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 \ 2
is odd and the other is even

Answer 1

Answer:

a and b are odd positive integers,and a greater than b

Using , euclid's division lemma , we can write that m=nq+r

taking, n=4

m=4q+r

when,r=1,m=4q+1

when,r=2,m=4q+2

when,r=3,m=4q+3

taking,a=4q+3&b=4q+1(as they are odd)

#a+b/2=(4q+3)+(4q+1)/2=8q+4/2=4q+2=even

#a-b/2=(4q+3)-(4q+1)/2=4q+3-4q-1/2=2/2=1=odd

Answer 2

Given:-

• If a and b are two odd positive integers such that a > b.

To prove:-

• That one of the two numbers (a + b)/2 and (a – b)/2 is odd and the other is even.

Proof:-

• Let a and b be any odd odd positive integer such that a > b

Since any positive integer is of the form q, 2q + 1

• Let a = 2q + 1 and b = 2m + 1,

Where, q and m are some whole numbes.

=> a + b/2 = (2q + 1) - (2m + 1) / 2

=> a + b/2 = 2(q + m) + 1 / 2

=> a + b/2 = (q + m + 1) / 2

Which is a positive integer.

Also,

=> a - b/2 = (2q + 1) - (2m + 1) / 2

=> a - b/2 = 2(q - m) / 2

=> a - b/2 = (q - m)

• Given a > b

Therefore,

=> 2q + 1 > 2m + 1

=> 2q > 2m

=> q = m

Therefore,

=> a + b / 2 = (q - m) > 0

Thus, (a + b)/2 is a positive integer.

Now,

Prove that one of the two numbers (a + b)/2 and (a – b)/2 is odd and the other is even.

=> (a + b)/2 -  (a – b)/2

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

=> 2b/2

=> b

b is odd positive integer.

Also,

The proof above that (a + b)/2 and (a - b)/2 are positive integers.

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