Let f: Z→ Z be defined as f(x) = 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 a - 7 = which simplifies to a = . Similarly if f(a) = f(b) is even then a and b are x+5 if x is odd X-7 if x is even' ---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 EZ that there is an XE Z such that f(x) = y. Let y E Z be odd. Then y + 7 is ---Select--- f(y + 7) = 7 Similarly, suppose y EZ is even. Then y- 5 is f(y - 5) = +5 -Select--- which simplifies to a = b. Therefore fis one-to-one. Therefore f is both onto and one-to-one and has a one-to-one correspondence.

Let f: Z→ Z be defined as f(x) = 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 a - 7 = which simplifies to a = . Similarly if f(a) = f(b) is even then a and b are x+5 if x is odd X-7 if x is even' ---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 EZ that there is an XE Z such that f(x) = y. Let y E Z be odd. Then y + 7 is ---Select--- f(y + 7) = 7 Similarly, suppose y EZ is even. Then y- 5 is f(y - 5) = +5 -Select--- which simplifies to a = b. Therefore fis one-to-one. Therefore f is both onto and one-to-one and has a one-to-one correspondence.

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter1: Line And Angle Relationships

Section1.4: Relationships: Perpendicular Lines

Problem 17E: Does the relation is a brother of have a reflexive property consider one male? A symmetric property...

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