Proper ideal and units proof Show that an ideal $I$ of $R$ is proper if and only if it does not contain $1$ iff and only if it does not contain any units.
($1$ is the identity element)
I'll need to show 3 parts:
(1) $\implies$ (2): Let $I$ be a proper ideal and assume $1 \in I$. I want to show that $1 \notin I$.
(2) $\implies$ (3): Let $I$ be a proper ideal that doesn't contain $1$ and assume there are units in $R$. By the definition, units of $R$ are elements with multiplicative inverses. So I need to show that there aren't multiplicative inverses.
(3) $\implies$ (1): Assume there are no units in $R$ and then show that $I$ is a proper ideal. For this part I was thinking of doing something like this- Let $u \in I$ such that $u$ has a multiplicative inverse. Let $x \in R$. Then
\begin{align}u^{-1}x \in R;\\
x=u(u^{-1}x) \in I,\end{align}
so $R \subseteq I$ which implies that $I = R$. But this is a contradition so therefore $I$ must be a proper ideal.
I just need help completing the proof.
 A: I assume that you are working with rings that have an identity. Note that in that case there are always units in $R$, simply because the identity is a unit.
If $1\in I$ then $r=r1\in I$ for each $r\in R$ wich means that $I$ is not proper.
Conversely if $1\notin I$ then $I\neq R$ hence it is proper.
If $u\in I$ and $u$ is a unit then $r=r1=r(u^{-1}u)=(ru^{-1})u\in I$ for each $r\in R$ wich means that $I$ is not proper.
Conversely if $u\notin I$ for some unit $u$ then again $I\neq R$ hence it is proper.
A: Hint $\ $ Contrapositively, it is the special case $\,j = 1\,$ of the following
$$\,I\supseteq (j)\iff j\in I\iff k\in I\ \ {\rm for\ some}\ \ k\mid j$$
A: (1) $\implies$ (2):

Show that if $1 \in I$, then every element of $R$ is in $I$. (Use the fact that $I$ is closed under multiplication by elements of $R$.)

(2) $\implies$ (3):

 Show that if a unit $u$ is in $I$, then $1$ is also in $I$. (Again, use the fact that $I$ is closed under multiplication by elements of $R$.)

(3) $\implies$ (1):

 If $I$ does not contain units, then it does not contain $1$, so it must be proper.

A: You actually proved $(1) \Longrightarrow (3)$, because you assume $I$ is a proper ideal and get a contradiction by assuming it has a unit. You can't just write "Assume there are no units and let $u$ be a unit", that's already a contradiction, but you prove nothing.
There's a lot of negations in your statements so I suggest you restate your problem to show equivalence between:
$(1')\ I = R$.
$(2')\ 1 \in I$.
$(3')$ there's a unit $u \in I$.
You can see that $(1') \Longrightarrow (2')$ and $(2') \Longrightarrow (3')$ are trivial.
The only remaining thing is to prove $(3') \Longrightarrow (2')$, but that's exactly what you did, for any $x \in R$ it must be $(x u^{-1})u \in I \Longrightarrow x \in I$.
