Prove some properties of ring homomorphism 
Let $R,S$ be rings and $\varphi : R\to S$ be a ring homomorphism. Verify that

*

*$\varphi(na) = n\varphi(a)$ for all $n\in\mathbb Z$ and $a\in R$.

*$\varphi(a^n) = (\varphi(a))^n$ for all $n\in\mathbb Z^+$ and all $a\in R$.

*If $A$ is a subring of $R$, then $\varphi(A) = \{\varphi(a):a\in A\}$ is a subring of $S$.


For (1) I have the following:
$$
\begin{split}
\varphi(na) &= \varphi((n-1)a+a) \\
&= \varphi((n-1)a)+\varphi(a) \\
&= (n-1)\varphi(a)+\varphi(a)\\
&= n\varphi(a).
\end{split}
$$
For (2): For $n=1$, we have $\varphi(a) = \varphi(a)$. Suppose for some $n\in\mathbb Z^+$ that $\varphi(a^n)=(\varphi(a))^n$. Observe:
$$
\begin{split}
\varphi(a^n) & =(\varphi(a))^n \\
\varphi(a)\cdot \varphi(a^n) & =\varphi(a)\cdot (\varphi(a))^n \\
\varphi(a \cdot a^n) & =(\varphi(a))^{n+1} \\
\varphi( a^{n+1}) & =(\varphi(a))^{n+1} \\
\end{split}
$$
For (3): I verify for $a,b\in\varphi(A)$ then $a-b\in\varphi(A)$ and $a\cdot b\in \varphi(A)$ (for brevity).

*

*Well, $0\in A \implies 0 \in \varphi(A) $

*For $a,b \in \varphi(A)$ we have $\varphi(a) - \varphi(b) = \varphi(a-b)$, thus, verified, because $a-b\in A$.

*For $a,b \in \varphi(A)$, we have $\varphi(a) \cdot \varphi(b) = \varphi (a\cdot b)$, thus, verified, because $a\cdot b \in A$.

Did I do these right? Feedback appreciated!
 A: For the first property,
use induction to show that $\phi(na) = n\phi(a)$ for all $a$ and $n\geq 0$.
This is actually what you have done.
For negative $n$ one must be careful.
For $n>0$ put
$(-n)a = -(na)$, which is the additive inverse of $na$.
This part is then true simply by definition.
A: I agree with Wuestenfux's answer, and have some feedback to give. There is some error in your answer to part $(1)$, which is similar to one in part $(2)$ as well. How do you know that $\phi((n-1)a)=(n-1)\phi(a)$? For $(2)$ you suppose that there exists some $n\in\mathbb{Z}^+$ satisfies $\phi(a^n) = (\phi(a))^n$, but it could be the case that
$\{n\in\mathbb{Z}^+|\phi(a^n) = (\phi(a))^n\}$ is empty. For both $(1)$ and $(2)$, one can use induction after showing the properties hold with base case $n=2$.
For problem $(3)$, remember that if $b = \phi(a)$ for some $a\in A$ then $b\in \phi(A)$. It is only once we write an element $b$ in the form $\phi(a)$ that the containment $b\in \phi(A)$ becomes clear. For example, given $0\in A$, we know by properties of ring homomorphisms that
$$\phi(0) =0.$$
Hence $0 = \phi(0) \in \phi(A)$.
"For $a,b\in \phi(A)$, we have $\phi(a) - \phi(b) = \phi(a-b)$." This does not show that $a-b \in \phi(A)$, because $a \neq \phi(a)$ and $b\neq \phi(a)$. Again, to say that $a\in \phi(A)$ means there exists some $x\in A$ for which $\phi(x) = a$.
