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-

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} \dots p_{j} p_{j + 1} p_{j + 2} \dots

o r n = 0 . (2) (3) (5) (7) (11) \dots (521) (523) (541) \dots

\Rightarrow n = 0.235711 \dots 521523541 \dots

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 = \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} + \dots + \frac{p_{j}}{10^{j}}

n = \frac{100 p_{1} + 10 p_{2}}{1000} + \frac{p_{3}}{1000} + \frac{p_{4}}{10000} + \dots

n = \frac{100000 p_{1} + 10000 p_{2} + 1000 p_{3}}{1000000} + \frac{p_{4}}{10000} + \dots

n = \frac{10^{j - 1} p_{1} + 10^{j - 2} p_{2} + \dots}{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