Question

8. Show that any positive odd integer is of the form, 6q + 1 or 6q+3,
or 6q + 5 where q is some integer.​

Answer 1

Solution:

As per Euclid's division lemma

if a and b are 2 positive intefers , then

a=bq+r

where 0<r<b

let positive integers be a

and b=6

hence a=6q+r

where(0<r<6)

r is an integer greter than or equal to 0 and less than 6

hence r can be either 0,1,2,3,4 or 5

if r=1

our equation becomes

a=6q+r

a=6q+1

this will always be an odd integer

if r=3

our equation becomes

a=6q+r

a=6q+3

this will always be an odd integers

if r=5

our equation becomes

a=6q+r

a=6q+5

this will always be an odd integers

therefore,any odd integers is of form 6q+1,6q+3 or6q+5

Hence proved.

Answer 2

Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b=6

Since 0 ≤ r < 6, the possible reminders are 0,1,2,3,4 and 5.

That is, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q+ 5 where q is the quotient. However, since a is odd, a can not be 6q or 6q+2 or 6q+4 (since they are divisible by 2). Therefore any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5