Suppose that a is a nonempty set, and f is a function that has a as its domain. let r be the relation on a consisting of all ordered pairs (x, y) such that f(x) = f(y). (a) show that r is an equivalence relation on a.

See proof below

Step-by-step explanation:

Let xRy denote the statement "(x,y)∈A". To prove that R is an equivalence relation on A, we must prove that R

For the first one, we know that equality is reflexive, and for all x, f(x) is unique because f is a function, then f(x)=f(x), that is, xRX. So R is reflexive.

For symmetry, suppose that xRy. Then, by definition of R, f(x)=f(y) (again, this value is unique for all x,y). Equality is symmetric, then f(y)=f(x), hence yRx. We have that R is symmetric.

For the last one, suppose that xRy and yRz. Then f(x)=f(y) and f(y)=f(z). f(x),f(y) and f(z) are uniquely determined by the function f. Moreover, equality is transitive, then f(x)=f(y)=f(z) implies that f(x)=f(z), that is, xRz. Hence R is transitive.

R is reflexive, symmetric and transitive, therefore R is a equivalence relation on A.

The hypothesis on f is important. If f wasn't a function, f(x) could represent than one element, so it would be ambiguous and the properties of equality wouldn't necessarily hold.

he can buy 4lbs. of live bait and 1lbs of natural bait

step-by-step explanation:

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -