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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.
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