Using Euclid's division lemma, show that the square of any positive integer is either of the
form 3m or (3m + 1) for some integer m.
Answers
AnSwEr :
- Let a be the any positive integer which is divided by 3. then we get quotient as q and remainder as r.
★ Then by Euclid's Division lemma we get :
→ a = 3q + r, 0 ≤ r < 3
⛬ r = 0, 1, 2
★ Now, putting the value of r we get :
- When r = 0
⇒a = 3q
★ Similarly, by taking r = 1 :
⇒a = 3q + 1
★ Also, for r = 2 :
⇒a = 3q + 2
⛬ Required form number = 3q, 3q + 1 and 3q + 2.
Now,
- a = 3q
★ Squaring both the sides we get :
⇒a² = (3q)²
⇒a² = 9q²
⇒a² = 3(3q²)
⇒a² = 3m [∵ 3q² = m]
When,
- a = 3q + 1
★ Squaring both the sides we get :
⇒a² = (3q + 1)²
⇒a² = (3q)² + 1 + 6q
⇒a² = 9q² + 1 + 6q
⇒a² = 3(3q² + 2q) + 1
⇒a² = 3m + 1 [∵ 3q² + 2q = m]
- Hence, square of any positive integer is of the form 3m and 3m + 1.
Solution:-
•Using Euclid's division lemma we can prove that the square of any positive integer is either in the form of
•Euclid's Division lemma: If two positive integer "a" and "b", and there exists unique integers "q" and "r" such that which satisfies the condition a = bq + r where 0 < r < b
•Let a be the positive integer
➊ Case:
★ a = bq + r ★
•Let b = 3 , r = 0
★ a = 3q + r ★
•Squaring on both sides,
➞ (a)² = (3q)² + 0
➞ (a)² = 9q²
➞ (a)² = 3(3q)²
•Let m stand's for 3q²
➥ a² = 3m ----------》(1)
➋ Case:
★ a = bq + r ★
•Let b = 3 , r = 1
★ a = 3q + 1 ★
•Squaring on both sides,
➞ (a)² = (3q)² +(1)²
➞ (a)² = (3q + 1)²
➞ (a)² = (3q)² + (1)² +2(3q)(1)
➞ (a)² = 9q² + 1 + 6q
➞ (a)² = 3(3q² + 2q) + 1
•Let m stand's for 3q² + 2q
➥ a² = 3m + 1 ----------》(2)
➌ Case:
★ a = bq + r ★
Let b = 3 , r = 2
★ a = 3q + 2 ★
•Squaring on both sides,
➞ (a)² = (3q)² + (2)²
➞ (a)² = (3q + 2)²
➞ (a)² = (3q)² + (2)² + 2(3q)(2)
➞ (a)² = 9q² + 4 + 12q
➞ (a)² = 9q² + 3 + 1 + 12q
➞ (a)² = 3(3q² + 4q + 1) + 1
•Let m stand's for 3q² + 4q + 1
➥ a² = 3m + 1 ----------》(3)
★Therefore , the square of any positive integer is either of the form 3m or 3m+1.