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.


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Question Number 5216 by Yozzii last updated on 01/May/16

$L e t p_{j} r e p r e s e n t t h e j - t h p r i m e n u m b e r .$ $N o w, d e f i n e t h e n u m b e r n w h o s e$ $d e c i m a l r e p r e s e n t a t i o n i s w r i t t e n o u t$ $i n t e r m s o f p_{j} (j \in N) i n t h e f o l l o w i n g$ $w a y :$ $n = 0 . p_{1} p_{2} p_{3} p_{4} p_{5} . . . p_{j} p_{j + 1} p_{j + 2} . . .$ $o r n = 0 . (2) (3) (5) (7) (11) . . . (521) (523) (541) . . .$ $\Rightarrow n = 0.235711 . . . 521523541 . . .$ $P r o v e o r d i s p r o v e t h a t n i s i r r a t i o n a l .$
	Answered by FilupSmith last updated on 01/May/16
	$If we assume n is rational, ∴ n=(a/b), a,b∈Z (non reducible) p_j =jth prime ∴n=(p_1 /(10))+(p_2 /(100))+(p_3 /(1000))+...+(p_j /(10^j )) n=((100p_1 +10p_2 )/(1000))+(p_3 /(1000))+(p_4 /(10000))+... n=((100000p_1 +10000p_2 +1000p_3 )/(1000000))+(p_4 /(10000))+... n=((10^(j−1) p_1 +10^(j−2) p_2 +...)/(10^j )) usure if this is correct thus continuing we can always reduce the fraction. ∴ irrational ??? is this enough to prove??? i dont think so. Please check over this I am usure of the reliability of this proof. I am sure I could be wrong!$
	$If we assume n is rational,$ $∴ n = \frac{a}{b}, a, b \in Z (non reducible)$ $p_{j} = j th prime$ $∴ n = \frac{p_{1}}{10} + \frac{p_{2}}{100} + \frac{p_{3}}{1000} + . . . + \frac{p_{j}}{10^{j}}$ $n = \frac{100 p_{1} + 10 p_{2}}{1000} + \frac{p_{3}}{1000} + \frac{p_{4}}{10000} + . . .$ $n = \frac{100000 p_{1} + 10000 p_{2} + 1000 p_{3}}{1000000} + \frac{p_{4}}{10000} + . . .$ $n = \frac{10^{j - 1} p_{1} + 10^{j - 2} p_{2} + . . .}{10^{j}} u s u r e i f t h i s i s c o r r e c t$ $thus continuing we can always reduce$ $the fraction .$ $∴ irrational ? ? ? is this enough to prove ? ? ?$ $i dont think so .$ $Please check over this I am usure of the$ $reliability of this proof . I am sure I$ $could be wrong!$
	Commented by Rasheed Soomro last updated on 03/May/16

	$I f p_{i} c o n s i s t o f n d i g i t s t h e n e a c h d i g i t$ $h a s i t s o w n d e n o m i n a t o r, n o t \frac{p_{i}}{10^{i}} .$ $F o r e x a m p l e p l a c e v a l u e o f 11 i n g i v e n n u m b e r$ $i s n o t e q u a l t o \frac{11}{10^{5}} b u t i s e q u a l t o \frac{1}{10^{5}} + \frac{1}{10^{6}} .$
	Commented by FilupSmith last updated on 04/May/16

	$ahh I see . My mistake$
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