Question

Show that every positive integer is either even or odd.

Answer 1

Let n is a positive integer . The basic principle is " when positive n is either odd or even then (n + 1) is also either even or odd .
Means if n is odd then (n +1) should be even and if n is even then (n+1) should odd.

Case 1 :- when n is odd e.g., n = 2k + 1 , where k is integer then, (n +1) = (2k+1)+ 1
= (2k +2) , divisible by 2 hence, (n +1) is even .

Case 2:- when n is even e.g., n = 2k , where k is integer then (n +1) = 2k +1
doesn't divisible by 2 , so, (n +1) is odd integer .

From case1 and case2 it is clear that if n is positive then it is either odd or even.

Answer 2

Step-by-step explanation:

let us assume that there exist a small positive integer that is neither odd or even, say n.

Since n is least positive integer which is neither even nor odd, n - 1 must be either or or even.

CASE 1 :

If n - 1 is even , then n - 1 = 2m for some integer m .

But , => n = 2m + 1 .

This implies n is odd .

CASE 2 :

If n - 1 is odd , then n - 1 = 2m + 1 for some integer m .

But, => n = 2m + 2 = 2( m + 1 ) .

This implies n is even .

In both cases , there is a contradiction .

Thus , every positive integer is either even or odd .

Hence, it is solved

THANKS

#BeBrainly.