Let f: Z-Z be defined as f(x)=x+7 if x is odd x-51 if x is even Show that is 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 is 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 b are --Select--- Substitutinga and b into f we have the equation a-5- which simplifies to a - which simplifies to a b. 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 y E Z be odd. Then y + 5 is ---Select--- fly+5)= Similarly, suppose y EZ is even. Then y-7 is ---Select--- f(y-7)= Therefore fis both onto and one-to-one and has a one-to-one correspondence.
Let f: Z-Z be defined as f(x)=x+7 if x is odd x-51 if x is even Show that is 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 is 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 b are --Select--- Substitutinga and b into f we have the equation a-5- which simplifies to a - which simplifies to a b. 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 y E Z be odd. Then y + 5 is ---Select--- fly+5)= Similarly, suppose y EZ is even. Then y-7 is ---Select--- f(y-7)= Therefore fis both onto and one-to-one and has a one-to-one correspondence.
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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Let f: z->z defined as
The options are all odd/even.
Transcribed Image Text:Let f: Z → Z be defined as
f(x) =
x+7 if x is odd
x-5 if x is even
Show that f is 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 - 5
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 f is 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 + 5 is ---Select---.
f(y + 5) =
5
Similarly, suppose y EZ is even. Then y-7 is ---Select---
f(y-7)=
+7
Therefore f is both onto and one-to-one and has a one-to-one correspondence.
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