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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

vidhan
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Lax city
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    I think you're going to find that when you ask people for help, telling them what they "must" do is unlikely to make them feel like helping you. This is a lesson that goes beyond math.stackexchange. – WillO May 28 '15 at 19:22
  • Really sorry, is is my first time using this site – Lax city May 28 '15 at 19:23
  • I copied the question from my paper exactly, that's why – Lax city May 28 '15 at 19:24
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    Hint: Convert the fractions to a common denominator. – callculus42 May 28 '15 at 19:24
  • Start with this: Here are two numbers: 5 and 9. Which is closer to 8? Now think about exactly how you figured that out, and apply the same method to your problem with the fractions. – WillO May 28 '15 at 19:30
  • In this situation I filled in the missing numbers , that is 5,6,7,8,9 . Here I can see clearly that 8 is closer to 9. I suppose You mean I should do the same with fractions but am going blank when it comes to doing the same with fractions. – Lax city May 28 '15 at 19:36
  • @JoshuaMbewe Did you read my comment ? – callculus42 May 28 '15 at 19:40

2 Answers2

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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.

Timbuc
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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.

T. Eskin
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