We appreciate your visit to Suppose tex f B rightarrow C tex and tex g A rightarrow B tex are functions Check all the statements that are true A If. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Certainly! Let's examine each statement one by one to determine which ones are true about functions:
Statement A: If [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are both onto, then so is [tex]\( f \circ g \)[/tex].
- Explanation: An onto function, also known as a surjective function, implies that every element in the codomain is mapped to by at least one element in the domain. If [tex]\( g: A \rightarrow B \)[/tex] is onto, it means that every element in [tex]\( B \)[/tex] is mapped to by some element in [tex]\( A \)[/tex]. Similarly, if [tex]\( f: B \rightarrow C \)[/tex] is onto, every element in [tex]\( C \)[/tex] is mapped to by some element in [tex]\( B \)[/tex]. Therefore, the composition [tex]\( f \circ g: A \rightarrow C \)[/tex] is also onto, as every [tex]\( c \in C \)[/tex] can trace back to elements in [tex]\( A \)[/tex] through [tex]\( g \)[/tex] and [tex]\( f \)[/tex].
This statement is true.
Statement B: If [tex]\( f \circ g \)[/tex] is onto and [tex]\( g \)[/tex] is not, then [tex]\( f \)[/tex] cannot be [tex]\(1-1\)[/tex].
- Explanation: If [tex]\( g \)[/tex] is not onto, there are some elements in [tex]\( B \)[/tex] that are not mapped by [tex]\( A \)[/tex]. However, if [tex]\( f \circ g \)[/tex] is still onto, every element in [tex]\( C \)[/tex] is being mapped by [tex]\( A \)[/tex] through [tex]\( f \circ g \)[/tex]. For this to be true, [tex]\( f \)[/tex] cannot be [tex]\(1-1\)[/tex] because there needs to be an overlapping in the codomain of [tex]\( f \)[/tex] to make up for the missing elements that [tex]\( g \)[/tex] doesn't reach.
This statement is true.
Statement C: If [tex]\( f \circ g \)[/tex] is onto, then so is [tex]\( f \)[/tex].
- Explanation: For [tex]\( f \circ g \)[/tex] to be onto, every element in [tex]\( C \)[/tex] must be reached, which means that [tex]\( f \)[/tex] must map [tex]\( B \)[/tex] onto every element in [tex]\( C \)[/tex]. Thus, [tex]\( f \)[/tex] must be onto for [tex]\( f \circ g \)[/tex] to be onto.
This statement is true.
Statement D: If [tex]\( f \circ g \)[/tex] is onto, then so is [tex]\( g \)[/tex].
- Explanation: While it's necessary for [tex]\( f \)[/tex] to be onto if [tex]\( f \circ g \)[/tex] is onto, it is not necessary for [tex]\( g \)[/tex] to be onto. [tex]\( g \)[/tex] might not map all of [tex]\( A \)[/tex] to [tex]\( B \)[/tex], but through specific mappings and [tex]\( f \)[/tex], [tex]\( f \circ g \)[/tex] can still cover all of [tex]\( C \)[/tex]. Therefore, [tex]\( g \)[/tex] is not required to be onto just because [tex]\( f \circ g \)[/tex] is onto.
This statement is false.
Statement E: If [tex]\( f \circ g \)[/tex] is onto and [tex]\( f \)[/tex] is not, then [tex]\( g \)[/tex] cannot be [tex]\(1-1\)[/tex].
- Explanation: If [tex]\( f \)[/tex] is not onto, not every [tex]\( c \in C \)[/tex] can be reached from [tex]\( B \)[/tex]. But if [tex]\( f \circ g \)[/tex] is onto, it implies every element in [tex]\( C \)[/tex] is reached from [tex]\( A \)[/tex] through mappings of [tex]\( f \)[/tex] and [tex]\( g \)[/tex]. For this to happen, [tex]\( g \)[/tex] cannot be [tex]\( 1-1 \)[/tex] because there needs to be a way to compensate for elements that [tex]\( f \)[/tex] misses, requiring additional inputs from [tex]\( A \)[/tex].
This statement is true.
Hence, the answer is that statements A, B, C, and E are true, while statement D is false.
Statement A: If [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are both onto, then so is [tex]\( f \circ g \)[/tex].
- Explanation: An onto function, also known as a surjective function, implies that every element in the codomain is mapped to by at least one element in the domain. If [tex]\( g: A \rightarrow B \)[/tex] is onto, it means that every element in [tex]\( B \)[/tex] is mapped to by some element in [tex]\( A \)[/tex]. Similarly, if [tex]\( f: B \rightarrow C \)[/tex] is onto, every element in [tex]\( C \)[/tex] is mapped to by some element in [tex]\( B \)[/tex]. Therefore, the composition [tex]\( f \circ g: A \rightarrow C \)[/tex] is also onto, as every [tex]\( c \in C \)[/tex] can trace back to elements in [tex]\( A \)[/tex] through [tex]\( g \)[/tex] and [tex]\( f \)[/tex].
This statement is true.
Statement B: If [tex]\( f \circ g \)[/tex] is onto and [tex]\( g \)[/tex] is not, then [tex]\( f \)[/tex] cannot be [tex]\(1-1\)[/tex].
- Explanation: If [tex]\( g \)[/tex] is not onto, there are some elements in [tex]\( B \)[/tex] that are not mapped by [tex]\( A \)[/tex]. However, if [tex]\( f \circ g \)[/tex] is still onto, every element in [tex]\( C \)[/tex] is being mapped by [tex]\( A \)[/tex] through [tex]\( f \circ g \)[/tex]. For this to be true, [tex]\( f \)[/tex] cannot be [tex]\(1-1\)[/tex] because there needs to be an overlapping in the codomain of [tex]\( f \)[/tex] to make up for the missing elements that [tex]\( g \)[/tex] doesn't reach.
This statement is true.
Statement C: If [tex]\( f \circ g \)[/tex] is onto, then so is [tex]\( f \)[/tex].
- Explanation: For [tex]\( f \circ g \)[/tex] to be onto, every element in [tex]\( C \)[/tex] must be reached, which means that [tex]\( f \)[/tex] must map [tex]\( B \)[/tex] onto every element in [tex]\( C \)[/tex]. Thus, [tex]\( f \)[/tex] must be onto for [tex]\( f \circ g \)[/tex] to be onto.
This statement is true.
Statement D: If [tex]\( f \circ g \)[/tex] is onto, then so is [tex]\( g \)[/tex].
- Explanation: While it's necessary for [tex]\( f \)[/tex] to be onto if [tex]\( f \circ g \)[/tex] is onto, it is not necessary for [tex]\( g \)[/tex] to be onto. [tex]\( g \)[/tex] might not map all of [tex]\( A \)[/tex] to [tex]\( B \)[/tex], but through specific mappings and [tex]\( f \)[/tex], [tex]\( f \circ g \)[/tex] can still cover all of [tex]\( C \)[/tex]. Therefore, [tex]\( g \)[/tex] is not required to be onto just because [tex]\( f \circ g \)[/tex] is onto.
This statement is false.
Statement E: If [tex]\( f \circ g \)[/tex] is onto and [tex]\( f \)[/tex] is not, then [tex]\( g \)[/tex] cannot be [tex]\(1-1\)[/tex].
- Explanation: If [tex]\( f \)[/tex] is not onto, not every [tex]\( c \in C \)[/tex] can be reached from [tex]\( B \)[/tex]. But if [tex]\( f \circ g \)[/tex] is onto, it implies every element in [tex]\( C \)[/tex] is reached from [tex]\( A \)[/tex] through mappings of [tex]\( f \)[/tex] and [tex]\( g \)[/tex]. For this to happen, [tex]\( g \)[/tex] cannot be [tex]\( 1-1 \)[/tex] because there needs to be a way to compensate for elements that [tex]\( f \)[/tex] misses, requiring additional inputs from [tex]\( A \)[/tex].
This statement is true.
Hence, the answer is that statements A, B, C, and E are true, while statement D is false.
Thanks for taking the time to read Suppose tex f B rightarrow C tex and tex g A rightarrow B tex are functions Check all the statements that are true A If. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
