Knowee
Questions
Features
Study Tools

For each of the following functions, state whether it is injective, surjective, and/or bijective, and why.(a) The function f (n) = 2n, mapping from integers to integers.(b) The function q(ϕ), with codomain N≥0, which maps any formula of predicate logic to the number of quantifiersin that formula

Question

For each of the following functions, state whether it is injective, surjective, and/or bijective, and why.(a) The function f (n) = 2n, mapping from integers to integers.(b) The function q(ϕ), with codomain N≥0, which maps any formula of predicate logic to the number of quantifiersin that formula

🧐 Not the exact question you are looking for?Go ask a question

Solution

(a) The function f(n) = 2n, mapping from integers to integers, is injective but not surjective.

Injective: A function is injective (or one-to-one) if every element of the domain maps to a unique element in the codomain. In this case, for every integer n, there is a unique integer 2n. No two different integers will map to the same integer under this function, so it is injective.

Surjective: A function is surjective (or onto) if every element of the codomain is the image of at least one element of the domain. In this case, the codomain is the set of all integers, but the function only produces even integers. Therefore, it is not surjective because odd integers in the codomain do not have a preimage in the domain.

Bijective: A function is bijective if it is both injective and surjective. Since this function is not surjective, it is not bijective.

(b) The function q(ϕ), with codomain N≥0, which maps any formula of predicate logic to the number of quantifiers in that formula, is neither injective nor surjective without additional information.

Injective: Without additional information, we cannot say that this function is injective. Different formulas of predicate logic could have the same number of quantifiers, so they would map to the same number in the codomain.

Surjective: Without additional information, we cannot say that this function is surjective. It is possible that there are numbers in the codomain that are not the number of quantifiers in any formula of predicate logic.

Bijective: Since this function is neither injective nor surjective, it is not bijective.

This problem has been solved

Similar Questions

For each of the following, state whether it is possible to have a function meeting the criteria given, explain why orwhy not, and if it is possible, give two examples.(a) A function f : N≥0 → N≥0 which is not surjective.(b) A function f : N≥0 → Z which is injective.(c) A function f : Q → Q which is bijective.

5. Determine whether each of these functions from Z to Z is one-to-one (onto)a) f (n) = n − 1 b) f (n) = n2 + 1 c) f (n) = n3 d) 2nf n     6. Determine whether f : Z × Z → Z is onto ifa) f (m, n) = 2m − n b) f (m, n) = m2 − n2 c) f (m, n) = m + n + 1d) f (m, n) = |m| − |n| e) f (m, n) = m2 − 4 f) f (m, n) = m + n7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 18. Let S = {−1, 0, 2, 4, 7}. Find f (S) ifa) f (x) = 1 b) f (x) = 2x + 1 c) 5xf x     

Consider the function f: R→R defined by f(x)=sin(x)+cos(2x). Which of the following statements about f(x) is true? a. The function f(x) is one-to-one (injective). b. The function f(x) is onto (surjective). c. The function f(x) is both one-to-one and onto. d. The function f(x) is neither one-to-one nor onto.

4. Determine whether each of these functions from {a, b, c, d} to itself is one-to-one (onto)a) f (a) = b, f (b) = a, f (c) = c, f (d) = db) f (a) = b, f (b) = b, f (c) = d, f (d) = cc) f (a) = d, f (b) = b, f (c) = c, f (d) = d

Let f be a function such that f: N*N -> N such that:If (x,y) !=(0,0),then f(x,y) = gcd(x,y).if (x,y) = (0,0),then f(x,y) = 0.Is f surjective only,injective only,bijective,or none of these options?

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.