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[NEEDITNOW0819]

Determine which of the following function is an injective from the set of integers to the set of integers
A). f=n^2 + 1,
B). f=n^3
C). f=n^3-4
4). f=n/2

 

[🤖]

A function is said to be injective if each element in the codomain is the image of at most one element in the domain.

A) f = n^2 + 1 is not injective because, for example, f(-2) = f(2) = 5.

B) f = n^3 is not injective because, for example, f(-1) = f(1) = -1.

C) f = n^3 - 4 is injective because, for any two distinct integer values a and b, f(a) and f(b) will be different. That is, if f(a) = f(b), then a^3 - 4 = b^3 - 4, which implies a = b.

D) f = n/2 is not injective because, for example, f(1) = f(2) = 1/2.

Therefore, the only injective function from the given options is f = n^3 - 4.

 

[NEEDITNOW0819]

Determine which of the following function is an injective from the set of integers to the set of integers
A). f( n )=n^2 + 1,
B). f( n )=n^3
C). f( n )=n^3-4
4). f( n )=n/2

 

[🤖]

To determine whether a function is injective or not, we need to check whether different values in the domain map to the same value in the codomain. If this does not happen, then the function is injective.

A) f = n^2 + 1
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^2 + 1 = b^2 + 1.
Simplifying the above equation, we get (a^2 - b^2) = 0.
Factoring, we get (a-b)(a+b) = 0.
This implies that either a = b or a = -b.
Therefore, the function is not injective.

B) f = n^3
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^3 = b^3.
Taking the cube root of both sides, we get a = b.
Therefore, the function is injective.

C) f = n^3 - 4
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^3 - 4 = b^3 - 4.
Simplifying the above equation, we get (a^3 - b^3) = 0.
Factoring, we get (a-b)(a^2 + ab + b^2) = 0.
Since the discriminant of a^2 + ab + b^2 is negative, it cannot be zero.
Therefore, the only possibility is a = b.
Therefore, the function is injective.

D) f = n/2
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a/2 = b/2.
Simplifying the above equation, we get a = b.
Therefore, the function is injective.

In conclusion, the injective functions are B) f = n^3 and D) f = n/2.

 

[NEEDITNOW0819]

Let(x) = X^2/3, find f(S) if S= {-2,-1,0,1,2,3}

 

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To find f(S), we need to apply the function f(x) = x^(2/3) to every element in the set S = {-2, -1, 0, 1, 2, 3}.

f(-2) = (-2)^(2/3) = 2^(2/3) = 1.587401052
f(-1) = (-1)^(2/3) = 1
f(0) = 0^(2/3) = 0
f(1) = 1^(2/3) = 1
f(2) = 2^(2/3) = 1.587401052
f(3) = 3^(2/3) = 1.817120592

Therefore, f(S) = {1.587401052, 1, 0, 1, 1.587401052, 1.817120592}.

 

[NEEDITNOW0819]

Consider the following relations {1,2,3,4}. Which of the following relations is symmetric?
R = {(2,2),(2,3),(2,4)}
R = {(1,2),(1,3),(2,3)}
R = {(1,3),(2,2),(3,1)}
R = {(1,2),(3,4),(4,4)}

 

[🤖]

A relation R on a set A is called symmetric if (a, b) ∈ R whenever (b, a) ∈ R for all a, b ∈ A.

Now, let's check each of the given relations to see if it is symmetric or not.

R = {(2,2),(2,3),(2,4)}
Here, we see that (2,3) ∈ R, but (3,2) ∉ R. Therefore, R is not symmetric.

R = {(1,2),(1,3),(2,3)}
Here, we see that (1,2) ∈ R implies (2,1) ∈ R, (1,3) ∈ R implies (3,1) ∈ R, and (2,3) ∈ R implies (3,2) ∈ R. Therefore, R is symmetric.

R = {(1,3),(2,2),(3,1)}
Here, we see that (1,3) ∈ R implies (3,1) ∈ R, and (2,2) ∈ R is already symmetric. However, (3,1) ∈ R but (1,3) ∉ R. Therefore, R is not symmetric.

R = {(1,2),(3,4),(4,4)}
Here, we see that (1,2) ∈ R implies (2,1) ∈ R, (4,4) ∈ R is already symmetric, but (3,4) ∈ R but (4,3) ∉ R. Therefore, R is not symmetric.

Hence, only the relation R = {(1,2),(1,3),(2,3)} is symmetric.

 

[NEEDITNOW0819]

Consider the function from the set of teachers in a school if it assigns a teacher per bus of students?
A). Onto
B). Neither One-to-one nor Onto
C). One-to-one
D). One-to-one and onto

 

[🤖]

The function that assigns a teacher to a bus of students is likely a one-to-many function, as one teacher may be assigned to multiple buses of students. Therefore, this function is not one-to-one.

However, it is possible that every bus has been assigned a teacher, meaning that every element in the range is covered by the function. In this case, the function is onto.

Therefore, the answer is B) Neither One-to-one nor Onto.