Distance between 2 5 \frac{2}{5} 5 2 and 6 21 \frac{6}{21} 21 6 is d ( 2 5 , 6 21 ) = ∣ 2 5 − 6 21 ∣ = ∣ 2 5 − 2 7 ∣ = ∣ 2 ⋅ 7 − 2 ⋅ 5 5 ⋅ 7 ∣ = ∣ 4 5 ⋅ 7 ∣ = 4 5 ⋅ 7 d\left(\frac{2}{5},\frac{6}{21}\right)=\bigl|\frac{2}{5}-\frac{6}{21}\bigr|=\bigl|\frac{2}{5}-\frac{2}{7}\bigr|=\bigl|\frac{2\cdot 7-2\cdot 5}{5\cdot 7}\bigr|=\bigl|\frac{4}{5\cdot 7}\bigr|=\frac{4}{5\cdot 7} d ( 5 2 , 21 6 ) = ∣ ∣ 5 2 − 21 6 ∣ ∣ = ∣ ∣ 5 2 − 7 2 ∣ ∣ = ∣ ∣ 5 ⋅ 7 2 ⋅ 7 − 2 ⋅ 5 ∣ ∣ = ∣ ∣ 5 ⋅ 7 4 ∣ ∣ = 5 ⋅ 7 4
Distance between 2 5 \frac{2}{5} 5 2 and 13 29 \frac{13}{29} 29 13 is d ( 2 5 , 13 29 ) = ∣ 2 5 − 13 29 ∣ = ∣ 2 ⋅ 29 − 13 ⋅ 5 5 ⋅ 29 ∣ = ∣ − 7 5 ⋅ 29 ∣ = 7 5 ⋅ 29 d\left(\frac{2}{5},\frac{13}{29}\right)=\bigl|\frac{2}{5}-\frac{13}{29}\bigr|=\bigl|\frac{2\cdot 29-13\cdot 5}{5\cdot 29}\bigr|=\bigl|\frac{-7}{5\cdot 29}\bigr|=\frac{7}{5\cdot 29} d ( 5 2 , 29 13 ) = ∣ ∣ 5 2 − 29 13 ∣ ∣ = ∣ ∣ 5 ⋅ 29 2 ⋅ 29 − 13 ⋅ 5 ∣ ∣ = ∣ ∣ 5 ⋅ 29 − 7 ∣ ∣ = 5 ⋅ 29 7
We have d ( 2 5 , 6 21 ) = 4 5 ⋅ 7 = 4 ⋅ 29 5 ⋅ 7 ⋅ 29 > 7 ⋅ 7 5 ⋅ 7 ⋅ 29 = 7 5 ⋅ 29 = d ( 2 5 , 13 29 ) d\left(\frac{2}{5},\frac{6}{21}\right)=\frac{4}{5\cdot 7}=\frac{4\cdot 29}{5\cdot 7\cdot 29}>\frac{7\cdot 7}{5\cdot 7\cdot 29}=\frac{7}{5\cdot 29}=d\left(\frac{2}{5},\frac{13}{29}\right) d ( 5 2 , 21 6 ) = 5 ⋅ 7 4 = 5 ⋅ 7 ⋅ 29 4 ⋅ 29 > 5 ⋅ 7 ⋅ 29 7 ⋅ 7 = 5 ⋅ 29 7 = d ( 5 2 , 29 13 ) .
So 13 29 \frac{13}{29} 29 13 is closer to 2 5 \frac{2}{5} 5 2 than 6 21 \frac{6}{21} 21 6 .
Answer: 13 29 \frac{13}{29} 29 13
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