# If R = {(x,y)| y = 3x + 2}, then R -1 = (1) x = 3y + 2 (2) y = (x – 2)/3 (3) {(x,y)| y = 3x + 2} (4) {(x,y)| y = (x – 2)/3} (5) {(x,y)| y – 2 = 3x} (6)

## Presentation on theme: "If R = {(x,y)| y = 3x + 2}, then R -1 = (1) x = 3y + 2 (2) y = (x – 2)/3 (3) {(x,y)| y = 3x + 2} (4) {(x,y)| y = (x – 2)/3} (5) {(x,y)| y – 2 = 3x} (6)"— Presentation transcript:

If R = {(x,y)| y = 3x + 2}, then R -1 = (1) x = 3y + 2 (2) y = (x – 2)/3 (3) {(x,y)| y = 3x + 2} (4) {(x,y)| y = (x – 2)/3} (5) {(x,y)| y – 2 = 3x} (6) {(x,y)| y = x/3 – 2}

What is the first line of this proof? (1) Let x  R. (2) Let x  R -1. (3) Let (x,y)  R. (4) Let (x,y)  R -1.

What is the first line of this proof? (1) Let x  R.(2) Let x  R -1. (3) Let (x,y)  R.(4) Let (x,y)  R -1. (5) Let (x,y)  Dom(R) (6) Let (x,y)  Dom(R -1 ) (7) Let x  Dom(R) (8) Let x  Dom(R -1 )

Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then S  R = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then R  S = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then R  R = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then S  S = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,2), (2,3), (3,1)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

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