Function from Z to N: Onto N but Not One-to-One

  • Thread starter nicnicman
  • Start date
In summary, the conversation discussed a problem regarding finding a function from Z to N that is onto N but not one-to-one. The solution provided was f(x) = {x if x ≥ 0, -1x if x < 0, which is essentially the absolute value function with its domain restricted to the integers. The conversation also addressed the importance of following protocol when posting problems.
  • #1
nicnicman
136
0
Hello all,

This is tripping me up a bit an I just want to see if I on the right track. Here is the problem:

Give a function from Z to N that is onto N but not one-to-one.

Answer: f(x) = {x if x ≥ 0, -1x if x < 0

Seems simple, but I think it works. Note: in our book, 0 is included in the set of natural numbers.
 
Physics news on Phys.org
  • #2
nicnicman said:
Hello all,

This is tripping me up a bit an I just want to see if I on the right track. Here is the problem:

Give a function from Z to N that is onto N but not one-to-one.

Answer: f(x) = {x if x ≥ 0, -1x if x < 0

Seems simple, but I think it works. Note: in our book, 0 is included in the set of natural numbers.

That works. Your function is essentially the absolute value function, |x|, with its domain restricted to the integers.

Fair warning: The three parts of the homework template are there for a reason. In the future, when you post a problem, do not delete them.
 
  • #3
Thanks for the help. And, I'll be sure to follow protocol next time.

I guess I could just do this:

f(x) = |x|
 
  • #4
Yep.
 

1. What does it mean for a function to be onto N but not one-to-one?

A function from Z to N is considered onto N if every element in the set of natural numbers (N) has at least one preimage in the set of integers (Z). This means that all possible output values in N are being mapped to by at least one input value in Z. However, a function is not one-to-one if different input values can map to the same output value. Therefore, a function from Z to N that is onto N but not one-to-one means that all elements in N are being mapped to, but there are multiple elements in Z that map to the same element in N.

2. Can a function from Z to N be both onto N and one-to-one?

No, a function from Z to N cannot be both onto N and one-to-one. This is because for a function to be one-to-one, each element in the range (N) must have a unique preimage in the domain (Z). However, for a function to be onto N, all elements in the range must have at least one preimage in the domain, which means that some elements in the domain will have multiple preimages. Therefore, a function from Z to N cannot satisfy both conditions simultaneously.

3. How is a function from Z to N onto N but not one-to-one represented mathematically?

Mathematically, a function from Z to N that is onto N but not one-to-one can be represented using the notation f: Z → N, where f is the name of the function, Z is the domain (set of integers), and N is the range (set of natural numbers). Additionally, the statement "f is onto N but not one-to-one" can be written as ∀n∈N, ∃z1,z2∈Z such that f(z1) = f(z2) = n.

4. What is an example of a function from Z to N that is onto N but not one-to-one?

An example of a function from Z to N that is onto N but not one-to-one is f: Z → N, where f(z) = |z|. This function maps all integers (positive, negative, and zero) to their absolute values, which are all natural numbers. Therefore, all elements in N are being mapped to, but different input values in Z can map to the same output value in N (e.g. f(-2) = f(2) = 2).

5. How does a function from Z to N onto N but not one-to-one differ from a function from Z to N that is onto N and one-to-one?

The main difference between a function from Z to N onto N but not one-to-one and a function from Z to N that is onto N and one-to-one is that the latter is a bijection, which means it is a one-to-one correspondence between the domain and range. This means that each element in the range (N) has a unique preimage in the domain (Z), and vice versa. In contrast, a function that is onto N but not one-to-one does not have a one-to-one correspondence, as some elements in the domain may have multiple preimages in the range.

Similar threads

Multiplication -- prove this one by induction
    • Precalculus Mathematics Homework Help
Replies
5
Views
913
How to find z^n of a complex number
    • Precalculus Mathematics Homework Help
Replies
12
Views
969
Is ##f(x)=2^{x}-1## considered an exponential function?
    • Precalculus Mathematics Homework Help
Replies
12
Views
1K
Solving Plane Equation 3x + 2y -z = 4
    • Precalculus Mathematics Homework Help
Replies
8
Views
1K
Reducing one algebraic fraction to another
    • Precalculus Mathematics Homework Help
Replies
3
Views
1K
Markov processes (Weight Suspended equally by n Cables)
    • Precalculus Mathematics Homework Help
Replies
4
Views
740
Does there exist a surjection from the natural numbers to the reals?
    • Precalculus Mathematics Homework Help
Replies
4
Views
2K
Finding the domain (and range) of a logarithmic function
    • Precalculus Mathematics Homework Help
Replies
11
Views
492
If a polynomial is identically zero, then all its coefficients are 0
    • Precalculus Mathematics Homework Help
Replies
14
Views
2K
Exploring Equation of a Straight Line in 3D Space
    • Precalculus Mathematics Homework Help
Replies
17
Views
980

Hot Threads

Recent Insights

Back
Top