prove-that-0-101101110-irrational-number-

Question Number 4670 by malwaan last updated on 20/Feb/16

p r o v e t h a t

0.101101110 \to

i r r a t i o n a l n u m b e r

Commented by prakash jain last updated on 21/Feb/16

Non - recurring non - terminating decimals

are irrational .

Commented by malwaan last updated on 21/Feb/16

I s t h e r e m a t h e m a t i c a l p r o v e ?

Commented by 123456 last updated on 21/Feb/16

any proof would be like this, call

s = 0.101101110 \dots .

you must shown that no

(p, q) \in Z^{2}, q \neq 0, (p, q) = 1 such that

s = \frac{p}{q}

Commented by FilupSmith last updated on 22/Feb/16

a_{1} = 0.1 = \frac{1}{10}

a_{2} = 0.0011 = \frac{11}{10^{}}

a_{3} = 0.00000111 = \frac{111}{10^{8}}

a_{4} = 0.0000000001111 = \frac{1111}{10^{13}}

a_{5} = 0.0000000000000011111 = \frac{11111}{10^{19}}

a_{6} = 0.00000000000000000000111111 = \frac{111111}{10^{26}}

⋮

a_{n} = \frac{(\underset{n - ones}{111 \dots 1})}{10^{k}}, k_{n} = 0, 1, 4, 8, 13, 19, 26, \dots

According to Wolfram Alpha

k = - \frac{{(n - 2)}^{2} n}{{(n - 1)}^{3}}

A = \underset{i = 1}{\sum^{\infty}} a_{i}

A = 0.101101110 = \frac{1}{10} + \frac{11}{10^{4}} + \frac{111}{10^{8}} + \dots

Prove RHS is irrational

Commented by prakash jain last updated on 22/Feb/16

A number of \frac{p}{q} (p, q) is either a teminating

decimal or repeatitive decimal .

This can be proved easily .

So by contradiction the number in example

cannot be written in form \frac{p}{q} .