You can consider two methods to show that the function is
not bijective.
Method 1) Suppose that the function is
bijective. For a bijective function, if exists an image y1, then it must exist at least
one corresponding value x1.
f(x1)=y1/2 =>
f(x1)+f(x1+y1)=y1 => f(x1+y1)=y1/2 .
If the function
is bijective, then it is injective=> x1+y1=x1 => y1=0, which is an absurd
assumption because the range of function is
(0;infinite).
Method 2) Take x=y=1 =>
f(1)+f(2)=1(a)
Take other values: x=2 and y=1 =>
f(2)+f(3)=1(b)
Compare (a) and (b)=> f(1)=f(3)
=> the function is not injective => it is not
bijective.
Answer: The function is not injective and it is
not bijective.
No comments:
Post a Comment