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

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# If R = {(x,y)| y = 3x + 2}, then R -1 = x = 3y + 2 (2) y = (x â 2)/3 - PowerPoint PPT Presentation

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

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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}

What is the first line of this proof?

• Let x  R.
• Let x  R-1.
• Let (x,y)  R.
• Let (x,y)  R-1.

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)

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) 

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) 

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) 

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) 