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Question Number 71944 by aliesam last updated on 22/Oct/19

	Answered by mind is power last updated on 23/Oct/19
	$a={(−1)^k (1−(1/k));k∈Z} let a_n ∈a a_(2n) =1(1−(1/(2n)))→1 a_(2n+1) =−1(1−(1/(2n+1)))→−1 for a={−1,1} b:Z=Z let a bee acumulstion point of Z ⇒a∈adh(Z−{a})⇒a∈adh(Z)=Z ⇒a∈Z but adh(Z−{a})=Z−{a}⇒a∈Z−{a} absurd b={} empty set we will zhow that is R we see that ∀n∈IN ((√2))^(1/n) ∈IR−{Q} suppose inver⇒((√2))^(1/n) =(a/b)⇒(√2)=(a^n /b^n )∈Q aburde u_n s.((√2))^(1/n) .n∈IN^∗ .∈IR−{Q} ,s∈Q u_n →s∈Q⇒for C=IR−{Q}∪Q=IR for this we must show that ∀x∈IR ∃u_n ∈IR−{Q∪{x}} such that u_n →x if x∈Q just put u_n =((√2))^(1/n) .x→x and ∀n∈N^∗ un∉IR−{Q} x∉Q⇒x∈IR−{Q} mor difficult u_n = { ((x−1+((√2))^(1/n) ,if x−1+((√2))^(1/n) ∉Q)),((x−1+((√3))^(1/n) else)) :} suppose ∃_n for witch x−1+((√3))^(1/n) =(a/b)⇒ oand x−1+((√2))^(1/n) =(c/d)⇒(a/b)−((√3))^(1/n) +((√2))_ ^(1/n) =(c/d) ⇒∃_n ∈N^∗ such that ((√2))^(1/n) −((√3))^(1/n) ∈Q absurd U_(n ) well defind →x$
	$a = {{(- 1)}^{k} (1 - \frac{1}{k}); k \in Z}$ $let a_{n} \in a$ $a_{2 n} = 1 (1 - \frac{1}{2 n}) \to 1$ $a_{2 n + 1} = - 1 (1 - \frac{1}{2 n + 1}) \to - 1$ $for a = {- 1, 1}$ $b : Z = Z$ $let a bee acumulstion point of Z$ $\Rightarrow a \in adh (Z - {a}) \Rightarrow a \in adh (Z) = Z$ $\Rightarrow a \in Z but adh (Z - {a}) = Z - {a} \Rightarrow a \in Z - {a} absurd$ $b = {} empty set$ $we will zhow that is R$ $we see that \forall n \in IN {(\sqrt{2})}^{\frac{1}{n}} \in IR - {Q}$ $suppose inver \Rightarrow {(\sqrt{2})}^{\frac{1}{n}} = \frac{a}{b} \Rightarrow \sqrt{2} = \frac{a^{n}}{b^{n}} \in Q aburde$ $u_{n} s . {(\sqrt{2})}^{\frac{1}{n}} . n \in {IN}^{} . \in IR - {Q}, s \in Q$ $u_{n} \to s \in Q \Rightarrow for C = IR - {Q} \cup Q = IR$ $for this we must show that \forall x \in IR \exists u_{n} \in IR - {Q \cup {x}}$ $such that u_{n} \to x$ $if x \in Q$ $just put u_{n} = {(\sqrt{2})}^{\frac{1}{n}} . x \to x$ $and \forall n \in N^{} un \notin IR - {Q}$ $x \notin Q \Rightarrow x \in IR - {Q} mor difficult$ $u_{n} = {\begin{cases} x - 1 + {(\sqrt{2})}^{\frac{1}{n}}, if x - 1 + {(\sqrt{2})}^{\frac{1}{n}} \notin Q \\ x - 1 + {(\sqrt{3})}^{\frac{1}{n}} else \end{cases}$ $suppose \exists_{n} for witch x - 1 + {(\sqrt{3})}^{\frac{1}{n}} = \frac{a}{b} \Rightarrow$ $oand x - 1 + {(\sqrt{2})}^{\frac{1}{n}} = \frac{c}{d} \Rightarrow \frac{a}{b} - {(\sqrt{3})}^{\frac{1}{n}} + {(\sqrt{2})}_{}^{\frac{1}{n}} = \frac{c}{d}$ $\Rightarrow \exists_{n} \in N^{*} such that {(\sqrt{2})}^{\frac{1}{n}} - {(\sqrt{3})}^{\frac{1}{n}} \in Q$ $absurd U_{n} well defind \to x$
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