Without Definitions: Why $\displaystyle\frac{1}{3}+\frac{2}{5}=\frac{11}{15}?$

Introduction

I have recently run into a 2007 page of mine Why 1/3 + 1/4 = 7/12? The fact that that identity can be established with no resort to any properties of addition, let alone addition of fractions, is, I believe, of a high educational value. For this reason, I have posted the title question on twitter (which was in its infancy in 2007) in the hope of collecting more insights as to why fraction addition works as it does.

I was not disappointed.

Problem

Without Definitions: Why 1/3 + 2/5 = 11/15

Solution 1

Start with $11$ doughnuts. Cut $5$ into $3$ parts each (part $A)$ and the remaining $6$ do nuts into $5$ parts each (part $B).$ Hand each of $15$ kids one part $A$ and two parts $B.$ Thus each of the $15$ kids gets the same amount and all $11$ do nuts are equitably exhausted.

Solution 2

Arrange $15$ tiles in a $3\times 5$ grid and color as shown below:

Without Definitions: Why 1/3 + 2/5 = 11/15, by Daren Chapin

Looking at columns, clearly $\displaystyle\frac{1}{3}$ of the tiles are blue (because each column contain just one blue tile), and at rows, clearly $\displaystyle\frac{2}{5}$ are red (because each row contains exactly two red tiles.) Consider all colored tiles together: $\displaystyle \frac{11}{15}$ are colored.

Solution 3

Without Definitions: Why 1/3 + 2/5 = 11/15, by Sergi Pino

Solution 4

This could be possibly construed as an illustration for Solution 1.

Without Definitions: Why 1/3 + 2/5 = 11/15, by Kunihiko Chikaya

Solution 5

Without Definitions: Why 1/3 + 2/5 = 11/15, by Jim Henegan

Solution 6

Without Definitions: Why 1/3 + 2/5 = 11/15, by Max Friedrich Hartmann

Acknowledgment

This is a follow-up on a much-much older page. Solution 1 is by Amit Itagi; Solution 2 is by Daren Chapin; Solution 3 is by Sergi Pino; Solution 4 is by Kunihiko Chikaya; Solution 5 is by Jim Henegan; Solution 6 is by Max Friedrich Hartmann.

 

Fractions

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

72004047