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..
Answers
Let x be any positive integer and y = 3.
By Euclid’s division algorithm;
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
As per the given question, if we take the square on both the sides, we get;
x2 = (3q)2 = 9q2 = 3.3q2
Let 3q2 = m
Therefore,
x2 = 3m ………………….(1)
x2 = (3q + 1)2
= (3q)2 + 12 + 2 × 3q × 1
= 9q2 + 1 + 6q
= 3(3q2 + 2q) + 1
Substitute, 3q2+2q = m, to get,
x2 = 3m + 1 ……………………………. (2)
x2 = (3q + 2)2
= (3q)2 + 22 + 2 × 3q × 2
= 9q2 + 4 + 12q
= 3(3q2 + 4q + 1) + 1
Again, substitute, 3q2 + 4q + 1 = m, to get,
x2 = 3m + 1…………………………… (3)
Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer:
Solutions Given Below
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.
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