Prove by Contraposition: $~\forall ~r~ \in ~\mathbb Q~$, $~s~ \in ~\mathbb R~$, if $~s~$ is irrational then $~r + \frac{1}{s}~$ is irrational Need help proving that: $~\forall ~r~ \in ~\mathbb Q~$, $~s~ \in ~\mathbb R~$,  if $~s~$ is irrational then $~r + \frac{1}{s}~$ is irrational
The contrapositive is: $~\forall ~r~ \in \mathbb Q~$, $~s~ \in \mathbb R~$,  if $~r + \frac{1}{s}~$ is rational then $~s~$ is rational
I don't seem to be getting anywhere with it.
 A: Notice that for some choice of $p,q \in \mathbb{Z}$ we have $r + \frac{1}{s} = \frac{p}{q}$, hence 
\begin{align}
\frac{1}{s} = \frac{p}{q} - r
\end{align}
Verify that the sum of two rational numbers is also a rational number itself. Then, $\frac{1}{s}$ is rational, so $s$ is rational. (The nonzero rational number $a/b$ has multiplicative inverse $b/a$.)
REMARK: It is not appropriate when trying to prove the statement to assume $r/s$ is rational. Why? Put $r/s = t$. Then, $r/t = s$. See then that $s$ is rational if and only if $r/s$ is rational, thus assuming $r/s$ is rational is like assuming that we already have the desired result.
If $r = 0$, the statement says that the multiplicative inverse of a rational number is also rational.
A: $\Bbb Q$ is closed under $+,-, \times,$ and reciprocals. "Closed under" means that if $r,r'\in \Bbb Q$ then $r'+r,\,r'-r,\,rr' \in \Bbb Q$ and if $0\ne x\in \Bbb Q$ then $x^{-1}\in \Bbb Q.$
Suppose $r\in \Bbb Q$ and $s\not \in \Bbb Q$ and  $r+\frac {1}{s}\in \Bbb Q.$
Let $r'=r+\frac {1}{s}.$   Since $r,r'\in \Bbb Q$ we have 
$$r+\frac {1}{s}=r'\implies \frac {1}{s}=r'-r\in \Bbb Q$$ $$\implies \frac {1}{s}\in \Bbb Q$$ $$\implies 0\ne\frac {1}{s}\in \Bbb Q \quad\text {(because no $\frac {1}{s}$ can be $0$)}$$ $$\implies s=\left(\frac {1}{s}\right)^{-1}\in \Bbb Q$$ contrary to the assumption that $s\not \in \Bbb Q.$
A: If $r+\frac{1}{s}=r'$ would be rational, then $\frac{1}{s}=r'-r$ would be rational and so $s= \frac{1}{r'-r}$ would be rational. Contradiction.
Note that the rational numbers form a field and so closed under addition and multiplication.
Here if $r,r'$ are rational, then $r'-r = r'+(-r)$ and $\frac{1}{r} = r^{-1}$ are rational.
