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.

Look Before You Leap

Relate these algebraic expressions to geometrical diagrams.

All Tangled Up

Can you tangle yourself up and reach any fraction?

Can it Be

Age 16 to 18 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.