Understanding Floor and Ceiling Functions

This question asks us to evaluate the correctness of two statements involving the floor and ceiling functions. Let's analyze each statement separately.

Statement I Analysis: Floor Function Identity

Statement I is: \(\left\lfloor {2x} \right\rfloor = \left\lfloor x \right\rfloor + \left\lfloor {x + (1/2)} \right\rfloor\) for all real number x.

The floor function, denoted by \(\lfloor y \rfloor\), gives the greatest integer less than or equal to y. For example, \(\lfloor 3.7 \rfloor = 3\) and \(\lfloor -1.2 \rfloor = -2\).

Statement I is a known identity related to floor functions, often referred to as Hermite's Identity for \(n=2\). Let's prove this identity.

Let x be any real number. We can write x as \(x = n + \alpha\), where n is an integer (\(n = \lfloor x \rfloor\)) and \(\alpha\) is the fractional part of x (\(0 \le \alpha < 1\)).

Now, let's evaluate both sides of the equation using this representation of x.

Left-Hand Side (LHS) of Statement I:

LHS = \(\left\lfloor {2x} \right\rfloor\)

Substitute \(x = n + \alpha\):

LHS = \(\left\lfloor {2(n + \alpha)} \right\rfloor = \left\lfloor {2n + 2\alpha} \right\rfloor\)

Since 2n is an integer, we can take it out of the floor function:

LHS = \(2n + \left\lfloor {2\alpha} \right\rfloor\)

Right-Hand Side (RHS) of Statement I:

RHS = \(\left\lfloor x \right\rfloor + \left\lfloor {x + (1/2)} \right\rfloor\)

Substitute \(x = n + \alpha\):

RHS = \(\left\lfloor {n + \alpha} \right\rfloor + \left\lfloor {n + \alpha + 1/2} \right\rfloor\)

Since n is an integer, \(\lfloor n + \alpha \rfloor = n + \lfloor \alpha \rfloor\). Also, \(\lfloor \alpha \rfloor = 0\) because \(0 \le \alpha < 1\).

So, \(\left\lfloor {n + \alpha} \right\rfloor = n\).

For the second term, since n is an integer, \(\left\lfloor {n + \alpha + 1/2} \right\rfloor = n + \left\lfloor {\alpha + 1/2} \right\rfloor\).

RHS = \(n + n + \left\lfloor {\alpha + 1/2} \right\rfloor = 2n + \left\lfloor {\alpha + 1/2} \right\rfloor\)

Comparing LHS and RHS:

LHS = \(2n + \left\lfloor {2\alpha} \right\rfloor\)

RHS = \(2n + \left\lfloor {\alpha + 1/2} \right\rfloor\)

For the equality to hold, we need to show that \(\left\lfloor {2\alpha} \right\rfloor = \left\lfloor {\alpha + 1/2} \right\rfloor\) for \(0 \le \alpha < 1\).

• Case 1: \(0 \le \alpha < 1/2\)
  • In this case, \(0 \le 2\alpha < 1\). So, \(\left\lfloor {2\alpha} \right\rfloor = 0\).
  • Also, \(0 + 1/2 \le \alpha + 1/2 < 1/2 + 1/2\), which means \(1/2 \le \alpha + 1/2 < 1\). So, \(\left\lfloor {\alpha + 1/2} \right\rfloor = 0\).
  • In this case, \(\left\lfloor {2\alpha} \right\rfloor = \left\lfloor {\alpha + 1/2} \right\rfloor = 0\).
• Case 2: \(1/2 \le \alpha < 1\)
  • In this case, \(2 \times 1/2 \le 2\alpha < 2 \times 1\), which means \(1 \le 2\alpha < 2\). So, \(\left\lfloor {2\alpha} \right\rfloor = 1\).
  • Also, \(1/2 + 1/2 \le \alpha + 1/2 < 1 + 1/2\), which means \(1 \le \alpha + 1/2 < 3/2\). So, \(\left\lfloor {\alpha + 1/2} \right\rfloor = 1\).
  • In this case, \(\left\lfloor {2\alpha} \right\rfloor = \left\lfloor {\alpha + 1/2} \right\rfloor = 1\).

In both cases, \(\left\lfloor {2\alpha} \right\rfloor = \left\lfloor {\alpha + 1/2} \right\rfloor\). Therefore, \(2n + \left\lfloor {2\alpha} \right\rfloor = 2n + \left\lfloor {\alpha + 1/2} \right\rfloor\), which means LHS = RHS.

Thus, Statement I is true.

Statement II Analysis: Ceiling Function Property

Statement II is: \(\left\lceil {x + y} \right\rceil = \left\lceil x \right\rceil + \left\lceil y \right\rceil\) for all real numbers x and y.

The ceiling function, denoted by \(\lceil z \rceil\), gives the smallest integer greater than or equal to z. For example, \(\lceil 3.7 \rceil = 4\) and \(\lceil -1.2 \rceil = -1\).

Statement II claims that the ceiling function distributes over addition. Let's test this with a counterexample.

Let \(x = 0.5\) and \(y = 0.5\).

Left-Hand Side (LHS) of Statement II:

LHS = \(\left\lceil {x + y} \right\rceil\)

Substitute \(x = 0.5\) and \(y = 0.5\):

LHS = \(\left\lceil {0.5 + 0.5} \right\rceil = \left\lceil 1 \right\rceil = 1\)

Right-Hand Side (RHS) of Statement II:

RHS = \(\left\lceil x \right\rceil + \left\lceil y \right\rceil\)

Substitute \(x = 0.5\) and \(y = 0.5\):

RHS = \(\left\lceil 0.5 \right\rceil + \left\lceil 0.5 \right\rceil\)

\(\left\lceil 0.5 \right\rceil = 1\) (the smallest integer greater than or equal to 0.5 is 1)

So, RHS = \(1 + 1 = 2\)

Comparing LHS and RHS:

LHS = 1

RHS = 2

Since \(1 \neq 2\), the statement \(\left\lceil {x + y} \right\rceil = \left\lceil x \right\rceil + \left\lceil y \right\rceil\) is false for \(x = 0.5\) and \(y = 0.5\). A single counterexample is sufficient to prove that a statement is false for "all real numbers".

Thus, Statement II is false.

Conclusion on Floor and Ceiling Statements

Based on our analysis:

• Statement I: \(\left\lfloor {2x} \right\rfloor = \left\lfloor x \right\rfloor + \left\lfloor {x + (1/2)} \right\rfloor\) is true.
• Statement II: \(\left\lceil {x + y} \right\rceil = \left\lceil x \right\rceil + \left\lceil y \right\rceil\) is false.

Therefore, Statement I is true but Statement II is false.

Revision Table: Floor and Ceiling Functions

Additional Information on Floor and Ceiling Properties

Floor and ceiling functions are fundamental in discrete mathematics and computer science. They have several important properties:

• For any real number x, \(\lfloor x \rfloor \le x < \lfloor x \rfloor + 1\).
• For any real number x, \(\lceil x \rceil - 1 < x \le \lceil x \rceil\).
• \(\lceil x \rceil = -\lfloor -x \rfloor\).
• \(\lfloor x \rfloor = -\lceil -x \rceil\).
• For any integer n, \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) and \(\lceil x + n \rceil = \lceil x \rceil + n\).
• The property \(\lfloor x+y \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor\) holds for floor functions, but equality is not always true. Similarly, \(\lceil x+y \rceil \le \lceil x \rceil + \lceil y \rceil\) holds for ceiling functions, but equality is not always true (as shown by Statement II).
• Statement I, \(\left\lfloor {nx} \right\rfloor = \sum_{k=0}^{n-1} \left\lfloor {x + k/n} \right\rfloor\), is known as Hermite's Identity. Statement I is the specific case where \(n=2\).