Answer to Question #93911 in Abstract Algebra for Gboyega

Question

If p*q = p^2-q^2-2pq. Find the inverse of p under the operation.

Answer 1

By definition, the inverse element is

$$p*p^{-1}=p^{-1}*p=e$$

where $e$ is an identity element.

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

Then,

$$p*q=p^2-q^2-2pq=e$$

We will consider this as an equation for a variable $q$ .

$$q^2+2pq+(e-p^2)=0\\ D=(2p)^2-4\cdot(e-p^2)=4p^2+4p^2-4e=8p^2-4e\\ q_1=\frac{-2p+\sqrt{8p^2-4e}}{2}=-p+\sqrt{2p^2-e}\\ q_2=\frac{-2p-\sqrt{8p^2-4e}}{2}=-p-\sqrt{2p^2-e}$$

Check result

$$p*q_1=p^2-q_1^2-2pq_1=\\ =p^2-\left(\sqrt{2p^2-e}-p\right)^2-2p\left(\sqrt{2p^2-e}-p\right)=\\ =p^2-\left(2p^2-e-2p\sqrt{2p^2-e}+p^2\right)+2p^2-2p\sqrt{2p^2-e}=\\ =p^2-3p^2+e+2p\sqrt{2p^2-e}+2p^2-2p\sqrt{2p^2-e}=e$$

Conclusion,

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

ANSWER

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