Prove that induction to all $n$ it is natural that $a_n$ is irrational and positive number. Given 
$a_1$=$\sqrt 2$  and  $\\a_{n+1}= 1-\dfrac{5}{4a_n+8}$.
Prove that induction to all $n$ it is natural that  $a_n$ is irrational and positive number.
I have an introduction with induction but i have not seen proof like this,i don't know how to start .
I'd love to see a proof of this .
 A: The set of rationals is a field: closed under multiplication, addition, subtraction, and division. Thus, if $a$ is irrational, then so must be $a+q$ where $q$ is rational (otherwise if $a+q$ were rational then the subtraction of two rational numbers i.e., $(a+q)-q =a$ would be irrational). Similarly, if $a$ is irrational, then so must $\frac{1}{a}$.
So, assume that $a_n$ is irrational (it clearly is for $n=1$ i.e.. $a_1= \sqrt{2}$ is irrational), and use the above to show that if $a_n$ is irrational, then so must be $a_{n+1}$.
This is a (pretty strong) hint which should allow you to finish the proof.
(@Henry already said this--cross-posting on my part)
A: It is easy to prove a somewhat more general statement, also by induction:
Suppose
$a_1 = s \in \Bbb R \setminus \Bbb Q, \; s > 0, \tag 1$
that is, $s$ is a positive irrational number; then the sequence defined by
$a_{n + 1} = 1 - \dfrac{5}{4a_n + 8} \tag 2$
is positive and irrational for all $n \in \Bbb N$, the natural numbers.
The base case of the induction lies in (1), since it affirms that $a_1$ is positive and irrational.  
So we assume
$0 < a_k \in \Bbb R \setminus \Bbb Q \tag 3$
for some $k \in \Bbb N$; we note the fact that
$p, q \in \Bbb Q, \; p \ne 0, \; x \in \Bbb R \setminus \Bbb Q \Longrightarrow px + q \in \Bbb R \setminus \Bbb Q, \tag 4$
for if not,
$px + q \in \Bbb Q \Longrightarrow px \in \Bbb Q \Longrightarrow x \in \Bbb Q, \tag 5$
contradicting the hypothesis of (4); thus,
$a_k \in \Bbb R \setminus \Bbb Q \Longrightarrow 4a_k + 8 \in \Bbb R \setminus \Bbb Q \Longrightarrow \dfrac{5}{4a_k + 8} \in \Bbb R \setminus \Bbb Q, \tag 6$
where in (6) we have used another arithmetic property of rationals and reals, viz., 
$0 \ne p \in \Bbb Q, \; x \in \Bbb R \setminus \Bbb Q \Longrightarrow \dfrac{p}{x} \in \Bbb R \setminus \Bbb Q, \tag 7$
for
$0 \ne \dfrac{p}{x} = q \in \Bbb Q \Longrightarrow 0 \ne qx = p \in \Bbb Q \Longrightarrow x = \dfrac{p}{q} \in \Bbb Q \Rightarrow \Leftarrow x \in \Bbb R \setminus \Bbb Q; \tag 8$
continuing then from (6) we have
$\dfrac{5}{4a_k + 8} \in \Bbb R \setminus \Bbb Q \Longrightarrow a_{k + 1} = 1 - \dfrac{5}{4a_k + 8} \in \Bbb R \setminus \Bbb Q, \tag 9$
essentially another application of (4) with $-p = q = 1$; also,
$0 < a_k \Longrightarrow 0 < 8 < 4a_k + 8 \Longrightarrow 0 < \dfrac{5}{4a_k + 8} < \dfrac{5}{8} < 1, \tag{10}$
whence
$0 < 1 - \dfrac{5}{8} < 1 - \dfrac{5}{4a_k + 8} = a_{k + 1}; \tag{11}$
combining (9) and (11) we see that 
$0 < a_{k + 1} \in \Bbb R \setminus \Bbb Q, \tag{12}$
and the inductionion is complete.
Therefore,
$0 < a_n \in \Bbb R \setminus \Bbb Q \tag{13}$
for all $n \in \Bbb N$, and the induction is complete.
