Question

give the correct answer..??

Answer 1

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.

Answer 2

Question:-

☞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.

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Solution:-

☞Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

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

Now as per the question given, by squaring both the sides, we get,

$x^2\: =\: (3q)^2\: =\: 9q^2 \:= \:3 × 3q^2$

Let $3q^2$= m

Therefore, $x^2$= 3m _______(1)

$x^2 \:= \:(3q + 1)^2$

$=\: (3q)^2+1^2+2×3q×1 \:$

$= \:9q^2 + 1 +6q$

$=\: 3(3q^2+2q) +1$

Substitute, $3q^2+2q$= m, to get,

$x^2$= 3m + 1 _________ (2)

$x^2\:= \:(3q + 2)^2 \:$

$=\: (3q)^2+2^2+2×3q×2$

$= \:9q^2 + 4 + 12q \:$

$=\:3 (3q^2 + 4q + 1)+1$

Again, substitute, $3q^2+4q+1$= m, to get,

$x^2$= 3m + 1__________(3)

Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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Hope it helps you!☺️