Question

Show that any positive integer is of the form 3q or,3q+2 for some integer q. 

Answer 1

Let x be any +ve number.
 If x=3q on squaring both sides
(x)2 = (3q)2 =9q2 =3(3q2) =3m
where m=3q2 m is also an integer
so, x2=3m 
If x= 3q+1 on squaring both sides
x2= (3q+1)2 x2=9q2+1+2*3q*1 x2= 3(3q2+2q)+1 x2=3m+1
where m =3q2+2q and is an integer (3m+1)

Answer 2

To Show :

Any positive integer is of the form 3q or 3q+1 or 3q+2 .

Solution :

Let a be any positive integer .

Then b = 3

So by Euclid's Division lemma there exist integers q and r such that ,

a = bq+r

a = 3q+r (b = 3)

And now ,

As we know that according to Euclid's Division Lemma :

0 ≤ r < b

Here ,

0 ≤ r < 3

Here the possible values of r are = 0,1,2

=> 0 ≤ r < 1<2

=> r = 0 or r = 1 or r = 2

And then

a = 3q+r

a = 3q+0 = 3q

a = 3q+1

a = 3q+2

#Hence Proved !!