• ChrisS
Determine whether the given map $\phi$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? $<\mathbb{Q},*>\ with\ <\mathbb{Q},*>\ where\ \phi(x)=x^{2}\ for\ x \in \mathbb{Q}$ I have the one-to-one requirement $If\ \phi(x)=\phi(y) then\ x^{2}=y^{2}$ $Then\ \sqrt{x^{2}}=\sqrt{y^{2}}$ So x=y Thus the map is one-to-one. Now I need to show that it is onto and I need to show $\phi(x*y)=\phi(x)*\phi(y)$
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
• ChrisS
Determine whether the given map $\phi$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? $<\mathbb{Q},*>\ with\ <\mathbb{Q},*>\ where\ \phi(x)=x^{2}\ for\ x \in \mathbb{Q}$ I have the one-to-one requirement $If\ \phi(x)=\phi(y) then\ x^{2}=y^{2}$ $Then\ \sqrt{x^{2}}=\sqrt{y^{2}}$ So x=y Thus the map is one-to-one. Now I need to show that it is onto and I need to show $\phi(x*y)=\phi(x)*\phi(y)$
Mathematics

Looking for something else?

Not the answer you are looking for? Search for more explanations.