We assume that the Binary Operation #ast# is a function,
# ast : ZZxxZZ to ZZ : a ast b=a+b-7, AAa,b in ZZ.#
Now, to find the inverse of #2# w.r.t. #ast#, we have to first find the
Identity Element , say, #e# for #ast#.
By the Defn. of #e#, then, #aaste=easta=a, AA a.#
#aaste=a rArr a+e-7=a rArr e=7.#
Also, #7 ast a=a rArr 7+a-7=a.#
So, #7# is the Identity for #ast#.
Now, suppose that, #x# is an Inverse of #2" under "ast#.
Then, by Defn. of an inverse, we must have,
#x ast 2=2 ast x=e=7.#
#x ast 2=7 rArr x+2-7=7 rArr x=12.#
We also have, #2 ast 12=2+12-7=7=e.#
Thus, #12 ast 2= 2 ast 12=e(=7).#
Therefore, the Inverse of #2#, denoted by, #2^-1#, under the
operation #ast# is #12 in ZZ.#
N.B.: In fact, it is easy to see that, # AA a in ZZ, a^-1=14-a.#
Enjoy Maths.!
