Question

Show that Any positive integer is of the form 3q or 3q+1 or 3q+2 for some integer q.

Answer 1

Let x be the integer

x=3q

x²=9q²

x²=3(3q²)

x²=3q [let 3q² be q]

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x=3q+1

x²=(3q+1)²

x²=9q²+6q+1

x²=3(3q²+2q)+1

x²=(3q+1) [let 3q²+2q be q]

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x=3q+2

x²=(3q+2)²

x²=9q²+12q+4

x²=3(3q²+4q+1)+1

x²=(3q+1) [3q²+4q+1 be q]

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it proves q² is in the form of 3q and (3q+1) but not in the form of (3q+2)

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 !!