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Show that any positive odd integer is of the form 3m or 3m+1 or 3m+2 where |
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Answer» -STEP explanation:Proof:We know that from Euclid’s division lemma for b= 3Let us assume that any positive integer ‘n’ be of the form 3Q or, 3q+1 or 3q+2.If n= 3q,On squaring we get,⇒ n2= (3q)2 = 9q2⇒ n2= 3(3q2)⇒ n2= 3m, where m is some integer [m = 3q2]If n= 3q+1,On squaring we get,⇒ n2= (3q+1)2 = 9q2 + 6q + 1 { Solved using the IDENTITY (a+b) 2 = a2 + b2 + 2ab}⇒ n2= 3(3q2 +2q) + 1⇒ n2= 3m + 1, where m is some integer [m = 3q2 +2q]If n= 3q+2,On squaring we get,⇒ n2= (3q+2)2 = 9q2 + 12Q + 4 { Solved using the identity (a+b) 2 = a2 + b2 + 2ab}⇒ n2= 3(3q2 + 4q + 1) + 1⇒ n2= 3m, where m is some integer [m = 3q2 + 4q + 1]Therefore, the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.Hence PROVED |
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