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Magazine Chapter 2: Problem 15
Determine whether the function \(f : \mathbf{Z} \times \mathbf{Z} \rightarrow \mathbf{Z}\) is onto if a) \(f(m, n)=m+n .\) b) \(f(m, n)=m^{2}+n^{2}\) c) \(f(m, n)=m\) d) \(f(m, n)=|n|\) e) \(f(m, n)=m-n\)
Short Answer
Expert verified
a) Onto b) Not onto c) Onto d) Not onto e) Onto
Step by step solution
01
Determine if f(m, n) = m+n is onto
Check if for every value in \(\textbf{Z}\) (integers), there exist integers m and n such that m + n equals that value. Since for any integer k, you can choose m = k and n = 0, \(k = m + n = k + 0\). Therefore, there are always such m and n.
02
Conclusion for f(m, n) = m+n
The function \(f(m, n) = m + n\) is onto because for any integer k, m and n can be chosen to satisfy the equation.
03
Determine if f(m, n) = m^{2} + n^{2} is onto
Check if for every value in \(\textbf{Z}\) (integers), there exist integers m and n such that \(m^{2} + n^{2}\) equals that value. Since \(m^{2} + n^{2}\) is always non-negative and the smallest value is 0, \(f(m, n)\) cannot produce negative integers.
04
Conclusion for f(m, n) = m^{2} + n^{2}
The function \(f(m, n) = m^{2} + n^{2}\) is not onto because it cannot produce negative integers, which are part of \(\textbf{Z}\).
05
Determine if f(m, n) = m is onto
Check if for every value in \(\textbf{Z}\), there exists pairs (m, n) such that f(m, n) = m equals that value. Since \(f(m, n) = m\) is just the projection onto the first component, every integer value of m can be achieved.
06
Conclusion for f(m, n) = m
The function \(f(m, n) = m\) is onto because for any integer k, m = k satisfies the equation.
07
Determine if f(m, n) = |n| is onto
Check if for every value in \(\textbf{Z}\), there exist integers m and n such that \( |n| \) equals that value. Since \( |n| \) is always non-negative and the smallest value is 0, \(f(m, n)\) cannot produce negative integers.
08
Conclusion for f(m, n) = |n|
The function \( f(m, n) = |n| \) is not onto because it cannot produce negative integers, which are part of \(\textbf{Z}\).
09
Determine if f(m, n) = m-n is onto
Check if for every value in \(\textbf{Z}\), there exist integers m and n such that \(m - n\) equals that value. For any integer k, you can choose m = k and n = 0, thus achieving \(k = m - n = k - 0\). Therefore, there are always such m and n.
10
Conclusion for f(m, n) = m-n
The function \( f(m, n) = m - n \) is onto because for any integer k, m and n can be chosen to satisfy the equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
function analysis
In discrete mathematics, analyzing functions is a crucial skill. A function relates elements from one set, called the domain, to another set, known as the codomain. When determining if a function is 'onto' (or 'surjective'), we check if every element in the codomain is mapped by at least one element from the domain. In other words, every possible output should be achieved by some input. To illustrate this, let's analyze some functions:
- For the function \( f(m, n) = m + n \), we see that for any integer \( k \), setting \( m = k \) and \( n = 0 \) will result in \( m + n = k \). Thus, this function is onto.
- However, for \( f(m, n) = m^2 + n^2 \), the function can only produce non-negative outputs (since squares are always non-negative) and thus fails to be onto because it cannot produce negative integers.
integer sets
An integer set (denoted as \( \mathbf{Z} \)) includes all whole numbers, both negative and positive, and zero. In problems involving functions where the domain and codomain are sets of integers, understanding these sets becomes pivotal.
- In \( f(m, n) = |n| \), the absolute value of an integer \( n \) is always non-negative. Therefore, this function cannot map to negative integers. Consequently, \( f(m, n) = |n| \) is not onto.
- Similarly, in the function \( f(m, n) = m^2 + n^2 \), only non-negative outputs are possible, again excluding negative integers from the range. Thus, \( f(m, n) = m^2 + n^2 \) is not onto as well.
mapping functions
Mapping functions involve connecting elements of the domain with elements of the codomain. To see a function's behavior, we look at the specific 'mapping', or assignment of values. For example:
- For the function \( f(m, n) = m \), every integer value of \( m \) can be achieved regardless of \( n \). Thus, this function is onto the set of integers.
- In contrast, the function \( f(m, n) = m - n \) can also achieve any integer value, because for any integer \( k \), setting \( m = k \) and \( n = 0 \) satisfies \( m - n = k \). So, \( f(m, n) = m - n \) is onto.
surjective functions
Surjective functions, also known as 'onto' functions, are those where every element in the codomain has at least one pre-image in the domain. Here is how we determine this property:
- For the function \( f(m, n) = m + n \), any integer \( k \) in the codomain can be achieved by choosing suitable integers \( m \) and \( n \), proving that this function is surjective.
- Conversely, \( f(m, n) = |n| \) is not surjective, as negative integers in the codomain cannot be obtained from any input pair \( (m, n) \).
