If $p$ is a rational and $x$ is irrational, prove that $p+x$ and $px$ are irrational. If $p$ is a rational and $x$ is irrational, prove that $p+x$ and $px$ are irrational. 
Pf: Let $p$ be a rational such that $p= \dfrac{m}{n}$, $m, n \in \mathbb{Z}$ and $n \neq 0$ such that $m,n$ are not both even. An irrational number $x$, by definition, cannot be expressed in this form. 
$$p+x = \dfrac{m}{n}+x = \dfrac{m+xn}{n}$$
$$px = \dfrac{mx}{n}$$
We defined $m, n \in \mathbb{Z}$. We wish to show that $m+xn$ is not an integer. 
$+$ is an operation on $\mathbb{Z}$, which implies it must map pairs of elements of $\mathbb{Z}$ to an element of $\mathbb{Z}$. If  $nx = x+...+x = z$ ($n$ times) for some $z \in \mathbb{Z}$, $x$ must be an integer. In this case, $x$ is non integer so is $nx$. Therefore, $m+nx$ is also non-integer. This shows that $px$ and $p+x$ cannot be expresed in terms of division of two integers $\square$.
 A: Let $q:=p+x$, then $x=q-p$. If $q$ would have been rational then also $x$ is rational as difference of rational numbers and this is a contradiction.
Same for $r=px$ and we get $x=\frac{r}{p}$ (supposing clearly $p \neq 0$). A quotient of rational numbers is rational, so $r$ can't be rational.
A: We can show it is irrational using one of the most classic tricks for proving irrationality of a number.
Let us assume that $\, p+x$ is rational  , and is  of the form $\frac ab$.Then , 
$$p+x = \frac ab \implies \frac mn +x=\frac ab\implies \boxed{x = \frac ab - \frac mn}$$
Note that R.H.S is rational implying that $x$ is also rational . But this contradicts the fact the $x$ is actually irrational . Hence we conclude that $p+x$ is irrational.
You can similarly show $px$ is also  irrational .
A: 
If $nx=x+...+x=z$ ($n$ times) for some $z∈\mathbb Z$, x must be an integer.

This is incorrect. Consider $2\cdot\frac12=\frac12+\frac12=1$. This would actually tell us that $x=\frac{z}{n}$ and is rational. Personally, I think your proof is fine if you remedy this. But you must also take into account that $p\ne0$.

We defined $m, n \in \mathbb{Z}$. We wish to show that $m+xn$ is not an integer. 
  $+$ is an operation on $\mathbb{Z}$, which implies it must map pairs of elements of $\mathbb{Z}$ to an element of $\mathbb{Z}$. If  $nx$ $\color{red}{\text{is an integer and $n$ is nonzero}}$, $x$ must be $\color{red}{\text{rational}}$. In this case, $x$ is $\color{red}{\text{irrational}}$ so $\color{red}{\text{$nx$ cannot be an integer}}$. Therefore, $m+nx$ is also non-integer. This shows that ${\color{red}{\text{if $p\neq0$}\text{, then}}}$ $px$ and $p+x$ cannot be expressed in terms of division of two integers $\square$.

