 # RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.1 in Chapter 1 - Real Numbers

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

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. (From Euclid's division lemma for b= 3)

∴ q can be 3k or, 3k+1 or, 3k+2. (From Euclid's division lemma for b= 3)

If q= 3k, then

⇒ n= 6q+5

⇒ n= 6(3k)+5

⇒ n= 18q+5 = (18q+3)+ 2

⇒ n= 3(6q+1)+2

⇒ n= 3m+2, where m is some integer

If q= 3k+1, then

⇒ n= 6q+5

⇒ n= 6(3k+1)+5

⇒ n= 18q+6+5 = (18q+9)+ 2

⇒ n= 3(6q+3)+2

⇒ n= 3m+2, where m is some integer

If q= 3k+2, then

⇒ n= 6q+5

⇒ n= 6(3k+2)+5

⇒ n= 18q+12+5 = (18q+15)+ 2

⇒ n= 3(6q+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. (From Euclid's division lemma for b= 6)

So, now if q=6k then

⇒ n= 3q+2

⇒ n= 3(6k)+2

⇒ n= 6(3k)+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.

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