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use Euclid division lemma to show that square of any positive integer is either of form 3m or 3m+1 for some integer m |
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Answer» If a and b are 2 positive integers then, By USING Euclid's Division Lemma a = bq + r (where 0 ≤ r < b) Let a be the positive integer and b = 3 So, a = 3q + r (where 0 ≤ r < 3) Also, r is the integer which is greater or EQUAL to 0 and less than 3. So, r can be 0, 1 and 2. Case 1) If r = 0 then, a = 3q + 0 a = 3q Squaring both sides a² = (3q)² a² = 9q² a² = 3(3q²) Where 3q² = m → a² = 3m Case 2) If r = 1 a = 3q + 1 Squaring both sides a² = (3q + 1)² a² = (3q)² + (1)² + 2(3q)(1) a² = 9q² + 1 + 6q a² = 3(3q² + 2q) + 1 Where 3q² + 2q = m a² = 3(m) + 1 → a² = 3m + 1 Case 3) If r = 2 a = 3q + 2 Squaring both sides a² = (3q + 2)² a² = (3q)² + (2)² + 2(3q)(2) a² = 9q² + 4 + 12q a² = 9q² + 12q + (3 + 1) a² = 9q² + 12q + 3 + 1 a² = 3(3q² + 4q + 1) + 1 Where 3q² + 4q + 1 = m a² = 3(m) + 1 → a² = 3m + 1 Therefore, Squaring of any positive integer can be expressed in the form 3m or 3m + 1 for some integer m. Hence, proved |
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