A function f : BR that is injective. 3. Note that is not surjective because, for example, the vector cannot be obtained as a linear combination of the first two vectors of the standard basis (hence there is at least one element of the codomain that does not belong to the range of ). A not-injective function has a âcollisionâ in its range. a) Give an example of a function f : N ---> N which is injective but not surjective. 4. f(x) = 0 if x â¤ 0 = x/2 if x > 0 & x is even = -(x+1)/2 if x > 0 & x is odd. Thus when we show a function is not injective it is enough to nd an example of two di erent elements in the domain that have the same image. A function f :Z â A that is surjective. Proof. Whatever we do the extended function will be a surjective one but not injective. It is injective (any pair of distinct elements of the â¦ Give an example of a function â¦ c) Give an example of two bijections f,g : N--->N such that f g â g f. Give an example of a function F:Z â Z which is surjective but not injective. A function f : A + B, that is neither injective nor surjective. Example 2.6.1. Hope this will be helpful It is not injective, since \(f\left( c \right) = f\left( b \right) = 0,\) but \(b \ne c.\) It is also not surjective, because there is no preimage for the element \(3 \in B.\) The relation is a function. A function is a way of matching all members of a set A to a set B. Then, at last we get our required function as f : Z â Z given by. Example 2.6.1. Hence, function f is injective but not surjective. 2. 22. 2.6. But, there does not exist any element. Give an example of a function F :Z â Z which is injective but not surjective. Thus, the map is injective. A non-injective non-surjective function (also not a bijection) . Let the extended function be f. For our example let f(x) = 0 if x is a negative integer. 23. (v) f (x) = x 3. â´ f is not surjective. There is an important quality about injective functions that becomes apparent in this example, and that is important for us in defining an injective function rigorously. Injective and surjective are not quite "opposites", since functions are DIRECTED, the domain and co-domain play asymmetrical roles (this is quite different than relations, which in a sense are more "balanced"). Prove that the function f: N !N be de ned by f(n) = n2, is not surjective. The number 3 is an element of the codomain, N. However, 3 is not the square of any integer. Injective, Surjective, and Bijective tells us about how a function behaves. b) Give an example of a function f : N--->N which is surjective but not injective. f(x) = 10*sin(x) + x is surjective, in that every real number is an f value (for one or more x's), but it's not injective, as the f values are repeated for different x's since the curve oscillates faster than it rises. Now, 2 â Z. 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