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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

$$\mathrm{a}=\left\{\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}}\right);\mathrm{k}\in\mathrm{Z}\right\} \\ $$$$\mathrm{let}\:\mathrm{a}_{\mathrm{n}} \in\mathrm{a} \\ $$$$\mathrm{a}_{\mathrm{2n}} =\mathrm{1}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}}\right)\rightarrow\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{2n}+\mathrm{1}} =−\mathrm{1}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right)\rightarrow−\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{a}=\left\{−\mathrm{1},\mathrm{1}\right\} \\ $$$$\mathrm{b}:\mathrm{Z}=\mathrm{Z} \\ $$$$\mathrm{let}\:\mathrm{a}\:\mathrm{bee}\:\mathrm{acumulstion}\:\mathrm{point}\:\mathrm{of}\:\mathrm{Z} \\ $$$$\Rightarrow\mathrm{a}\in\mathrm{adh}\left(\mathrm{Z}−\left\{\mathrm{a}\right\}\right)\Rightarrow\mathrm{a}\in\mathrm{adh}\left(\mathrm{Z}\right)=\mathrm{Z} \\ $$$$\Rightarrow\mathrm{a}\in\mathrm{Z}\:\mathrm{but}\:\mathrm{adh}\left(\mathrm{Z}−\left\{\mathrm{a}\right\}\right)=\mathrm{Z}−\left\{\mathrm{a}\right\}\Rightarrow\mathrm{a}\in\mathrm{Z}−\left\{\mathrm{a}\right\}\:\mathrm{absurd} \\ $$$$\mathrm{b}=\left\{\right\}\:\mathrm{empty}\:\mathrm{set} \\ $$$$\mathrm{we}\:\mathrm{will}\:\mathrm{zhow}\:\mathrm{that}\:\mathrm{is}\:\mathbb{R} \\ $$$$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\forall\mathrm{n}\in\mathrm{IN}\:\:\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \in\mathrm{IR}−\left\{\mathrm{Q}\right\} \\ $$$$\mathrm{suppose}\:\mathrm{inver}\Rightarrow\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} =\frac{\mathrm{a}}{\mathrm{b}}\Rightarrow\sqrt{\mathrm{2}}=\frac{\mathrm{a}^{\mathrm{n}} }{\mathrm{b}^{\mathrm{n}} }\in\mathrm{Q}\:\mathrm{aburde}\: \\ $$$$\mathrm{u}_{\mathrm{n}} \:\mathrm{s}.\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} .\mathrm{n}\in\mathrm{IN}^{\ast} .\in\mathrm{IR}−\left\{\mathrm{Q}\right\}\:,\mathrm{s}\in\mathrm{Q} \\ $$$$\mathrm{u}_{\mathrm{n}} \rightarrow\mathrm{s}\in\mathrm{Q}\Rightarrow\mathrm{for}\:\mathrm{C}=\mathrm{IR}−\left\{\mathrm{Q}\right\}\cup\mathrm{Q}=\mathrm{IR} \\ $$$$\mathrm{for}\:\mathrm{this}\:\mathrm{we}\:\mathrm{must}\:\mathrm{show}\:\mathrm{that}\:\forall\mathrm{x}\in\mathrm{IR}\:\exists\mathrm{u}_{\mathrm{n}} \in\mathrm{IR}−\left\{\mathrm{Q}\cup\left\{\mathrm{x}\right\}\right\} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{u}_{\mathrm{n}} \rightarrow\mathrm{x} \\ $$$$\mathrm{if}\:\mathrm{x}\in\mathrm{Q} \\ $$$$\mathrm{just}\:\mathrm{put}\:\mathrm{u}_{\mathrm{n}} =\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} .\mathrm{x}\rightarrow\mathrm{x} \\ $$$$\mathrm{and}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \mathrm{un}\notin\mathrm{IR}−\left\{\mathrm{Q}\right\} \\ $$$$\mathrm{x}\notin\mathrm{Q}\Rightarrow\mathrm{x}\in\mathrm{IR}−\left\{\mathrm{Q}\right\}\:\mathrm{mor}\:\mathrm{difficult} \\ $$$$\mathrm{u}_{\mathrm{n}} =\begin{cases}{\mathrm{x}−\mathrm{1}+\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} ,\mathrm{if}\:\mathrm{x}−\mathrm{1}+\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \notin\mathrm{Q}}\\{\mathrm{x}−\mathrm{1}+\left(\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\mathrm{else}}\end{cases} \\ $$$$\mathrm{suppose}\:\exists_{\mathrm{n}} \mathrm{for}\:\mathrm{witch}\:\mathrm{x}−\mathrm{1}+\left(\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} =\frac{\mathrm{a}}{\mathrm{b}}\Rightarrow \\ $$$$\mathrm{oand}\:\mathrm{x}−\mathrm{1}+\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} =\frac{\mathrm{c}}{\mathrm{d}}\Rightarrow\frac{\mathrm{a}}{\mathrm{b}}−\left(\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} +\left(\sqrt{\mathrm{2}}\right)_{} ^{\frac{\mathrm{1}}{\mathrm{n}}} =\frac{\mathrm{c}}{\mathrm{d}} \\ $$$$\Rightarrow\exists_{\mathrm{n}} \in\mathbb{N}^{\ast} \mathrm{such}\:\mathrm{that}\:\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} −\left(\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \in\mathrm{Q} \\ $$$$\mathrm{absurd}\:\mathrm{U}_{\mathrm{n}\:} \mathrm{well}\:\mathrm{defind}\:\rightarrow\mathrm{x} \\ $$

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