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Question Number 5216 by Yozzii last updated on 01/May/16
Let p_j represent the j−th prime number. Now, define the number n whose decimal representation is written out in terms of p_j (j∈N) in the following way: n=0.p_1 p_2 p_3 p_4 p_5 ...p_j p_(j+1) p_(j+2) ... or n=0.(2)(3)(5)(7)(11)...(521)(523)(541)... ⇒n=0.235711...521523541... Prove or disprove that n is irrational.
$${Let}\:{p}_{{j}} \:{represent}\:{the}\:{j}−{th}\:{prime}\:{number}. \\ $$$${Now},\:{define}\:{the}\:{number}\:{n}\:{whose} \\ $$$${decimal}\:{representation}\:{is}\:{written}\:{out} \\ $$$${in}\:{terms}\:{of}\:{p}_{{j}} \:\left({j}\in\mathbb{N}\right)\:{in}\:{the}\:{following} \\ $$$${way}: \\ $$$${n}=\mathrm{0}.{p}_{\mathrm{1}} {p}_{\mathrm{2}} {p}_{\mathrm{3}} {p}_{\mathrm{4}} {p}_{\mathrm{5}} …{p}_{{j}} {p}_{{j}+\mathrm{1}} {p}_{{j}+\mathrm{2}} … \\ $$$${or}\:{n}=\mathrm{0}.\left(\mathrm{2}\right)\left(\mathrm{3}\right)\left(\mathrm{5}\right)\left(\mathrm{7}\right)\left(\mathrm{11}\right)…\left(\mathrm{521}\right)\left(\mathrm{523}\right)\left(\mathrm{541}\right)… \\ $$$$\Rightarrow{n}=\mathrm{0}.\mathrm{235711}…\mathrm{521523541}… \\ $$$${Prove}\:{or}\:{disprove}\:{that}\:{n}\:{is}\:{irrational}. \\ $$$$ \\ $$$$ \\ $$
Answered by FilupSmith last updated on 01/May/16
If we assume n is rational, ∴ n=(a/b), a,b∈Z (non reducible) p_j =jth prime ∴n=(p_1 /(10))+(p_2 /(100))+(p_3 /(1000))+...+(p_j /(10^j )) n=((100p_1 +10p_2 )/(1000))+(p_3 /(1000))+(p_4 /(10000))+... n=((100000p_1 +10000p_2 +1000p_3 )/(1000000))+(p_4 /(10000))+... n=((10^(j−1) p_1 +10^(j−2) p_2 +...)/(10^j )) usure if this is correct thus continuing we can always reduce the fraction. ∴ irrational ??? is this enough to prove??? i dont think so. Please check over this I am usure of the reliability of this proof. I am sure I could be wrong!
$$\mathrm{If}\:\mathrm{we}\:\mathrm{assume}\:{n}\:\mathrm{is}\:\mathrm{rational}, \\ $$$$\therefore\:{n}=\frac{{a}}{{b}},\:\:{a},{b}\in\mathbb{Z}\:\:\:\:\left(\mathrm{non}\:\mathrm{reducible}\right) \\ $$$${p}_{{j}} ={j}\mathrm{th}\:\mathrm{prime} \\ $$$$ \\ $$$$\therefore{n}=\frac{{p}_{\mathrm{1}} }{\mathrm{10}}+\frac{{p}_{\mathrm{2}} }{\mathrm{100}}+\frac{{p}_{\mathrm{3}} }{\mathrm{1000}}+…+\frac{{p}_{{j}} }{\mathrm{10}^{{j}} } \\ $$$${n}=\frac{\mathrm{100}{p}_{\mathrm{1}} +\mathrm{10}{p}_{\mathrm{2}} }{\mathrm{1000}}+\frac{{p}_{\mathrm{3}} }{\mathrm{1000}}+\frac{{p}_{\mathrm{4}} }{\mathrm{10000}}+… \\ $$$${n}=\frac{\mathrm{100000}{p}_{\mathrm{1}} +\mathrm{10000}{p}_{\mathrm{2}} +\mathrm{1000}{p}_{\mathrm{3}} }{\mathrm{1000000}}+\frac{{p}_{\mathrm{4}} }{\mathrm{10000}}+… \\ $$$${n}=\frac{\mathrm{10}^{{j}−\mathrm{1}} {p}_{\mathrm{1}} +\mathrm{10}^{{j}−\mathrm{2}} {p}_{\mathrm{2}} +…}{\mathrm{10}^{{j}} }\:\:\:\:\:\:\:{usure}\:{if}\:{this}\:{is}\:{correct} \\ $$$$\mathrm{thus}\:\mathrm{continuing}\:\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{reduce} \\ $$$$\mathrm{the}\:\mathrm{fraction}. \\ $$$$ \\ $$$$\therefore\:\mathrm{irrational}\:\:\:\:\:\:\:???\:\mathrm{is}\:\mathrm{this}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{prove}??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{i}\:\mathrm{dont}\:\mathrm{think}\:\mathrm{so}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{check}\:\mathrm{over}\:\mathrm{this}\:\mathrm{I}\:\mathrm{am}\:\mathrm{usure}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{reliability}\:\mathrm{of}\:\mathrm{this}\:\mathrm{proof}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{sure}\:\mathrm{I}\: \\ $$$$\mathrm{could}\:\mathrm{be}\:\mathrm{wrong}! \\ $$
Commented by Rasheed Soomro last updated on 03/May/16
If p_(i ) consist of n digits then each digit has its own denominator, not (p_i /(10^i )) . For example place value of 11 in given number is not equal to ((11)/(10^5 )) but is equal to (1/(10^5 ))+(1/(10^6 )) .
$${If}\:{p}_{{i}\:} {consist}\:{of}\:{n}\:{digits}\:{then}\:{each}\:{digit} \\ $$$${has}\:{its}\:{own}\:{denominator},\:{not}\:\frac{{p}_{{i}} }{\mathrm{10}^{{i}} }\:. \\ $$$${For}\:{example}\:{place}\:{value}\:{of}\:\mathrm{11}\:{in}\:{given}\:{number} \\ $$$${is}\:{not}\:{equal}\:{to}\:\frac{\mathrm{11}}{\mathrm{10}^{\mathrm{5}} }\:{but}\:{is}\:{equal}\:{to}\:\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{6}} }\:. \\ $$
Commented by FilupSmith last updated on 04/May/16
ahh I see. My mistake
$$\mathrm{ahh}\:\mathrm{I}\:\mathrm{see}.\:\mathrm{My}\:\mathrm{mistake} \\ $$

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