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### Problem 1

Determine whether the given maps $\phi$ determine an isomorphism of the first binary structure with the second.If it isn't an isomorphism, why not?

(a)$\langle\mathbb{Z},+\rangle$ with $\langle\mathbb{Z},+\rangle$ where $\phi(n)=3n+4$.

(b) $\langle \mathbb {R}, + \rangle$ with $\langle \mathbb {R}^+ , * \rangle$ where $\phi (r) = 0.5 ^r$ for $r \in \mathbb {R}$.

(c) $\langle F, + \rangle$ with $\langle F, + \rangle$ where $F$ is the set of all functions $f: \mathbb {R} \rightarrow \mathbb {R}$ that have derivatives of all orders and $\phi (f) = f'$, the derivative of $f$.

**Solution**

(a)

**Step1** The obvious function $\phi:\mathbb{Z}\rightarrow\mathbb{Z}$ to try is given by $\phi(n)=3n+4$ for $n\in\mathbb{Z}$.

**Step2** Suppose $\phi(m)=\phi(n)$, then $3m+4=3n+4$. So $m=n$ .Thus $\phi$ is one to one.

**Step3** If $n\in\mathbb{Z}$, then $n=3m+4$ for $m=\frac{n-4}{3}$ which is not in $\mathbb{Z}$ absolutely. So $\phi$ is not onto $\mathbb{Z}$.

**Step4** Let $m,n\in\mathbb{Z}$. The equation $\phi(m+n)=3(m+n)+4=3m+3n+4\neq\phi(m)+\phi(n)$.

So $\phi$ is not an isomorphism.

(b)

Suppose $\phi (x) = \phi (y)$, so $0.5^x = 0.5^y$, then $xln|0.5| = y ln|0.5|$ and $x=y$. So, $\phi$ is one-to-one.

If $r \in \mathbb {R}^+$, then $log_{0.5} (r) \in \mathbb {R}$, $\phi (log_{0.5} (r)) = 0.5^{log_{0.5} r} = r$. Thus $\phi$ is onto.

For $x,y \in \mathbb {R}$ we have $\phi (x+y) = 0.5^{x+y}=0.5^x 0.5^y= \phi (x) \phi (y)$. Thus we see that $\phi$ is indeed an isomorphism.

(c)

This is not an isomorphism, since it is not one-to-one. $\phi (2x+1) = 2$ and $\phi (2x) = 2$.