To solve an equation of the form |\text{greatest integer}=\text{number},| it is necessary to understand the definition of the integer part of a number.
Here's a reminder.
The greatest integer of a number, denoted |[x],| corresponds to the unique integer such that |[x]\leq x<[x]+1.| This symbol is also called the greatest integer less than or equal to |x.| These two terms are synonymous.
Note: If |[x]=a| where |a| must be an integer, then |a\leq x<a+1|. So |x| belongs to the interval |[a,a+1[.|
|[2.3]=2,| so we are looking for the greatest integer less than or equal to |2.3.| Also, |2\leq 2.3<3.|
|[-2.3]=-3,| so we are looking for the greatest integer less than or equal to |-2.3.| In addition, |-3\leq -2.3<-2.|
|[45]=45,| so we are looking for the greatest integer less than or equal to |45.| Moreover, |45\leq45<46.|
This definition must therefore be used when solving.
Consider the equation |\dfrac{1}{2}[-(x+1)]-1=3.|
First, isolate the greatest integer bracket.||\begin{align}\dfrac{1}{2}\big[\!-(x+1)\big] - 1 &= 3\\[3pt] \dfrac{1}{2} \big[\!-(x+1)\big] &= 4\\[3pt] \big[\!-(x+1)\big] &= 8\end{align}||As the integer is equal to a whole number, the equation can now be solved.
Next, apply the definition, that is |8\leq-(x+1)<8+1.|
Then, solve the double inequality.
| 8 \leq -(x+1) < 9|
|-8 \geq x + 1 > -9| (dividing by a negative number reverses the direction of inequality symbols)
|-8 - 1 \geq x > -9 -1|
|-9 \geq x > -10|
Thus, the solution set corresponds to the interval |]-10,-9].| The brackets of the interval are determined by the last inequality. In addition, the values inside the square brackets must always be in ascending order.
Consider the equation |2[x-1]-3=4.|
First, isolate the greatest integer bracket.||\begin{align}2[x-1]-3&=4\\[3pt] 2[x-1]&=7\\[3pt] [x-1]&=\dfrac{7}{2} = 3.5\end{align}||Here, it is necessary to stop solving since a greatest integer bracket cannot be equal to a decimal number. Therefore, the equation has no solution.