Question

Prove that if m and n are positive integers and x is a real number, then

$⌊\frac{\lfloor x\rfloor +n}{m}⌋=\left[\frac{x+n}{m}\right.⌋$

Short answer

For m and n are positive integers and x is a real number, then

$⌊\frac{\lfloor x\rfloor +n}{m}⌋=\left[\frac{x+n}{m}\right.⌋$

Step-by-step solution

Step: 1

Let us define the real number x as the sum of an integer ‘a’ and a positive real number ‘b’ which is lesser than 1.Therefore,$x=a+b$ , where a is an integer and b is a real number lesser than 1.

Step: 2

Now taking the R.H.S,

$$\begin{gathered} ⌊\frac{\lfloor x\rfloor +n}{m}⌋=⌊\frac{\lfloor a+b\rfloor +n}{m}⌋ \\ =⌊\frac{\lfloor b\rfloor +a+n}{m}⌋ \\ =⌊\frac{1+a+n}{m}⌋ \\ =⌊\frac{1}{m}+\frac{a+n}{m}⌋\dots .\left(1\right) \end{gathered}$$

Step: 3

Now taking the L.H.S,

$$\begin{gathered} ⌊\frac{x+n}{m}⌋=⌊\frac{a+b+n}{m}⌋ \\ =⌊\frac{x}{m}+\frac{n}{m}⌋ \\ =⌊\frac{b}{m}+\frac{a+n}{m}⌋\dots (2) \end{gathered}$$

From 1 and 2, we get$t⌊\frac{\lfloor x\rfloor +n}{m}⌋=⌊\frac{x+n}{m}⌋$