The group (Q*,o) where x o y = xy/7 G1 closure hold G2 Idenitiy - I have found the identity to be 7 I just need to prove G3 inversers and am not sure how to do this. Can anyone help? Many thanks. Arron
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We have $\displaystyle x\circ x^{-1}=e$ ($\displaystyle e$ is the unit), so $\displaystyle x\cdot x^{-1}/7=7$. Thus, $\displaystyle x^{-1}=49/x$.
Originally Posted by emakarov $\displaystyle x\cdot x^{-1}/7=7$. Thus, $\displaystyle x^{-1}=49/x$. What? $\displaystyle \displaystyle x \circ x^{-1} = e \implies \frac{xx^{-1}}{7} = 1$ doesn't it? -Dan
The OP has already established that the unit is 7.
No. The OP said that e=7 in his original post.
(Ahem!) Missed that. Thanks. -Dan
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