Some tricky problems about the floor and ceiling of X

1.Definition of [x](Here I use the [x] for the floor of x.):

x R, y = [x] y  Z and y <=x and z Z, z <= x z <= y.

1.Prove: x R, [x] > x - 1
Proof:
Assume x is a generic real number,
Then [x] N, 
Then [x] +1 N,
Then [x] + 1 > [x]
Then [x] +1 > x    # By contrapositive and according to the definition of [x]
Then [x] > x - 1.

2.Define a function: n in N, a(n) = [n/2]
Disprove: i N, j N, i < j a(i) = a(j).
Proof: To disprove this statement, we need to prove the negation of the statement is true.
That is: For any i in N, there exists a j in N, such that i < j and a(i) a(j).
Assume i is a generic natural number,
Pick j = i + 2,
Then a(j) = [j/2] = [(i+2)/2] = [i/2 + 1] = [i/2] +1
Then a(i) = [i/2]
Then a(i) a(j).
For any i in N, there exists a j in N, such that i < j and a(i) a(j).

Pay attention to two places:

1. About the definition of floor  of x: [x], when using the second part of the definition, we have to make sure the z that is used for comparison is an integer.

Here we use the contra-positive of the second part of the definition, that is [x] if [x]+1 > [x], then [x] +1 > x.

(I made this mistake during the test2.!!!)

2. When we try to disprove some statement, we use the negation of the original statement, for the quantifiers we were using, we have to negate them and for the boolean, we need to make sure it follows:

negation of p q is (p and not q)


2.Now talking about the question about the ceiling of x(using [x] as well).
 

Definition:
x R, y = [x] y  Z and y >=x and z Z, z >= x z >= y.

Prove: for any x in R, [x] < x+1
Proof: Assume x is a generic real number,
              Then [x] < [x +1]
            Then [x] < x + 1      # According to the definition and contra-positive.
           Then for any x in R, [x] < x+1.

This is the same type of question which tests on the definition of ceiling of x.