Question

show that every positive integer is either even or odd

Answer 1

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

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Answer 2

Answer:

yes , every positive integer is either even or odd.

Step-by-step explanation:

let n be a positive integer.

If n is odd then (n+1) is even and if n is even

then (n+1) is odd.

so we will take the cases ,

Case 1:- When n is odd

n= (2m+1) +1

n+1=2m+2

n+1=2(m+1)

which is divisible by 2

therefore, n+1 is even..

Case 2:- when n is even

n=2m

n+1=2m+1

which is odd

therefore, n+1 is odd ....

From Case 1 and Case 2 it is clear that if n is positive it is either odd or even.

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