Modulo 7

A week or two ago I noticed some students in the math lounge solving these type of equations $2x=3\mod{7}$ and $x^2=-3\mod{7}$. Abstract algebra is above my pay grade but I spent a couple of lunch hours exploring equations $\mod{7}$.

If we solve $2x=3\mod{7}$, we get $5$ so $5 = \frac{3}{2}\mod{7}$. I built this table of all possible fractions $\mod{7}$.

Fractions Modulo 7

This is simply a rearrangement of the multiplication table $\mod{7}$.  It  shows that all rational numbers can be put into seven equivalence classes not just the integers.  That’s pretty cool!

Some of these equations have no solution, so let’s extend the field like we do with complex numbers where we define $i$ to be a solution to $x^2+1=0$.  Let one solution to $x^2-3=0$ be $\sqrt{3}$.  Then the other solution would be $-\sqrt{3} = -1\sqrt{3}=6\sqrt{3}\mod{7}$.  The solutions to $x^2-5=0$ would be $2\sqrt{3}$ and $5\sqrt{3}$ and the solutions to $x^2-6=0$ would be $3\sqrt{3}$ and $4\sqrt{3}$.  Also cool.