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Question Number 124192 by Algoritm last updated on 01/Dec/20

Commented by MJS_new last updated on 01/Dec/20

2

$$\mathrm{2} \\ $$

Commented by Algoritm last updated on 01/Dec/20

proof

$$\mathrm{proof} \\ $$

Answered by MJS_new last updated on 01/Dec/20

2<(√3)^(√3) <3 2<3^((√3)/2) <3 4<3^(√3) <9 4^(√3) <27<9^(√3) (3/2)<(√3)<2 ⇒ {_(9^(3/2) <9^(√3) <9^2 ⇔ 27<9^(√3) <81 ⇒ 27<9^(√3) ) ^(4^(3/2) <4^(√3) <4^2 ⇔ 8<4^(√3) <16 ⇒ 4^(√3) <27) ⇒ 2<(√3)^(√3) <3 ⇒ ⌊(√3)^(√3) ⌋=2

$$\mathrm{2}<\sqrt{\mathrm{3}}\:^{\sqrt{\mathrm{3}}} <\mathrm{3} \\ $$$$\mathrm{2}<\mathrm{3}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} <\mathrm{3} \\ $$$$\mathrm{4}<\mathrm{3}^{\sqrt{\mathrm{3}}} <\mathrm{9} \\ $$$$\mathrm{4}^{\sqrt{\mathrm{3}}} <\mathrm{27}<\mathrm{9}^{\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}<\sqrt{\mathrm{3}}<\mathrm{2}\:\Rightarrow\:\left\{_{\mathrm{9}^{\frac{\mathrm{3}}{\mathrm{2}}} <\mathrm{9}^{\sqrt{\mathrm{3}}} <\mathrm{9}^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{27}<\mathrm{9}^{\sqrt{\mathrm{3}}} <\mathrm{81}\:\Rightarrow\:\mathrm{27}<\mathrm{9}^{\sqrt{\mathrm{3}}} } ^{\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} <\mathrm{4}^{\sqrt{\mathrm{3}}} <\mathrm{4}^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{8}<\mathrm{4}^{\sqrt{\mathrm{3}}} <\mathrm{16}\:\Rightarrow\:\mathrm{4}^{\sqrt{\mathrm{3}}} <\mathrm{27}} \right. \\ $$$$\Rightarrow\:\mathrm{2}<\sqrt{\mathrm{3}}\:^{\sqrt{\mathrm{3}}} <\mathrm{3}\:\Rightarrow\:\lfloor\sqrt{\mathrm{3}}\:^{\sqrt{\mathrm{3}}} \rfloor=\mathrm{2} \\ $$

Commented by MJS_new last updated on 01/Dec/20

same procedure 6<3^(√3) <7 6^(√3) <27<7^(√3) ((17)/(10))<(√3)<((18)/(10)) ⇒ {_(7^((17)/(10)) <7^(√3) <7^((18)/(10)) ) ^(6^((17)/(10)) <6^(√3) <6^((18)/(10)) ) 6^((18)/(10)) <27 ⇔ 6^(18) <27^(10) ⇔ 2^(18) 3^(18) <3^(12) 3^(18) ⇔ ⇔ 2^(18) <3^(12) ⇔ 2(√2)<3 true 27<7^((17)/(10)) ⇔ 27^(10) <7^(17) ⇔ 205891132094649<232630513987207 true (sorry I found no better way) ⇒ 6^(√3) <6^((18)/(10)) <27<7^((17)/(10)) <7^(√3) ⇔ 6<3^(√3) <7

$$\mathrm{same}\:\mathrm{procedure} \\ $$$$\mathrm{6}<\mathrm{3}^{\sqrt{\mathrm{3}}} <\mathrm{7} \\ $$$$\mathrm{6}^{\sqrt{\mathrm{3}}} <\mathrm{27}<\mathrm{7}^{\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{17}}{\mathrm{10}}<\sqrt{\mathrm{3}}<\frac{\mathrm{18}}{\mathrm{10}}\:\Rightarrow\:\left\{_{\mathrm{7}^{\frac{\mathrm{17}}{\mathrm{10}}} <\mathrm{7}^{\sqrt{\mathrm{3}}} <\mathrm{7}^{\frac{\mathrm{18}}{\mathrm{10}}} } ^{\mathrm{6}^{\frac{\mathrm{17}}{\mathrm{10}}} <\mathrm{6}^{\sqrt{\mathrm{3}}} <\mathrm{6}^{\frac{\mathrm{18}}{\mathrm{10}}} } \right. \\ $$$$\mathrm{6}^{\frac{\mathrm{18}}{\mathrm{10}}} <\mathrm{27}\:\Leftrightarrow\:\mathrm{6}^{\mathrm{18}} <\mathrm{27}^{\mathrm{10}} \:\Leftrightarrow\:\mathrm{2}^{\mathrm{18}} \mathrm{3}^{\mathrm{18}} <\mathrm{3}^{\mathrm{12}} \mathrm{3}^{\mathrm{18}} \:\Leftrightarrow \\ $$$$\Leftrightarrow\:\mathrm{2}^{\mathrm{18}} <\mathrm{3}^{\mathrm{12}} \:\Leftrightarrow\:\mathrm{2}\sqrt{\mathrm{2}}<\mathrm{3}\:\mathrm{true} \\ $$$$\mathrm{27}<\mathrm{7}^{\frac{\mathrm{17}}{\mathrm{10}}} \:\Leftrightarrow\:\mathrm{27}^{\mathrm{10}} <\mathrm{7}^{\mathrm{17}} \:\Leftrightarrow\:\mathrm{205891132094649}<\mathrm{232630513987207} \\ $$$$\mathrm{true}\:\left(\mathrm{sorry}\:\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{better}\:\mathrm{way}\right) \\ $$$$\Rightarrow\:\mathrm{6}^{\sqrt{\mathrm{3}}} <\mathrm{6}^{\frac{\mathrm{18}}{\mathrm{10}}} <\mathrm{27}<\mathrm{7}^{\frac{\mathrm{17}}{\mathrm{10}}} <\mathrm{7}^{\sqrt{\mathrm{3}}} \\ $$$$\Leftrightarrow\:\mathrm{6}<\mathrm{3}^{\sqrt{\mathrm{3}}} <\mathrm{7} \\ $$

Commented by Algoritm last updated on 01/Dec/20

Brilliant

$$\mathrm{Brilliant} \\ $$

Commented by Algoritm last updated on 01/Dec/20

Commented by MJS_new last updated on 01/Dec/20

what do you mean? 3^(√3) ≈6.70499

$$\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}? \\ $$$$\mathrm{3}^{\sqrt{\mathrm{3}}} \approx\mathrm{6}.\mathrm{70499} \\ $$

Commented by Algoritm last updated on 01/Dec/20

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