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Answer 1

Answer:

Step-by-step explanation:

Given :

To show that any positive odd integer is of the form,

6q + 1 , 6q + 3 (or) 6q + 5.

Where q ∈ Z

Solution :

We define an odd integer is always of the form,

a = 2q + 1 ; q ∈ Ζ

__

GIVEN : ' q is some integer '

Case (1) : Let q be an odd integer,

6 is an even integer,

We know that product of an odd integer & even number (6) is even ,.

Eg :

6 × (-3) = -18 ( -18 is even integer)

6 × 1 = 6 ( 6 is an even integer)

So,

6q is an even integer,

Sum of an even number & odd number is odd number,

⇒ 6q + 1 is an odd number (6q is even , 1 is odd)

⇒ 6q + 3 is an odd number (6q is even , 3 is odd)

⇒ 6q + 5 is an odd number (6q is even , 5 is odd)

Case (2) : Let q be an even integer,

6 is an even integer,

We know that product of an even integer & another  even number (6) is even ,.

Eg :

6 × (-6) = -36 ( -36 is even integer)

6 × 4 = 24 ( 24 is an even integer)

So,

6q is an even integer,

Sum of an even number & odd number is odd number,

⇒ 6q + 1 is an odd number (6q is even , 1 is odd)

⇒ 6q + 3 is an odd number (6q is even , 3 is odd)

⇒ 6q + 5 is an odd number (6q is even , 5 is odd)

∴ Any positive odd integer is of the form,

6q + 1 , 6q + 3 (or) 6q + 5.

Where q ∈ Z

Hence, proved,.

Answer 2

Answer:

let n be an positive odd integer.

On dividing n by 6 q is the quotient and r is the remainder,

n=6q+r where r is greater than 0 but lesser than 6.

So r=0,1,2,3,4,5.

Step-by-step explanation:

for odd integer r=1,3,5.

for even integer r=0,2,4.

Case1.

If r=1,n=6q+1

Case2.

If r=3,n=6q+3

Case3.

If r=5,n=6q+5

Hence any positive odd integer is in the form of 6q+1,6q+3,6q+5.

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