1. Suppose that we have the binary relations R1 C A× B, R2 C Bx C, and R3 C C x D. Then the relations R3 o (R2 o R1) and (R3 o R2) o R1 are both relations from A to D. In this exercise, you want to show that they are in fact the same relation, i.e. R3 0 (R2 o R1) (R3 o R2) o R1. Recall that • R2o R1 is the set of all ordered pairs (a, c) E A × C such that, for each of them, there exists a b e B such that (a, b) E Rị and (b, c) e R2, • R3o (R20 R1) is the set of all ordered pairs (a, d) E Ax D such that, for each of them, there exists a c E C such that (a, c) E R2 0 R1 and (c, d) E R3, • R3 o R2 is the set of all ordered pairs (b, d) EBx D such that, for each of them, there exists a c E C such that (b, c) E Bx C and (c, d) E C x D, (R3o R2) o Rị is the set of all ordered pairs (a, d) E Ax D such that, for each of them, there exists a be B such that (a, b) E A × B and (b, d) E R2 × R3. Use these four summary points to show that R3 o (R2 0 R1) = (R3 o R2) o R1. This can be shown by showing two things: first, show that each point in R3 o (R2 0 R1) is also in (R3 0 R2) o R1, and second, the converse, where each point in (R3 o R2) o R1 is also in R3 o (R2 o R1), is also true. (R2 0 R1) C (R3 o R2) o R1 and the second part shows that (R3 0 R2) o R C R3 o (R2 o R1), (This is often the way to show that two sets are equal. The first part shows that R3 o whence (R3 o R2) o R1 = R3 0 (R2o R1).)

The fractional value of 5- + 6– is 5 (The answers form 3 the three parts of a mixed number. The first blank corresponds to the whole number in the mixed number, the second blank to the numerator, and the third part to the denominator.)