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. About the author Kaylee
Step-by-step explanation: Sol:– Let a be a positive integer q be the quotient and r be the remainder Dividing a by 3 using the Euclid’s Division Lemma we have a = 3q + r, where a = ≤ r < 3 Putting r = 0, 1 and 2, we get: a = 3q → a^2 = 9q^2 = 3 × 3q^2 = 3m (Assuming m = q^2) Then, a = 3q + 1 a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m +1 (Assuming m = 3q^2 + 2q) Next, a = 3q + 2 → a² = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q +3+1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1. (Assuming m = 3q^2 + 4q+1) Therefore, the square of any positive integer (say a^2) is always of the form 3m or 3m + 1 Hence, proved Reply
Answer:
https://brainly.in/question/40248511
Step-by-step explanation:
Step-by-step explanation:
Sol:– Let a be a positive integer q be the quotient and r be the remainder
Dividing a by 3 using the Euclid’s Division Lemma
we have
a = 3q + r,
where a = ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q
→ a^2 = 9q^2
= 3 × 3q^2
= 3m (Assuming m = q^2)
Then, a = 3q + 1
a^2 = (3q + 1)^2
= 9q^2 + 6q + 1
= 3(3q^2 + 2q) + 1
= 3m +1 (Assuming m = 3q^2 + 2q)
Next, a = 3q + 2
→ a² = (3q + 2)^2
= 9q^2 + 12q + 4
= 9q^2 + 12q +3+1
= 3(3q^2 + 4q + 1) + 1
= 3m + 1. (Assuming m = 3q^2 + 4q+1)
Therefore, the square of any positive integer (say a^2) is always of the form 3m or 3m + 1
Hence, proved