Here are two fractions, $\frac{2}{3}$,$\frac{7}{8}$, which of these fractions are closer to $\frac{3}{4}$?

Question

I've been throwing this question around my family. No one has a clue, therefore can someone help?

I'm pretty sure this will be easy to do

Answer 1

Calculus suggestion is straight on the money: write the three fractions involved in this problem with a common denominator, say $\;24\; $:

$$\begin{align}&\frac23=\frac{16}{24}\\{}\\ &\frac78=\frac{21}{24}\\{}\\ &\color{red}{\frac34=\frac{18}{24}}\end{align}$$

Well, so which fraction is the closest one to the red one and why? Complete the reasoning.

Answer 2

The distance between real numbers $x$ and $y$ is $|x-y|$. So the distance from $\frac{2}{3}$ to $\frac{3}{4}$ is \begin{align*} \Big|\frac{2}{3}-\frac{3}{4}\Big|=\Big|\frac{8}{12}-\frac{9}{12}\Big|=\frac{1}{12}. \end{align*} And the distance from $\frac{7}{8}$ to $\frac{3}{4}$ is \begin{align*} \Big|\frac{7}{8}-\frac{3}{4}\Big|=\Big|\frac{7}{8}-\frac{6}{8}\Big|=\frac{1}{8}. \end{align*} Since $\frac{1}{12}<\frac{1}{8}$, then $\frac{2}{3}$ is closer to $\frac{3}{4}$ than $\frac{7}{8}$ is.