Question

show that every positive even integer is of the form to give and that every positive odd integer is of the form 2q + 1 where Q is some integer​

Answer 1

Step-by-step explanation:

Solution :-

Let a be any positive integer .

and b = 2

We know that

By Euclid's Division Algorithm

Given positive integers a and b there exist unique pair of integers q and r satisfying a = bq+r , 0≤r<b .

Now,

a = 2q +r ,0≤ r < 2 ------------(1)

The possible values of r = 0,1

Case -1:-

Put r = 0 in (1) then

=> a = 2q+0

=> a = 2q ----------------------(2)

2q is an even integer .

Case -2:-

Put r = 1 in (1) then

=> a = 2q+1 ------------------(3)

2q+1 is an odd integer.

From (2) & (3)

We conclude that

Every positive integer can be either even or odd.

Therefore, " Any positive odd integer is of the form 2q+1 , where q is some integer.

Used Property:-

Euclid's Division Algorithm:-

"Given positive integers a and b there exist unique pair of integers q and r satisfying a = bq+r , 0≤r<b " .

Answer 2

Answer:

here is the answer of your question:-

let 'a' be any positive integer and b=2

By euclid's division algorithm

a=bq+r       0≤r<b

a=2q+r       0≤r<2

(i.e) r =0,1

r=0 , a=2q+0=> a=2q

r=1, a=2q+1

if a is the form of 2m then 'a' is an even integer and positive odd integer is of the form 2m+1..

Step-by-step explanation: