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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.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.7: Relations
Problem 21E: 21. A relation on a nonempty set is called irreflexive if for all. Which of the relations in...
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Let f: z->z defined as 

The options are all odd/even. 

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.
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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