Q.2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution
Let a be any odd positive integer we need to prove that a is of the form 6q + 1 , or 6q + 3 , or 6q + 5 , where q is some integer.
Since a is an integer consider b = 6 another integer applying Euclid's division lemma we get
a = 6q + r for some integer q
0, and r = 0, 1, 2, 3, 4, 5 since
0
r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)
Also, 6q + 1 = 2 x 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 odd numbers.
Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer
Concept Insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q + 3, 6q + 5.
Basic definition of even and odd numbers and the fact that addition and , multiplication of integers is always an integer are applicable here.
Solution
Let a be any odd positive integer we need to prove that a is of the form 6q + 1 , or 6q + 3 , or 6q + 5 , where q is some integer.
Since a is an integer consider b = 6 another integer applying Euclid's division lemma we get
a = 6q + r for some integer q
0, and r = 0, 1, 2, 3, 4, 5 since 0
r < 6.Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)
Also, 6q + 1 = 2 x 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 odd numbers.
Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer
Concept Insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q + 3, 6q + 5.
Basic definition of even and odd numbers and the fact that addition and , multiplication of integers is always an integer are applicable here.
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