Question

If a and b are two odd positive integers such that a > b, then prove that one of the two numbers (a+b)/2 and (a-b)/2is odd and the other is even.

Answer 1

Let a and b be any odd positive integer such that a > b. Since any positive integer is of the form q, 2q + 1

suppose , a = 2q + 1 & b = 2m + 1, where, q and m are some whole numbers

(a+b)÷ 2 = (2q+1)+(2m+1)÷2

(a+b)÷ 2 = (2q + 2m + 1+1)÷2

(a+b)÷ 2 = (2q + 2m + 2 )÷2

(a+b)÷2 = 2((q+m)+1) ÷ 2

(a+b) ÷ 2 = (q + m + 1)  which is a positive integer.  

Also,

(a–b) ÷ 2 = (2q+1) - (2m+1) ÷ 2

(a–b) ÷ 2 = (2q+1 - 2m - 1) ÷ 2

(a–b) ÷ 2 = (2q - 2m +1 - 1) ÷ 2

(a–b) ÷ 2 = (2q - 2m ) ÷ 2

(a–b) ÷ 2 = 2(q - m) ÷ 2

(a–b) ÷ 2 = (q – m)

a > b

[Given]

2q + 1 > 2m + 1

2q > 2m

q > m

Therefore,( a–b) ÷ 2 = (q – m)  > 0

Hence,( a–b) ÷ 2 is a positive integer.  

Now, we need to prove that one of the two numbers (a+b)/2 and (a–b)/2  is even and other is odd.

Now , (a+b)/2 – (a - b)/2  

=[ (a+b) - (a–b)]/2

= (a + b - a + b)/2

= 2b / 2

= b

(a+b)/2 and (a–b) / 2 are positive integers.

[Proved above that they are positive integers]

If one of the integer is odd and another is even,the difference of two integers is an odd number and the difference between two odd and two even integers is even.

Hence , it is proved that if a and b are two odd positive integers is even.

Hence, it is proved that if a and b are two odd positive integers such that a > b then one of the two number (a+b)/2 and (a–b)/2 is odd and the other is even.

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