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Question Number 112583 by mohammad17 last updated on 08/Sep/20

Commented by Aziztisffola last updated on 08/Sep/20

$$1.\mathrm{let}\sqrt{7}=\frac{\mathrm{a}}{\mathrm{b}}\mathrm{with}\gcd (\mathrm{a};\mathrm{b})=1(\mathrm{a}\mathrm{\&}\mathrm{b}\in \mathbb{Z}\mathrm{b}\ne 0)$$$$\Rightarrow 7=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}\Rightarrow {\mathrm{a}}^2={7\mathrm{b}}^2\Rightarrow 7\mid {\mathrm{a}}^2\stackrel{7\mathrm{is}\mathrm{prime}}{\Rightarrow }7\mid \mathrm{a}$$$$\Rightarrow \mathrm{a}=7\mathrm{k}(\mathrm{k}\in \mathbb{Z})$$$$\Rightarrow {\left(7\mathrm{k}\right)}^2={7\mathrm{b}}^2\Rightarrow {\mathrm{b}}^2={7\mathrm{k}}^2\Rightarrow 7\mid {\mathrm{b}}^2$$$$\Rightarrow 7\mid \mathrm{b}$$$$\mathrm{then}7\mid \mathrm{a}\mathrm{\&}7\mid \mathrm{b}\Rightarrow \gcd (\mathrm{a};\mathrm{b})\ne 1$$$$\Rightarrow \mathrm{contradiction}$$$$\mathrm{hence}\sqrt{7}\mathrm{is}\mathrm{irrational}.\sqrt{7}\notin \mathbb{Q}$$

Commented by mohammad17 last updated on 08/Sep/20

$$thankyousircanyouhelpmeinotherquestion$$

Commented by Aziztisffola last updated on 08/Sep/20

$$2.]0;1[\subset \mathbb{R}$$$$\mathrm{Suppose}\mathrm{we}\mathrm{can}\mathrm{list}\mathrm{all}\mathrm{number}\mathrm{in}]0;1[$$$${\mathrm{x}}_1=0.{\mathrm{x}}_{11}{\mathrm{x}}_{12}{\mathrm{x}}_{13}...$$$${\mathrm{x}}_2=0.{\mathrm{x}}_{21}{\mathrm{x}}_{22}{\mathrm{x}}_{23}...$$$${\mathrm{x}}_3=0.{\mathrm{x}}_{31}{\mathrm{x}}_{32}{\mathrm{x}}_{33}...$$$$.....$$$${\mathrm{x}}_{\mathrm{ij}}=0,...9$$$$\mathrm{Consider}\mathrm{a}=0.{\mathrm{x}}_{11}{\mathrm{x}}_{22}{\mathrm{x}}_{33}....$$$$\mathrm{construct}\mathrm{the}\mathrm{number}\mathrm{x}=0.{\mathrm{a}}_1{\mathrm{a}}_2{\mathrm{a}}_3...$$$$\mathrm{with}{\mathrm{a}}_{\mathrm{i}}\ne {\mathrm{x}}_{\mathrm{ii}}(\mathrm{and}{\mathrm{a}}_{\mathrm{i}}\ne 9)$$$$\mathrm{then}\mathrm{x}\ne {\mathrm{x}}_1;\mathrm{x}\ne {\mathrm{x}}_2....$$$$\mathrm{then}\mathrm{x}\mathrm{is}\mathrm{not}\mathrm{in}\mathrm{the}\mathrm{list}$$$$\mathrm{hence}\mathbb{R}\mathrm{is}\mathrm{uncountable}.$$

Answered by mathmax by abdo last updated on 09/Sep/20

$$\mathrm{nature}\mathrm{of}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{\sin \left(\mathrm{n}\right)}{\mathrm{n}(\mathrm{n}+1)}\mathrm{we}\mathrm{have}\mid \frac{\sin \left(\mathrm{n}\right)}{\mathrm{n}(\mathrm{n}+1)}\mid ⩽\frac{1}{\mathrm{n}(\mathrm{n}+1)}⩽\frac{1}{{\mathrm{n}}^2}$$$$\mathrm{the}\mathrm{serie}\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^2}\mathrm{converges}\Rightarrow \mathrm{this}\mathrm{serie}\mathrm{converges}$$$$\mathrm{nsture}\mathrm{of}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+\sqrt{\mathrm{n}}}\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}+\sqrt{\mathrm{x}}}\mathrm{with}\mathrm{x}⩾1\mathrm{we}\mathrm{hsve}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-\frac{1+\frac{1}{2\sqrt{\mathrm{x}}}}{{(\mathrm{x}+\sqrt{\mathrm{x}})}^2}<0\Rightarrow \mathrm{f}\mathrm{is}\mathrm{decreazing}\mathrm{so}\mathrm{this}\mathrm{serie}\mathrm{hsve}\mathrm{nature}$$$$\mathrm{of}{\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}}}\mathrm{changement}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{give}$$$${\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}}}={\int }_1^{+\mathrm{\infty }}\frac{2\mathrm{tdt}}{{\mathrm{t}}^2+\mathrm{t}}=2{\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{t}+1}={2\lim }_{\mathrm{a}\to +\mathrm{\infty }}{\int }_1^{\mathrm{a}}\frac{\mathrm{dt}}{\mathrm{t}+1}$$$$={2\lim }_{\mathrm{a}\to +\mathrm{\infty }}{\left[\ln (\mathrm{t}+1)\right]}_1^{\mathrm{a}}={2\lim }_{\mathrm{a}\to +\mathrm{\infty }}\ln (\mathrm{a}+1)-\ln 2=+\mathrm{\infty }\Rightarrow$$$$\mathrm{this}\mathrm{serie}\mathrm{is}\mathrm{divergent}$$

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