isomorphism swapping zero and unity

Let  $(R,+,\cdot )$  be a ring with unity 1.  Define two new binary operations of $R$ as follows:

$a\oplus b=:a+b-1,a\odot b=:a+b-a\cdot b$(1)

Then we see that

$a\oplus 1=a=\mathrm{\hspace{0.33em}1}\oplus a,a\odot 0=a=\mathrm{\hspace{0.33em}0}\odot a.$(2)

But moreover, the algebraic system  $(R,\oplus ,\odot )$  is a unitary ring, too, and isomorphic with the original ring.

In fact, we may define the bijective mapping

$f:x\mapsto \mathrm{\hspace{0.17em}1}-x$(3)

from $R$ to $R$ and verify that it is homomorphic:

$$f(a)\oplus f(b)=(1-a)\oplus (1-b)=(1-a)+(1-b)-1=\mathrm{\hspace{0.33em}1}-a-b=f(a+b),$$

$$f(a)\odot f(b)=(1-a)\odot (1-b)=(1-a)+(1-b)-(1-a)\cdot (1-b)=\mathrm{\hspace{0.33em}1}-a\cdot b=f(a\cdot b)$$

Thus  $(R,\oplus ,\odot )$  as a homomorphic image (http://planetmath.org/HomomorphicImageOfGroup) of the ring  $(R,+,\cdot )$  is a ring, it’s a question of two isomorphic rings.