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.
Age
16 to 18
Challenge level
Image
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In arithmetic modulo 7 ($Z_7$) one integer is equal to another if the difference between the two integers is a multiple of 7. Rather like the days of the week, in $Z_7$ we only need seven numbers and they are usually named 0, 1, 2, 3, 4, 5 and 6. |
If there is a solution in $Z_7$ to the equation $ax=1$ then we
call this solution the inverse (or reciprocal) of $a$ and write it
as $a^{-1}$ or ${1\over a}$. For example the fraction one half in
arithmetic modulo 7 is the inverse of 2, that is the solution of
$2x=1 \pmod 7$, namely the number 4 because $2\times 4 = 1 \pmod
7$.
What are the fractions one third, one quarter, one fifth and
one sixth in arithmetic modulo 7?
Explain why all fractions in arithmetic modulo 7 are
equivalent to one of the following set of numbers $\{0, 1, 2, 3, 4,
5, 6\}$.
Show that in $Z_7$ there are six different solutions to the
equation
$${1\over x} + {1\over y} = {1\over {x+y}}.$$
Show that, by way of contrast, when working with real numbers
this equation has no real solutions.
$$3^{-1}+6^{-1} = 5 + 6 = 11 = 4 \pmod 7$$
and
$$(3+6)^{-1} = 9^{-1} = 2^{-1} = 4 \pmod 7$$
so $x=3,\ y=6$ is one solution. Now find the other solutions.
Congratulations to Wei Zhang, age 18 from Merchiston Castle School for the superb solution given below. Andrei Lazanu, age 14, School No. 205, Bucharest, Romania submitted a similar solution. \par Suppose there is a fraction $1/n$ , where $n$ belongs to $\{1,2,3,4,5,6\}$.
$$\eqalign{ 1/1: 1x &=1\ {\rm so}\ x=1 \cr 1/2: 2x &=1\ {\rm so}\ x=4 \cr 1/3: 3x &=1\ {\rm so}\ x=5 \cr 1/4: 4x &=1\ {\rm so}\ x=2 \cr 1/5: 5x &=1\ {\rm so}\ x=3 \cr 1/6: 6x &=1\ {\rm so}\ x=6 }$$
As I showed above, $1/n$ is in the set $\{1,2,3,4,5,6\}$.
Now we have $a/n = a \times 1/n$ , where $a$ belongs to $\{0,1,2,3,4,5,6\}$, this is a number from the set $\{0,1,2,3,4,5,6\}$ times a number from the set of $\{1,2,3,4,5,6\}$. Therefore we will get a set of numbers for $a/n$ , which is $\{0,1,2,3,4,5,6\}$. Solving the equation
$${1\over x} + {1\over y}= {1\over{x+y}}$$
is equivalent to solving $x^2+xy+y^2=0$.
| when | $x=1$ | $y^2+y+1=0$ | so | $y^2+y=6$ | $y=2$ | or | $y=4$ |
| when | $x=2$ | $y^2+2y+4=0$ | so | $y^2+2y=3$ | $y=1$ | or | $y=4$ |
| when | $x=3$ | $y^2+3y+2=0$ | so | $y^2+3y=5$ | $y=6$ | or | $y=5$ |
| when | $x=4$ | $y^2+4y+2=0$ | so | $y^2+4y=5$ | $y=1$ | or | $y=2$ |
| when | $x=5$ | $y^2+5y+4=0$ | so | $y^2+5y=3$ | $y=3$ | or | $y=6$ |
| when | $x=6$ | $y^2+6y+1=0$ | so | $y^2+6y=6$ | $y=3$ | or | $y=5$ |
We have found the six solution pairs $(1,2)$, $(1,4)$, $(2,4)$, $(3,6)$, $(3,5)$, $(5,6)$. The equation is symmetric in $x$ and $y$ so there are six corresponding solutions when we exchange $x$ and $y$.
We don't allow negative solutions (e.g. $x=1,\ y=-3$) because we are working entirely in the set $\{0,1,2,3,4,5,6\}$.
If we work with real numbers, we think of solving the quadratic equation $x^2+yx+y^2=0$ as an equation in $x$. The formula for the solution of this quadratic equation involves $y^2-4y^2=-3y^2< 0$ which has no real values, therefore there is no real solution for $x$.