Prove that if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.
Answers
Step-by-step explanation:
Let n=6q+5 , where q is a positive integer.
We know that any positive integer is of the form 3k , 3k+1 , 3k+2.
Now , if q=3k then,
n=6(3k)+5=18q+5=18q+3+2
n=3(6q+1)+2
n=3m+2 where m=6q+1
Now, if q=(3k+1)
n=6(3k+1)+5
n=18q+6+5
n=18q+9+2
n=3(6q+3)+2
n=3m+2 , where m=6q+3
Now , if q=3k+2
n=6(3k+2)+5
n=18q+12+5
n=3(6q+5)+2
n=3m+2 , where m=(6q+5)
Therefore , if a positive integer is of the form 6q+5 then it is of the form 3q+2.
Now let n=3q+2 , where q is a positive integer.
We know that any positive integer is of the form 6q , 6q+2 , 6q+3 , 6q+4 , 6q+5
Now, if q=6q
n=3q+2
n=3(6q)+2
n=18q+2
n=2(9q+1)
n=2m
Here clearly we can observe that 3q+2 is not in the form of 6q+5.
Hence we can conclude that if a positive integer is of the form 6q+5 , then it is of the form 3q+2 but not conversely.
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.