Answer to Question #93911 in Abstract Algebra for Gboyega

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


"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\\\\\nD=(2p)^2-4\\cdot(e-p^2)=4p^2+4p^2-4e=8p^2-4e\\\\\nq_1=\\frac{-2p+\\sqrt{8p^2-4e}}{2}=-p+\\sqrt{2p^2-e}\\\\\nq_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=\\\\\n=p^2-\\left(\\sqrt{2p^2-e}-p\\right)^2-2p\\left(\\sqrt{2p^2-e}-p\\right)=\\\\\n=p^2-\\left(2p^2-e-2p\\sqrt{2p^2-e}+p^2\\right)+2p^2-2p\\sqrt{2p^2-e}=\\\\\n=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"


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