Respuesta :
Answer: The correct option is
(a) f(n) = n - 1.
Step-by-step explanation: We are given to determine whether the given functions are one-to-one or not.
We know that a function y = f(x) is one-to-one if and only if
f(x) = f(y) ⇔ x = y.
That is, any two distinct elements cannot have the same image.
(a) The given function is
[tex]f(n)=n-1.[/tex]
Let us consider that
[tex]f(n_1)=f(n_2)\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow n_1=n_2.[/tex]
Similarly,
[tex]n_1=n_2\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow f(n_1)=f(n_2).[/tex]
So, this function is one-to-one.
(b) The given function is
[tex]f(n)=n^2+1.[/tex]
Let us consider that
[tex]f(n_1)=f(n_2)\\\\\Rightarrow n_1^2+1=n_2^2+1\\\\\Rightarrow n_1^2=n_2^2\\\\\Rightarrow n_1=\pm n_2.[/tex]
That is, there may be two unequal elements having same image.
For example, f(-1)=(-1)²+1=1+1=2, f(1)=(1)²+1=1+1=2.
It implies that f(-1)=f(1) but 1 ≠ -1.
So, the given function is not one-to-one.
(c) The given function is
[tex]f(n)=n^0.[/tex]
Here, the image of all the elements is 1.
For example, [tex]f(2)=2^0=1,~~f(3)=3^0=1.[/tex]
f(2)=f(3) but 2≠3.
So, more than one element is having the same image and so the function cannot be one-to-one.
(d) The given function is
[tex]f(n)=\left[\dfrac{n}{2}\right].[/tex]
Here, we see that
[tex]f(2)=\left[\dfrac{2}{2}\right]=[1]=1,\\\\\\f(3)=\left[\dfrac{3}{2}\right]=[1.5]=1.[/tex]
So, f(2)=f(3) but 2≠3.
So, the given function is not one-to-one.
Thus, the correct option is (a).
Functions are mappings having each input mapped to only one output. The function from [tex]\mathbb Z[/tex] to [tex]\mathbb Z[/tex] which is one-to-one in given option is: Option
What is a one-to-one function?
Suppose that a function be [tex]f : A \rightarrow B[/tex] (A is domain and B is range and f is the function).
Then, by definition, we know that each element of A is mapped to single element of B.
But, when it is known that the each single output is connected to only single input (backwards), then such functions is called to have one-to-one relation.
See the diagram attached below.
For the given cases, checking them for being one-to-one function:
It is given that [tex]f: \mathbb Z \rightarrow \mathbb Z[/tex] (set of integers is domain and range of f)
- Case 1: [tex]f(n) = n - 1[/tex]
Let a-1 ∈ [tex]\mathbb Z[/tex] , then
f(a) = a-1
a = input, and a-1 is output.
a = f(a) + 1 = a unique value + 1 = a single unique value.
Thus, for each single output a-1, there is only one input responsible, by mapping f, which is 'a'
Thus, this function is one-to-one
- Case 2: [tex]f(n) = n^2 +1[/tex]
Let y be an output which is obtained by an input value 'n'
Then,
[tex]y = n^2 + 1\\n^2 = y-1\\n = \pm \sqrt{y-1}[/tex]
Thus, two inputs [tex]n = \sqrt{y-1}[/tex] and [tex]n = -\sqrt{y-1}[/tex] are connecting to single output y.
Thus, this function is not a one-to-one function.
- Case 3: [tex]f(n) = n^0[/tex]
[tex]a^0 = 1 \: \forall \: a \in \mathbb R[/tex]
Thus,
f(n) = 1
Any input 'n' will map to single output 1.
Thus, this function is not a one-to-one function.
- Case 4: f(n) = [n/2]
[x] = greatest integer less than or equal to x.
Now, if we take n = 2, and n = 3, we will get:
f(2) = [2/2] = [1] = 1
f(3) = [3/2] = [1.5] = 1
Thus, two input mapped to single output, thus, not a one-to-one function.
Learn more about one-to-one functions here:
https://brainly.com/question/12860292
