Question

n7.3, & n7.4
use benchmarks and a number line to order this set of numbers from least to greatest.
a) \\(\frac{11}{12}\\), \\(\frac{1}{3}\\), \\(\frac{7}{6}\\), \\(\frac{5}{4}\\)
from least to greatest:
number line with 0, \\(\frac{1}{2}\\), 1, \\(1\frac{1}{2}\\) marked

Explanation:

Step1: Convert to twelfths

Convert each fraction to twelfths:
$\frac{1}{3} = \frac{4}{12}$, $\frac{7}{6} = \frac{14}{12}$, $\frac{5}{4} = \frac{15}{12}$, $\frac{11}{12}$ remains.

Step2: Compare numerators

Now compare numerators (denominator 12):
$4 < 11 < 14 < 15$ → $\frac{4}{12} < \frac{11}{12} < \frac{14}{12} < \frac{15}{12}$.

Step3: Map back to original

Substitute back: $\frac{1}{3} < \frac{11}{12} < \frac{7}{6} < \frac{5}{4}$.

Answer:

$\frac{1}{3}$, $\frac{11}{12}$, $\frac{7}{6}$, $\frac{5}{4}$