Question

show that any positive odd integer is of the form 6q+1 or 6q+3 for 6q+5 where q is some integer​

Answer 1

Answer:

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Answer 2

$\fbox \red{Solution:}$

Let ‘a’ be any positive integer.

Then from Euclid's division lemma,

a = bq+r ; where 0 < r < b

Putting b=6 we get,

⇒ a = 6q + r, 0 < r < 6

For r = 0,

We get a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]

For r = 1,

We get a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]

For r = 2,

We get a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]

For r = 3,

We get a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]

For r = 4,

We get a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]

For r = 5,

We get a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]

Thus, from the above it can be seen that any positive odd integer can be of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.