Question #93911
If p*q = p^2-q^2-2pq. Find the inverse of p under the operation.
1
Expert's answer
2019-09-13T10:20:52-0400

By definition, the inverse element is


pp1=p1p=ep*p^{-1}=p^{-1}*p=e

where ee is an identity element.

For the convenience of the solution, we will denote p1qp^{-1}\equiv q .

Then,


pq=p2q22pq=ep*q=p^2-q^2-2pq=e

We will consider this as an equation for a variable qq .


q2+2pq+(ep2)=0D=(2p)24(ep2)=4p2+4p24e=8p24eq1=2p+8p24e2=p+2p2eq2=2p8p24e2=p2p2eq^2+2pq+(e-p^2)=0\\ D=(2p)^2-4\cdot(e-p^2)=4p^2+4p^2-4e=8p^2-4e\\ q_1=\frac{-2p+\sqrt{8p^2-4e}}{2}=-p+\sqrt{2p^2-e}\\ q_2=\frac{-2p-\sqrt{8p^2-4e}}{2}=-p-\sqrt{2p^2-e}

Check result


pq1=p2q122pq1==p2(2p2ep)22p(2p2ep)==p2(2p2e2p2p2e+p2)+2p22p2p2e==p23p2+e+2p2p2e+2p22p2p2e=ep*q_1=p^2-q_1^2-2pq_1=\\ =p^2-\left(\sqrt{2p^2-e}-p\right)^2-2p\left(\sqrt{2p^2-e}-p\right)=\\ =p^2-\left(2p^2-e-2p\sqrt{2p^2-e}+p^2\right)+2p^2-2p\sqrt{2p^2-e}=\\ =p^2-3p^2+e+2p\sqrt{2p^2-e}+2p^2-2p\sqrt{2p^2-e}=e

Conclusion,



pp1=2p2ep\boxed{p\longrightarrow p^{-1}=\sqrt{2p^2-e}-p}

ANSWER



pp1=2p2epp\longrightarrow p^{-1}=\sqrt{2p^2-e}-p


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