Let f: Z-Z be defined as 1x)-{xxis even x+5 if xis odd Show that is a one-to-one correspondence. To show is a one-to-one correspondence we need to show that fis one-to-one and onto. First show that fis one-to-one. Notice that if f(x) is even, then x must be -Select-. Also, if f(x) is odd, then x must be -Select-. This is because in both cases of the function definition for f f(x) differs from x by an -Select- integer. An odd integer plus an odd integer is-Select- while an odd integer plus an even integer is-Select--. Now show that f(a) f(b) implies a - b for both cases. Suppose f(a) f(b) is odd. Then a and bare -Select--. So . Similarly if f(a) f(b) is even then a and bare -Select-. Substituting a and b into f we have the equation a-7- which simplifies to a - which simplifies to ab. Therefore fis one-to-one. To show that fis onto we need to show that for some arbitrary y E Z that there is an x EZ such that f(x)= y. Let yE Z be odd. Then y + 7 is-Select-- f(y + 7) = Similarly, suppose y EZ is even. Then y- 5 is-Select-. fly-5) = Therefore fis both onto and one-to-one and has a one-to-one correspondence.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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answer the following, the drop down menu choices are odd, even 

Let f: Z→ Z be defined as
f(x) =
x + 5
if x is odd
if x is even
Show that fis a one-to-one correspondence.
To show f is a one-to-one correspondence we need to show that fis one-to-one and onto. First show that fis one-to-one. Notice that if f(x) is even, then x must be ---Select--. Also, if f(x) is odd, then x must be ---Select--- . This is because in both cases of the function definition for f,
f(x) differs from x by an ---Select--- integer. An odd integer plus an odd integer is ---Select--- while an odd integer plus an even integer is ---Select---. Now show that f(a) = f(b) implies a = b for both cases. Suppose f(a) = f(b) is odd. Then a and b are ---Select--. So
which simplifies to a = b. Therefore fis one-to-one.
a-7=
which simplifies to a =
Similarly if f(a) = f(b) is even then a and b are ---Select--- Substituting a and b into f we have the equation
To show that fis onto we need to show that for some arbitrary y E Z that there is an XE Z such that f(x) = y. Let y EZ be odd. Then y + 7 is ---Select---.
f(y + 7) =
- 7
Similarly, suppose y EZ is even. Then y- 5 is ---Select---
f(y-5) =
+ 5
Therefore fis both onto and one-to-one and has a one-to-one correspondence.
Transcribed Image Text:Let f: Z→ Z be defined as f(x) = x + 5 if x is odd if x is even Show that fis a one-to-one correspondence. To show f is a one-to-one correspondence we need to show that fis one-to-one and onto. First show that fis one-to-one. Notice that if f(x) is even, then x must be ---Select--. Also, if f(x) is odd, then x must be ---Select--- . This is because in both cases of the function definition for f, f(x) differs from x by an ---Select--- integer. An odd integer plus an odd integer is ---Select--- while an odd integer plus an even integer is ---Select---. Now show that f(a) = f(b) implies a = b for both cases. Suppose f(a) = f(b) is odd. Then a and b are ---Select--. So which simplifies to a = b. Therefore fis one-to-one. a-7= which simplifies to a = Similarly if f(a) = f(b) is even then a and b are ---Select--- Substituting a and b into f we have the equation To show that fis onto we need to show that for some arbitrary y E Z that there is an XE Z such that f(x) = y. Let y EZ be odd. Then y + 7 is ---Select---. f(y + 7) = - 7 Similarly, suppose y EZ is even. Then y- 5 is ---Select--- f(y-5) = + 5 Therefore fis both onto and one-to-one and has a one-to-one correspondence.
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