For each of the following relations, prove if the relation is reflexive, symmetric, antisymmetric, and transitive. All of these relations are taken over the set of positive integers.
For integers a and b, a R b if and only if a < b+333.
It is reflexive because a < a+333.
It is not symmetic because 100 R 500 but
~(500 R 100).
It is not antisymmetric because 100 R 300 and 300 R 100.
It is not transitive because 500 R 300 and
300 R 100 but ~500 R 100.
For integers a and b, a R b if and only if a+333 < b.
It is not reflexive because ~(400 R 400).
It is not symmetic because 300 R 700 but
~(700 R 300).
It is antisymmetric because if a+333 < b
then b+333 cannot be less than a.
It is transitive because if a+333 < b and
b+333 < c then
a+333 < b < b+333 < c.
For integers a and b, a R b if and only if a equals 333.
It is not reflexive because ~(400 R 400).
It is not symmetric because 333 R 251
and ~(251 R 333).
It is antisymmetric because if aRb and
bRa, then both a and
b are equal to 333.
It is transitive because if aRb and
bRc, then a must
equal 333 and consequently aRc.
For integers a and b, a R b if and only if a+b equals 333.
It is not reflexive because ~(100 R 100).
It is symmetric because if a+b equals 333
then b+a equals 333.
It is not antisymmetric because 100 R 233
and 233 R 100.
It is not transitive because 100 R 233 and
233 R 100 but ~(100 R 100).
Use induction to prove that the function defined by the following recurrence relation:
has the closed form solution
In the base case, we see that T(0) = 1 = 2*0+1.
In the induction case, we assume T(n) = 2 n + 1. By the recurrence relation T(n+1) = T(n) + 2. By induction this means that T(n+1) = (2 n + 1) + 2 which can be rearranged to obtain T(n+1) = 2 (n + 1) + 1.