Use Euclid’s division Lemma to show that the square of any positive integer is either of the
form 3 m or 3m + 1 for some integer m
Answers
Step-by-step explanation:
Answer
Let a be the positive integer and b=3.
We know a=bq+r, 0≤r<b
Now, a=3q+r, 0≤r<3
The possibilities of remainder is 0,1, or 2.
Case 1 : When a=3q
a
2
=(3q)
2
=9q
2
=3q×3q=3m where m=3q
2
Case 2 : When a=3q+1
a
2
=(3q+1)
2
=(3q)
2
+(2×3q×1)+(1)
2
=3q(3q+2)+1=3m+1 where m=q(3q+2)
Case 3: When a=3q+2
a
2
=(3q+2)
2
=(3q)
2
+(2×3q×2)+(2)
2
=9q
2
+12q+4=9q
2
+12q+3+1=3(3q
2
+4q+1)+1=3m+1
where m=3q
2
+4q+1
Hence, from all the above cases, it is clear that square of any positive integer is of the form 3m or 3m+1.
Here is ur answer:
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.