1. ## Abstract Algebra

Show that $\displaystyle (\mathbb{R}, \; +)$ is not isomorphic to $\displaystyle (\mathbb{R}^*, \; \cdot)$, where $\displaystyle \mathbb{R}^*$ is the group of units in $\displaystyle \mathbb{R}$. i.e. $\displaystyle \mathbb{R}^* = \mathbb{R}\backslash \{0\}$

2. Suppose that $\displaystyle (\mathbf{R},+)$ is isomorphic to $\displaystyle (\mathbf{R}^*,\cdot)$

Let $\displaystyle f\mathbf{R},+)\to (\mathbf{R}^*.\cdot)$ an isomorphism.

Then $\displaystyle f(x+y)=f(x)f(y), \ \forall x,y\in\mathbf{R}$

For $\displaystyle y=x\Rightarrow f(2x)=(f(x))^2\Rightarrow f(x)=\left(f\left(\frac{x}{2}\right)\right)^2\Righ tarrow f(x)\geq 0, \ \forall x\in\mathbf{R}$

Then f is not a surjection.