Math Problem Solver

\(1.75\) rounded to the nearest \(\frac{1}{100}\) is \(1.75\).

Rounding \(1.75\) to the nearest \(\frac{1}{100}\) is equivalent to finding the integer multiple of \(\frac{1}{100}\) nearest to \(1.75\).

The two consecutive multiples of \(\frac{1}{100}\) that \(1.75\) falls between are:\[175 \cdot \frac{1}{100} = 1.75\] and \[(175 + 1) \cdot \frac{1}{100} = 1.76\] So:\[1.75 \le 1.75 \le 1.76\]

The distance from \(1.75\) to \(1.75\) is \(\left| 1.75 - 1.75 \right| = 0\).

The distance from \(1.75\) to \(1.76\) is \(\left| 1.75 - 1.76 \right| = 0.01\).

\(1.75\) is closer to \(1.75\) than it is to \(1.76\), so \(1.75\) rounded to the nearest \(\frac{1}{100}\) is \(1.75\).

\(175\) from the above statement about consecutive multiples of \(\frac{1}{100}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{100} \le 1.75\): \[ n = \operatorname{floor}\left(1.75 \div \frac{1}{100} \right) = \operatorname{floor}\left(175 \right) = 175\]