Let O be the set of all odd integers, and let 2Z be the set of all even integers. Prove that O has the same cardinality as 2z. Proof: In order to show that O has the same cardinality as 22 we must show that there is a well-defined function f: 0 -→ 2Z that is both one-to-one and onto. We will show that the following is a function from 0 to 2Z that satisfies these requirements. (Choose one definition for f and use it for the rest of the proof.) O f(n) = ", for each odd integer n in o O f(n) = Inl, for each odd integer n in o O f(n) = n - 1, for each odd integer n in o O f(n) = n+ 2, for each odd integer n in o O f(n) - 3n, for each odd integer n in o Well-Defined One-to-One Onto Proof that f is onto: To show that f is onto, let m be any even integer in 2z. By definition of even, there exists an integer k such that m = 2k. On a separate piece of scratch paper, find an odd integer in 0, written as an expression using the variable k, with the property that when fis applied to it, the result is 2k. Write the expression in the box below. By construction, the quantity in the box is an odd integer, and when the function fis applied to it, the result is the even integer 2k, which equals m. Thus, we have shown that there exists an element of O that is sent to m by f, and so f is onto. Conclusion: Since f is a well-defined function from O to 2Z that is one-to-one and onto, we conclude that O and 22 have the same cardinality.
Let O be the set of all odd integers, and let 2Z be the set of all even integers. Prove that O has the same cardinality as 2z. Proof: In order to show that O has the same cardinality as 22 we must show that there is a well-defined function f: 0 -→ 2Z that is both one-to-one and onto. We will show that the following is a function from 0 to 2Z that satisfies these requirements. (Choose one definition for f and use it for the rest of the proof.) O f(n) = ", for each odd integer n in o O f(n) = Inl, for each odd integer n in o O f(n) = n - 1, for each odd integer n in o O f(n) = n+ 2, for each odd integer n in o O f(n) - 3n, for each odd integer n in o Well-Defined One-to-One Onto Proof that f is onto: To show that f is onto, let m be any even integer in 2z. By definition of even, there exists an integer k such that m = 2k. On a separate piece of scratch paper, find an odd integer in 0, written as an expression using the variable k, with the property that when fis applied to it, the result is 2k. Write the expression in the box below. By construction, the quantity in the box is an odd integer, and when the function fis applied to it, the result is the even integer 2k, which equals m. Thus, we have shown that there exists an element of O that is sent to m by f, and so f is onto. Conclusion: Since f is a well-defined function from O to 2Z that is one-to-one and onto, we conclude that O and 22 have the same cardinality.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Proof that f is onto
Transcribed Image Text:Let O be the set of all odd integers, and let 2Z be the set of all even integers. Prove that O has the same cardinality as 2Z.
Proof: In order to show that O has the same cardinality as 2Z we must show that there is a well-defined function f: 0 → 2Z that is both one-to-one and onto.
We will show that the following is a function from 0 to 2Z that satisfies these requirements. (Choose one definition for f and use it for the rest of the proof.)
O f(n) =
for each odd integer n in O
f(n) = |n|, for each odd integer n in O
O f(n) = n – 1, for each odd integer n in O
O f(n) = n + 2, for each odd integer n in O
O f(n) = 3n, for each odd integer n in O
Well-Defined
One-to-One
Onto
Proof that f is onto:
To show that fis onto, let m be any even integer in 2z. By definition of even, there exists an integer k such that m = 2k.
On a separate piece of scratch paper, find an odd integer in 0, written as an expression using the variable k, with the property that when f is applied to it, the result is 2k. Write the expression in the box below.
By construction, the quantity in the box is an odd integer, and when the function f is applied to it, the result
the even integer 2k, which equals m.
Thus, we have shown that there exists an element of O that is sent tom by f, and so f is onto.
Conclusion: Since f is a well-defined function from 0 to 2Z that is one-to-one and onto, we conclude that O and 2Z have the same cardinality.
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