Question

Prove that if n is an integer, then$[n/2]=n/2$ if n is even and$(n-1)/2$ if nis odd

Short answer

The proof is

If n is even, then$[n/2]=n/2$

If n is odd, then$[n/2]=(n-1)/2$

Step-by-step solution

Step 1:

Floor function [x] : largest integer that is less than or equal to x.

Step 2:To prove:

If n is even, then$[n/2]=n/2$

If n is odd, then$[n/2]=(n-1)/2$

FIRST CASE

Let n be even. Then there exist an integer k such that n = 2k .

$\frac{n}{2}=\frac{2k}{2}=k$

The ceiling function of an integer is the integer itself.

$\lfloor n/2\rfloor =\lfloor k\rfloor =k=n/2$

Step 3:

SECOND CASE

Let n be odd. Then there exist an integer k such that$n=2k+1$ .

$\frac{n}{2}=\frac{2k+1}{2}=k+\frac{1}{2}$

since k is an integer$k+\frac{1}{2}$ is not an integer and thus$\frac{n}{2}$ is not an integer.

$k<k+\frac{1}{2}<k+1$

Since$\frac{n}{2}=k+\frac{1}{2}$

$k<\frac{n}{2}<k+1$

The floor function of x is then the smaller integer from the inequality

$⌊\frac{n}{2}⌋=k$

Since or equivalently$k=\frac{n-1}{2}$

$⌊\frac{n}{2}\mid =\frac{n-1}{2}$

Hence, the solution is

If n is even, then$[n/2]=n/2$

If n is odd, then$[n/2]=(n-1)/2$