Question

Show that \(\left\lfloor {x - \frac{1}{2}} \right\rfloor \) is the closest integer to the number x, except when x is midway between two integers, when it is the large of these two integers.

Short answer

When x is not midway between the two integers, \(\left\lfloor {x - \frac{1}{2}} \right\rfloor \) is equal to the smaller of the two integers.

Step 1:

Step-by-step solution

Step 1:

Ceiling function \(\left\lfloor x \right\rfloor \): smallest integer that is greater than or equal to x.

Step 2:To prove:

When x is not midway between the two integers,\(\left\lfloor {x - \frac{1}{2}} \right\rfloor \)is equal to the smaller of the two integers.

When x is midway between the two integers,\(\left\lfloor {x - \frac{1}{2}} \right\rfloor \)is equal to the closet integer to the number x.

PROOF

Let x be a real number. Any real number x is lies between two consecutive integers(inclusive, as the real number can also be an integer)

\(\exists n \in {\rm Z}:n \le x \le n + 1\)

FIRST CASE

Let x be midway between the two integers n and n+1. Then x is n increased by half the distance between n and n+1

\(x = n + \frac{{(n + 1) - n}}{2} = n + \frac{1}{2}\)

Then we obtain:\(\left\lfloor {x - \frac{1}{2}} \right\rfloor \)=\(\left\lfloor {n + \frac{1}{2} - \frac{1}{2}} \right\rfloor \)=\(\left\lfloor n \right\rfloor \)=\(n\).

When x is midway between the two integers, \(\left\lfloor {x - \frac{1}{2}} \right\rfloor \) is equal to the smaller of the two integers.

Step 3:Part (b)

SECOND CASE

Let x be midway between the two integers n and n+1. Then x is smaller than the value\(n + \frac{1}{2}\) midway between the two integers.

\(n < x < n + \frac{1}{2}\)

Add\(\frac{1}{2}\)to each side of the inequality

\(n - 1 < n - \frac{1}{2} \le x - \frac{1}{2} < n\)

Since\(n - 1 < x + \frac{1}{2} < n\), the floor function of\(x - \frac{1}{2}\)is equal to the lower integers

Then we obtain:\(\left\lfloor {x - \frac{1}{2}} \right\rfloor = n\)

N was also the closet integer to x, thus we have proven the statement for this case.

Step 4:

THIRD CASE

Let x be midway between the two integers n and n+1. Then x is smaller than the value\(n + \frac{1}{2}\) midway between the two integers.

\(n + \frac{1}{2} < x < n + 1\)

Add\(\frac{1}{2}\)to each side of the inequality

\(n < x - \frac{1}{2} \le n + \frac{1}{2} < n + 1\)

Since\(n < x - \frac{1}{2} < n + 1\), the floor function of\(x - \frac{1}{2}\)is equal to the lower integers

Then we obtain:\(\left\lfloor {x - \frac{1}{2}} \right\rfloor = n + 1\)

\(n + 1\)was also the closet integer to x, thus we have proven the statement for this case.

Hence, the solution is

When x is not midway between the two integers,\(\left\lfloor {x - \frac{1}{2}} \right\rfloor \)is equal to the smaller of the two integers.

When x is midway between the two integers, \(\left\lfloor {x - \frac{1}{2}} \right\rfloor \) is equal to the closet integer to the number x.