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If R = {(x,y)| y = 3x + 2}, then R -1 = x = 3y + 2 (2) y = (x – 2)/3 PowerPoint Presentation
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If R = {(x,y)| y = 3x + 2}, then R -1 = x = 3y + 2 (2) y = (x – 2)/3

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

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

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  1. If R = {(x,y)| y = 3x + 2}, then R-1 = • x = 3y + 2 (2) y = (x – 2)/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}

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

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

  4. 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, 2, 2, 3} (2) { 3, 2, 3, 1} • {(2,1), (3,3)} (4) {(1,1), (2,1)} • {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} • (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

  5. 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, 2, 2, 3} (2) { 3, 2, 3, 1} • {(2,1), (3,3)} (4) {(1,1), (2,1)} • {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} • (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

  6. 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, 2, 2, 3} (2) { 3, 2, 3, 1} • {(2,1), (3,3)} (4) {(1,1), (2,1)} • {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} • (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 

  7. 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, 2, 2, 3} (2) { 3, 2, 3, 1} • {(2,1), (3,3)} (4) {(1,1), (2,1)} • {(1,2), (2,3), (3,1)} (6) {(1,3), (2,2)} • (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8) 