Question

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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Answer 1

Step-by-step explanation:

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

a2=(3q)2=9q2=3q×3q=3m where m=3q2

Case 2 : When a=3q+1

a2=(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

a2=(3q+2)2=(3q)2+(2×3q×2)+(2)2=9q2+12q+4=9q2+12q+3+1=3(3q2+4q+

Answer 2

Answer:

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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.