If $\xi$ is irrational, $\xi+X$ is also irrational where X is rational number. I am stuck while reading book FOUNDATIONS OF ANALYSIS by EDMUND LANDAU. I can't understand that how the number $\xi+X$ is always irrational whenever $\xi$ is irrational and $X$ is rational. 
The book says:
Please how the auther proves that  $\xi+X$ is irrational.
Thank you.
 A: The proof you cite is a proof by contradiction.  I'm going to insert way more words of explanation  than you will encounter in this kind of proof:  
Premises:  $\xi$ is irrational and $X$ is rational.  Assume $\xi + X$ is rational, then $\xi + X = Y$ where $Y$ is rational.  By arithmetic in real numbers, $\xi = Y-X$.  But $Y-X$ is the difference of two rationals, hence also rational (have you seen the proof of this fact?).  This contradicts the premise that $\xi$ is irrational.  Therefore $\xi + X$ must be irrational.
Hope this helps!
A: 
Field axiom (A1) states $x, y \in F \rightarrow x + y \in F$. It was used here to add $(-r)$ to the rational on each side.
Field axiom (A5) states that $\forall x \in F, \exists -x \in F : x + (-x) = 0$. It was used here to show $r + -r = 0.$
Field axiom (M5) states $x \in F, x \neq 0 \rightarrow \exists \frac{1}{x} \in F : x(\frac{1}{x}) = 1$.} 
Field axiom (M4) states $F$ contains an element $1 \neq 0 : 1x = x$ for all $x \in F$.


If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove $r + x$ and $rx$ are irrational.
(Proof by contradiction) Suppose $r + x$ is rational. Then 
$$
\exists a, b \in \mathbb{Q} : \frac{a}{b} = r + x
$$
Since $\mathbb{Q}$ is an (ordered) field, the field axioms are applicable. First using field axiom (A1) and then using field axiom (A5), the above equation can be re-written as 
$$
\frac{a}{b} + (-r) = (-r) + r + x \rightarrow \frac{a}{b} + (-r) = x
$$
The absurdity is clear now: although the LHS is in $\mathbb{Q}$, the RHS isn't. RAA.
Similarly, suppose $rx$ is rational. Then
$$
\exists a, b \in \mathbb{Q} : \frac{a}{b} = rx
$$
which, by (M5) and (M4) should give us 
$$
(\frac{1}{r})ab = (\frac{1}{r})rx \rightarrow (\frac{1}{r})ab = 1x = x
$$
The absurdity is clear now, just as before. RAA.
A: Rational numbers are (at least) closed under "+", "-" and "*", which is easy to prove.
See Wiki.
That means, the sum, difference, and product of two rational numbers are also rational.
If $\xi+X$ were rational, then $\xi+X - X=\xi$ must also be rational, contradiction!
A: Suppose it were true. Then let $\xi=Y-X$ and $Y=p/q$, $X=r/s$, then we can rewrite the equation as:
$\xi=p/q-r/s=\frac{ps-qr}{qs}$, a rational number, which contradicts the claim that $\xi$ is irrational.
