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 ≡ b if and only if a+b equals 10.
It's not reflexive because ~(3 ≡ 3).
If's not antisymmetric because 3 ≡ 7
and 7 ≡ 3.
It's not transitive because 3 ≡ 7 and
7 ≡ 3 but ~(3 ≡ 3).
It is symmetric because if a+b equals 10
then b+a equals 10.
For integers a and b, a ≡ b if and only if a+b is greater than 10.
It's not reflexive because ~(3 ≡ 3).
If's not antisymmetric because 3 ≡ 8
and 8 ≡ 3.
It's not transitive because 3 ≡ 8 and
8 ≡ 4 but ~(3 ≡ 4).
It is symmetric because if a+b is greater
than 10 then b+a is also greater than 10.
For integers a and b, a ≡ b if and only if a < b+3.
It's not symmetic because 3 ≡ 7 but
~(7 ≡ 3).
It's not antisymmetric 3 ≡ 4 and 4 ≡ 3.
It's not transitive because 9 ≡ 7 and
7 ≡ 5 but ~9 ≡ 5.
It is reflexive because a < a+3.
For integers a and b, a ≡ b if and only if a+b is divisible by 3.
It's not reflexive because ~(4 ≡ 4).
If's not antisymmetric because 4 ≡ 8
and 8 ≡ 4.
It's not transitive because 4 ≡ 8 and
8 ≡ 10 but ~(4 ≡ 10).
If is symmetric because if a+b is divisible by 3
the equivalent b+a.
For integers a and b, a ≡ b if and only if a equals 10.
It's not reflexive because ~(4 ≡ 4).
If's not symmetric because 10 ≡ 4
and ~(4 ≡ 10).
It is antisymmetric because if a≡b and
b≡a, then both a and
b are equal to 10.
It is transitive because if a≡b and
b≡c, then a must
equal 10 and consequently a≡c.
For integers a and b, a ≡ b if and only if a*b < a+b.
It's not reflexive because ~(4 ≡ 4).
It's not antisymmetric because 1 ≡ 5
and 5 ≡ 1.
It's not transitive because 10 ≡ 1 and
1 ≡ 3 but ~(10 ≡ 3).
It is symmetric.
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.