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Question Number 216572 by Samuel12 last updated on 10/Feb/25

	Answered by maths2 last updated on 11/Feb/25

	$s i α \notin I \Rightarrow α = \frac{a}{b} \in I Q b a = 1 o n v a m o n t r e r Q u e b ∣ a$ $\Rightarrow a^{n} - \underset{k = 0}{\sum^{n - 1}} c_{k} a^{k} b^{n - k} = 0$ $\Rightarrow \underset{k = 0}{\sum^{n - 1}} c_{k} a^{k} b^{n - k} = a^{n}; b ∣ \underset{k = 0}{\sum^{n - 1}} c_{k} a^{k} b^{n - k}; c a r n - k ⩾ 1; \forall k \in [0, n - 1]$ $\Rightarrow b ∣ a^{n} c o m m e a e t b s o n t p r e m i e r \Rightarrow b ∣ a \Rightarrow α \in Z$ $p o u r c o n c l u r e R = I Q \cup I$ $α \in R \Leftrightarrow (α \in I \lor (α \in I Q \Rightarrow α \in Z))$ $2$ $\sqrt{2} + \sqrt{3} = α \Rightarrow α^{2} = 5 + 2 \sqrt{6} \Rightarrow {(a^{2} - 5)}^{2} - 24 = 0$ $a^{4} - 10 a^{2} + 1 = 0; a e s t [r a c i n e d e P (x) = x^{4} - 10 x^{2} + 1$ $u t i l i s a n t l e l e m m e$ $s u f f i t d e p r o u v e r Q u e e s t s a n R a c i n e d e Z$ $a^{4} - 10 a^{2} = - 1 \Rightarrow a^{2} (a^{2} - 10) = - 1 \Rightarrow a^{2} = 1 & a^{2} - 10 = - 1$ $a^{2} = 1 & a^{2} = 9 a b s u r d e$ $α \in I$ $\sqrt{2} + \sqrt{3} + \sqrt{5} = α$ ${(a - \sqrt{5})}^{2} = 5 + 2 \sqrt{6}$ $a^{2} - 2 a \sqrt{5} = 2 \sqrt{6}$ $a^{4} = (24 + 20 a^{2}) + 4 a \sqrt{5}$ ${(a^{4} - 20 a^{2} - 24)}^{2} - 80 a^{2} = 0$ $m o d (5) \Rightarrow a^{4} - 4 \equiv 0 [5]$ $a \equiv 0, \underline{+} 1; \underline{+} 2 [5] \Rightarrow a^{4} \equiv 0, 1; 16 [5] a b s u r d$ $a \notin Z \Rightarrow a \in I$
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