solve the equation $8x+7 = 5$ in the field $(\mathbb{Z}_{11}, +, \cdot)$ solution $$8x+7=5$$ $$8x=5+(-7)$$ $$8x=-2$$ $$8x=9$$
since 7 is the multiplicative inverse of 8, multiply both sides by 7
$$x=63=8$$
the solution is $$x=8$$
$$8 \cdot 8 +7 = 71 =5$$
im having alot of trouble understanding finite fields, can someone please explain how $8x=-2$ turned into $8x=9$? and i dont understand why $8 \cdot 8+7=71=5$
In the world of $\mathbb{Z}_{11}$, $11$ is equivalent to $0$.
$$8x = -2 = 0-2 =11-2 = 9$$
$$71 = 6(11)+5=6(0)+5=5$$
More formally,
$$8x \equiv -2 \equiv 0-2 \equiv 11-2 \equiv 9 \pmod{11}$$
$$71 \equiv 6(11)+5 \equiv6(0)+5 \equiv5 \pmod{11}$$
We say two numbers are equivalent $\mod{11}$ if $a-b$ is divisible by $11$.