Question

1.Show that every positive integer is in the form of 3m, 3m+1,3m+2,where m is
any whole number.​

Answer 1

Answer:

hope you understand..

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

a^2 = (3b)^2

a^2 = 9b^2

a^2 = 3 × 3b^2

a^2 = 3m

Where m = 3b^2

Case 2: Let r = 1

Equation (1) becomes

a = 3b + 1

Squaring on both the side we get

a^2 = (3b + 1)^2

a^2 = (3b)^2 + 1 + 2 × (3b) × 1

a^2 = 9b^2 + 6b + 1

a^2 = 3(3b^2 + 2b) + 1

a^2 = 3m + 1

Where m = 3b^2 + 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides we get

a^2 = (3b + 2)^2

a^2 = 9b^2 + 4 + (2 × 3b × 2)

a^2 = 9b^2 + 12b + 3 + 1

a^2 = 3(3b^2 + 4b + 1) + 1

a^2 = 3m + 1

where m = 3b^2 + 4b + 1

∴ square of any positive integer is of the form 3m or 3m+1.

Hence proved.