Find the error in the "proof" of the following "theorem." "theorem": let r be a relation on a set a that is symmetric and transitive. then r is reflexive. "proof ": let a ∈ a. take an element b ∈ a such that (a, b) ∈ r. because r is symmetric, we also have (b, a) ∈r. now using the transitive property, we can conclude that (a, a) ∈ r because (a, b) ∈ r and (b, a) ∈ r.

The proof is wrong. The part that says 'take an element b ∈ A such that (a, b) ∈ R' is assuming that such an element b exists, and that may not be the case.  We can give the 'theorem' validation by adding the hypothesis ' each element of A is related to any other element of A', otherwise the 'theorem' is false. An example for that is the following relation on the natural numbers:

'a R b if both numbers are even'

This relation if crearly symmetric and transitive, but it is not reflexive, an odd number does not relate with itself. In fact, odd numbers dont relate with any number.

6 cars because if you have 8 and subtract 2 youll have 6 left over

2 equivalent expressions to represent the area of a rectangle and the top is n-3 and the side is 2

area of the rectangle = length * width

given length = n - 3

width = 2

replace it in the area

so area of the rectangle = (n-3) *2

= 2(n-3)

one of the expression is 2(n-3)

to get another expression we distribute 2 inside the parenthesis

2(n-3) is 2n - 6

so another expression is 2n - 6

two equivalent expression are

2(n - 3)   , 2n - 6