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Question Number 141997 by jlewis last updated on 25/May/21

find two irrational numbers between 0.333.... and 0.444...

$$\mathrm{find}\:\mathrm{two}\:\mathrm{irrational}\:\mathrm{numbers}\: \\ $$$$\mathrm{between}\:\mathrm{0}.\mathrm{333}....\:\mathrm{and}\:\mathrm{0}.\mathrm{444}... \\ $$$$ \\ $$

Answered by MJS_new last updated on 25/May/21

.33312345678910111213... (1/e)≈.367879441...

$$.\mathrm{33312345678910111213}... \\ $$$$\frac{\mathrm{1}}{\mathrm{e}}\approx.\mathrm{367879441}... \\ $$

Commented by MJS_new last updated on 25/May/21

.333000333300003333300000...

$$.\mathrm{333000333300003333300000}... \\ $$

Commented by jlewis last updated on 25/May/21

wah! i have not gotten how u arrived to this

$$\mathrm{wah}!\:\mathrm{i}\:\mathrm{have}\:\mathrm{not}\:\mathrm{gotten}\:\mathrm{how}\:\mathrm{u} \\ $$$$\:\mathrm{arrived}\:\mathrm{to}\:\mathrm{this} \\ $$

Commented by MJS_new last updated on 25/May/21

easy: invent an infinite non−periodic pattern .01 001 0001 00001 000001... .0 1 2 3 4 5 6 7 8 9 10 11 12... .1 4 9 16 25 36 49 64 81 100... .2 4 8 16 32 64 128 256 512... .0 1 1 2 3 5 8 13 21 34 55 89...

$$\mathrm{easy}:\:\mathrm{invent}\:\mathrm{an}\:\mathrm{infinite}\:\mathrm{non}−\mathrm{periodic}\:\mathrm{pattern} \\ $$$$.\mathrm{01}\:\mathrm{001}\:\mathrm{0001}\:\mathrm{00001}\:\mathrm{000001}... \\ $$$$.\mathrm{0}\:\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}\:\mathrm{8}\:\mathrm{9}\:\mathrm{10}\:\mathrm{11}\:\mathrm{12}... \\ $$$$.\mathrm{1}\:\mathrm{4}\:\mathrm{9}\:\mathrm{16}\:\mathrm{25}\:\mathrm{36}\:\mathrm{49}\:\mathrm{64}\:\mathrm{81}\:\mathrm{100}... \\ $$$$.\mathrm{2}\:\mathrm{4}\:\mathrm{8}\:\mathrm{16}\:\mathrm{32}\:\mathrm{64}\:\mathrm{128}\:\mathrm{256}\:\mathrm{512}... \\ $$$$.\mathrm{0}\:\mathrm{1}\:\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{5}\:\mathrm{8}\:\mathrm{13}\:\mathrm{21}\:\mathrm{34}\:\mathrm{55}\:\mathrm{89}... \\ $$

Commented by jlewis last updated on 25/May/21

i get it now.thanks

$$\:\mathrm{i}\:\mathrm{get}\:\mathrm{it}\:\mathrm{now}.\mathrm{thanks} \\ $$

Answered by MJS_new last updated on 25/May/21

you could also do this: .333...=(3/9) .444...=(4/9) ((((3/9))^2 ×(4/9)))^(1/3) =((4/(81)))^(1/3) ≈.366880805...∉Q (√((3/9)×(4/9)))=((2(√3))/9)≈.384900179...∉Q (((3/9)×((4/9))^3 ))^(1/4) =(((192))^(1/4) /9)≈.413602159...∉Q

$$\mathrm{you}\:\mathrm{could}\:\mathrm{also}\:\mathrm{do}\:\mathrm{this}: \\ $$$$.\mathrm{333}...=\frac{\mathrm{3}}{\mathrm{9}} \\ $$$$.\mathrm{444}...=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{3}}{\mathrm{9}}\right)^{\mathrm{2}} ×\frac{\mathrm{4}}{\mathrm{9}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{81}}}\approx.\mathrm{366880805}...\notin\mathbb{Q} \\ $$$$\sqrt{\frac{\mathrm{3}}{\mathrm{9}}×\frac{\mathrm{4}}{\mathrm{9}}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\approx.\mathrm{384900179}...\notin\mathbb{Q} \\ $$$$\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}}{\mathrm{9}}×\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{3}} }=\frac{\sqrt[{\mathrm{4}}]{\mathrm{192}}}{\mathrm{9}}\approx.\mathrm{413602159}...\notin\mathbb{Q} \\ $$

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