Question
Complete the addition-subtraction box.
✓
Answer
Here,
$\frac{1}{2}+\frac{1}{3}=\frac{(1 \times 3)+(1 \times 2)}{6}=\frac{3+2}{6}=\frac{5}{6}$
$\frac{1}{3}+\frac{1}{4}=\frac{(1 \times 4)+(1 \times 3)}{12}=\frac{4+3}{12}=\frac{7}{12}$
$\frac{1}{2}-\frac{1}{3}=\frac{(1 \times 3)-(1 \times 2)}{6}=\frac{3-2}{6}=\frac{1}{6}$
$\frac{1}{3}-\frac{1}{4}=\frac{(1 \times 4)-(1 \times 3)}{12}=\frac{4-3}{12}=\frac{1}{12}$
Also,
$\frac{1}{6}+\frac{1}{12}=\frac{(1 \times 2)+1}{12}=\frac{2+1}{12}=\frac{3}{12}=\frac{1}{4}$
Hence, the above-given table can be completed as follows:
$\frac{1}{2}+\frac{1}{3}=\frac{(1 \times 3)+(1 \times 2)}{6}=\frac{3+2}{6}=\frac{5}{6}$
$\frac{1}{3}+\frac{1}{4}=\frac{(1 \times 4)+(1 \times 3)}{12}=\frac{4+3}{12}=\frac{7}{12}$
$\frac{1}{2}-\frac{1}{3}=\frac{(1 \times 3)-(1 \times 2)}{6}=\frac{3-2}{6}=\frac{1}{6}$
$\frac{1}{3}-\frac{1}{4}=\frac{(1 \times 4)-(1 \times 3)}{12}=\frac{4-3}{12}=\frac{1}{12}$
Also,
$\frac{1}{6}+\frac{1}{12}=\frac{(1 \times 2)+1}{12}=\frac{2+1}{12}=\frac{3}{12}=\frac{1}{4}$
Hence, the above-given table can be completed as follows:
| $\frac{1}{2}$ | $\frac{1}{3}$ | $\frac{5}{6}$ |
| $\frac{1}{3}$ | $\frac{1}{4}$ | $\frac{7}{12}$ |
| $\frac{1}{6}$ | $\frac{1}{12}$ | $\frac{1}{4}$ |
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