Question

PROVE THAT IF A POSITIVE INTEGER IS OF THE FORM 6q+5 THEN IT IS OF THE FORM 3q+2 FOR SOME INTEGER q BUT NOT CONVERSELY
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Answer 1

Answer:

Step-by-step explanation:

Let, n = 6q + 5, when q is a positive integer ... Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q. Conversely ... Now, this is not of the form 6m + 5

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.