
a=6q + 4,even no
when r=5,
a= 6q + 5 , odd no
...Any positive odd integer is of the form 6q+1,6q+3 or 6q+5
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Answer:
Step-by-step explanation:
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and
r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly,
6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or
6q + 3, or 6q + 5
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