Homework help for discrete

Question

Suppose f: integer(z) to integer(z) is a function with the property that f(a+b) = f(a) + f(b) for every two integers a and b. Prove that if f(c) is even for some odd integer c then f(x) is even for ever integer x.

Answer 1

Let c = a+b,  c is odd.

We know a is odd and b is even, or a is even and b is odd.  Without loss of generality, assign a=odd, b=even.

 

f(c) = f(a+b) = f(a) + f(b)     

We know f(c) is even.   This leaves two possibilities:

(i)  f(a) and f(b) are both odd

(ii)  f(a) and f(b) are both even

 

Can (i) hold true?   Let's see:

   Assume (i) is true.   

   Then f(a) is odd for odd a.  

   However, notice if we let b=0 then a=c and f(a) = f(c) which was given as even.

   Hence assuming case(i) leads to a contradiction.

   Thus only case(ii) can be true.

 

Since case(ii) is true,  then f(a) is even for odd a and f(b) is even for even b (remember a and b can be any values that sum to c, we just assign the odd one to a and the even one to b).

 

∴  f(x) is even for all x.

 

 

 

Answer 2

Here's a hint....  If a number is even, it can be broken down as 2* something.  You have f(2something) .  That is f( something+something)