Question

Prove that a positive integer is of the form 6q+5, then it is of the form 3q + 2 for some integer, q; but not conversely

Answer 1

For the first part, lets assume that A is a positive integer of the form (3q+5) for some integer q
therefore,
A = (6q + 5)
= (3*2q + 3 + 2)
= (3*(2q+1) + 2)
As q is an integer, then (2q +1) is also an integer, lets call it p.
= (3p + 2)
=> every positive integer of the form (6q + 5) can also be expressed as (3p + 2) for some integers q and p.

For the converse part, a simple contradiction of the claim will suffice the proof.
Consider 2 (which is a positive integer), it can be expressed in the form (3q +2) for q=0, however it can't be expressed in the form (6q + 5) for any integer q as -
2 = 6q + 5
=> q = -(1/2) which is not an integer.

Answer 2

Let n= 6q+5 be a positive integer for some integer q.

We know that any positive integer can be of the form 3k, or 3k+1, or 3k+2.

∴ q can be 3k or, 3k+1 or, 3k+2.

If q= 3k, then

⇒ n= 6q+5

⇒ n= 6(3k)+5

⇒ n= 18k+5 = (18k+3)+ 2

⇒ n= 3(6k+1)+2

⇒ n= 3m+2, where m is some integer.

If q= 3k+1, then

⇒ n= 6q+5

⇒ n= 6(3k+1)+5

⇒ n= 18k+6+5 = (18k+9)+ 2

⇒ n= 3(6k+3)+2

⇒ n= 3m+2, where m is some integer

If q= 3k+2, then

⇒ n= 6q+5

⇒ n= 6(3k+2)+5

⇒ n= 18k+12+5 = (18k+15)+ 2

⇒ n= 3(6k+5)+2

⇒ n= 3m+2, where m is some integer

Hence, if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely,

Let n= 3q+2

And we know that a positive integer can be of the form 6k, or 6k+1, or 6k+2, or 6k+3, or 6k+4, or

6k+5.

So, now if q=6k+1 then

⇒ n= 3q+2

⇒ n= 3(6k+1)+2

⇒ n= 18k + 5

⇒ n= 6m+5, where m is some integer

So, now if q=6k+2 then

⇒ n= 3q+2

⇒ n= 3(6k+2)+2

⇒ n= 18k + 6 +2 = 18k+8

⇒ n= 6 (3k + 1) + 2

⇒ n= 6m+2, where m is some integer

Now, this is not of the form 6q + 5.

Therefore, if n is of the form 3q + 2, then is necessary won’t be of the form 6q + 5.