4. 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.
[Hint: Let x be any positive integer then it is of the form 34,34 + 1 or 34+2. Now square
each of these and show that they can be rewritten in the form 3m or 3m + 1.1
Answers
Using Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Let a be any positive integer and b = 3.
=) a = 3q + r, r = 0 or 1 or 2.
(By Euclid's lemma)
=) a = 3q or 3q + 1 or 3q + 2 for positive integer q.
1st case,
If a = 3q :
=) a² = (3q)²
= 9q²
= 3(3q²)
= 3m, where m= 3q².
2nd case,
If a = 3q+1,
=) a² = (3q+1)²
= (3q)² + 2(3q)(1) + 1²
= 9q² + 6q + 1
= 3(3q² + 2q) + 1
= 3m + 1, where m = 3q² + 2q.
3rd case,
If a = 3q+2:
=) a² = (3q+2)²
= (3q)² + 2(3q)(2) + 2²
= 9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q + 1) + 1
= 3m + 1, where m = 3q² + 4q + 1.
Hence the square of any positive integer is either of the form 3m or 3m+1 for some integer m
Answer:
hey mate here is ur loooonnng answer...OwO
Step-by-step explanation:
Let us consider a positive integer a
Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that
a = 3b + r…(1)
where r = 0,1,2,3…..
Case 1: Consider r = 0
Equation (1) becomes
a = 3b
On squaring both the side
a2 = (3b)2
a2 = 9b2
a2 = 3 × 3b2
a2 = 3m
Where m = 3b2
Case 2: Let r = 1
Equation (1) becomes
a = 3b + 1
Squaring on both the side we get
a2 = (3b + 1)2
a2 = (3b)2 + 1 + 2 × (3b) × 1
a2 = 9b2 + 6b + 1
a2 = 3(3b2 + 2b) + 1
a2 = 3m + 1
Where m = 3b2 + 2b
Case 3: Let r = 2
Equation (1) becomes
a = 3b + 2
Squaring on both the sides we get
a2 = (3b + 2)2
a2 = 9b2 + 4 + (2 × 3b × 2)
a2 = 9b2 + 12b + 3 + 1
a2 = 3(3b2 + 4b + 1) + 1
a2 = 3m + 1
where m = 3b2 + 4b + 1
∴ square of any positive integer is of the form 3m or 3m+1.
Hence proved.
- Thank you
please mark as brainliest.....UwU