Question: Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3. Therefore, x = 3q, 3q+1 and 3q+2 Now as per the question given, by squaring both the sides, we get, 2 = (3) 2= 9 2 = 3 × 3 2 Let 3 2 = m Therefore, 2 = 3m ……………………..(1) 2 = (3 + 1) 2 = (3) 2+1 2+2 × 3q × 1 = 9 2 + 1 + 6q = 3(3 2+2q) + 1 Substitute, 3 2+2q = m, to get, 2 = 3m + 1 ……………………………. (2) NCERT Solution for Class 10 Maths Chapter 1 Real Numbers 2 = (3 + 2) 2 = (3) 2+2 2+2 × 3q × 2 = 9 2 + 4 + 12q = 3 (3 2 + 4q + 1) + 1 Again, substitute, 3 2+4q+1 = m, to get, 2 = 3m + 1…………………………… (3) Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.