### Look Before You Leap

Relate these algebraic expressions to geometrical diagrams.

### Modular Fractions

We only need 7 numbers for modulus (or clock) arithmetic mod 7 including working with fractions. Explore how to divide numbers and write fractions in modulus arithemtic.

### Stretching Fractions: A Discussion

This article extends and investigates the ideas in the problem "Stretching Fractions".

# Can it Be

##### Stage: 5 Challenge Level:

Often people who don't know how to add fractions do so as if the rule is $${a\over b} + {c\over d} = {a+c\over b+d}$$ where $a$ and $b$ are coprime and $c$ and $d$ are coprime. Does this ever give the right answer?

As Andrei Lazanu from Tudor Vianu National College, Bucharest, Romania says "The only solution for the problem is $a = -c$ and $b = d$ so that the sum $${a\over b} + {c\over d}$$ is zero and $${a+c\over b+d}=0,$$ so the relation is true".

Why is this the only possibility? The relation
$${a\over b} + {c\over d} = {a+c\over b+d}$$ holds if and only if $${ad+bc \over bd} = {a+c \over b+d}$$ that is $$(ad+bc)(b+d) = (a+c)bd$$ which holds if and only if $$ad^2 + b^2c = 0,$$ or equivalently $$ad^2=-b^2c.$$ We know that any whole number can be written as the unique product of prime factors. As $a$ and $b$ are coprime and $c$ and $d$ are coprime $ad^2=-b^2c$ is true only if $a$ divides $c$ and $c$ divides $a$ and they are of opposite sign, that is $a = -c$. Thus the original formula holds if and only if $d = \pm b$ and $b+d \neq 0$ that is $b=d$ so the formula holds if and only if ${a\over b} = {-c\over d}$ and their sum is zero.