Greatest and least integer solutions

Many questions ask for the greatest or least value of an unknown that must be an integer, a positive integer, a prime number or a perfect square. Solve the inequality first, then choose from its solutions the numbers of the required kind.

Why: if \(x\le3\tfrac{2}{3}\), every number up to \(3\tfrac{2}{3}\) is a solution. The integers among them are \(3\), \(2\), \(1\), \(0\), \(-1\), and so on. The next integer, \(4\), lies beyond the boundary, so the greatest integer solution is \(3\).

Method

1. Solve the inequality and write the boundary as a mixed number, so you can see which two integers it lies between.
2. For \(x<a\) or \(x\le a\), the greatest integer is the integer just below \(a\). It is \(a\) itself only when \(a\) is an integer and the sign is \(\le\).
3. For \(x>a\) or \(x\ge a\), the least integer is the integer just above \(a\), or \(a\) itself when \(a\) is an integer and the sign is \(\ge\).
4. For prime numbers, perfect squares, perfect cubes or positive integers, list the numbers of that kind from the boundary until you find the one asked for.
5. For a rational value (any fraction or decimal is allowed), the boundary itself is the answer when it is included: \(x\le3\tfrac{2}{3}\) has greatest rational value \(3\tfrac{2}{3}\). For \(x<3\tfrac{2}{3}\) there is no greatest rational value, because there is always another fraction closer to the boundary.

Example

Solve \(3x\le11\). Hence find the greatest integer value of \(x\), the greatest prime value of \(x\), and the number of positive integer solutions.

\(\begin{aligned} 3x&\le11 \\ x&\le3\tfrac{2}{3} \end{aligned}\)

The greatest integer is \(3\). The primes up to \(3\tfrac{2}{3}\) are \(2\) and \(3\), so the greatest prime is \(3\). The positive integer solutions are \(1\), \(2\) and \(3\): there are three of them.

Second example: \(2x>9\) gives \(x>4\tfrac{1}{2}\), so the least integer is \(5\). But \(2x\ge10\) gives \(x\ge5\), and the least integer is \(5\) because \(5\) itself is included.

Check: for \(x<4\) the greatest integer is \(3\), not \(4\). With negative boundaries, look at the number line: \(x>-2\tfrac{1}{2}\) has least integer \(-2\), not \(-3\).