Boolean ring is commutative

The Mathematician — Wed, 28 Sep 2022 13:00:00 +0000

Show that a Boolean ring is commutative

By definition, a ring

R

is Boolean if

x^2 = x

\forall x \in R

.

Proof. We need to show that

xy = yx

for all

x,y \in R

. So first, we have:

\begin{align*} (x + y)^2 = (x + y) &\iff x^2 + xy + yx + y^2 = x + y \\ &\iff x + xy + yx + y = x + y \\ &\iff xy + yx = 0 \end{align*}

Now we have

xy = -yx

. We would like to prove that

-yx = yx

. We can check that by finding its inverse:

\begin{align*} (y + y) &= (y + y)^2 \\ &= y^2 + 2y + y^2 \\ &= y + y + y + y \\ &= 0 \end{align*}

which implies that

y + y = 0

. Now we get

-y = y

and therefore we have that

xy = -yx = yx

, which implies that Boolean ring

R

is indeed a commutative ring.

boolean ring is commutative Archives - Epsilonify