Irrationality question. (2 Viewers)

[seanieg89]

"You can find your phone number in any irrational number"

Prove this statement or provide a counter-example.

 

[seanieg89]

Related: Prove or disprove that a real number is rational if and only if its decimal expansion eventually consists of a repeating string.

 

[farnsworth]

"You can find your phone number in any irrational number"

.

Seriously, who cares?

 

[lolcakes52]

The moderators do? and I don't know how to approach this but it reminds me of this comic

 

[deswa1]

The first statement isn't true. A counterexample is Liouville's constant: http://mathworld.wolfram.com/LiouvillesConstant.html

 

[Carrotsticks]

Related: Prove or disprove that a real number is rational if and only if its decimal expansion eventually consists of a repeating string.

Suppose we have a real number R with a repeating string of a FINITE number of decimals (infinite string would imply irrationality, though I'm sure proof is needed). We denote the digits of this string to be b_1, b_2, b_3, ..., b_{m-n} such that the number (we don't know if rational or not yet) is in the form:



Where [R] is the integer component of R, a_1 + a_2 + ... + a_n is the finite decimal component leading up to the repeating string b_1 + b_2 + ... + b_{m-n} (the bar on top denotes its repetition).

For simplicity, we will call this repeating string B_{m-n} and the finite decimal component A_n such that R decomposes to become:



Even if R < 0, we can still do the exact same things, except replace all the + with - (equivalent to multiplying both sides by -1), so we will only consider the positive case.

[R] is an integer by definition, so we need not consider it.

A is a finite string of RATIONAL terms by definition again.

B consists of a string of repeating digits, which we will define to be b (ie for B = 0.123123..., b = 123)such that



And we recognise this to be a GP with the ratio |r| < 1, implying that it is convergent and using the limiting sum formula, we acquire a rational term ie: B is rational.

So now we have R = [R] + A + B, where A is rational and B is rational and since the field of Rational Numbers (Q) is closed under addition (is an integer considered to be rational?), we conclude that R is rational.

I probably could have communicated this in a clearer fashion but essentially, my argument is as follows:

1. The repeating string of digits form a GP with |r| < 1 such that we can use the Limiting Sum formula.

2. This formula yields a rational number.

3. Rational + Rational --> Rational since Q is closed under addition.

4. Therefore the number is rational.

 

[seanieg89]

Yep, for finite repeating strings a GP approach will work or the old (10^k)x-1 trick to obtain a number with finitely many nonzero digits to the right of it's decimal point.

I think the harder/more tedious thing is showing that this is a NECESSARY condition for rationality. I can think of a way of doing it using long division and induction, but it's not particularly pretty.

As a corollary of this result, we get that Liouville's constant is irrational and hence resolve the first question in the negative.

 

[Shadowdude]

iirc, in Discrete Maths my tutor used the pigeonhole principle to prove that a number is rational if it has a decimal expansion that repeats

 

[Carrotsticks]

iirc, in Discrete Maths my tutor used the pigeonhole principle to prove that a number is rational if it has a decimal expansion that repeats

Yep!

Suppose we have a rational number p/q where p,q are integers and p E [1,q) and q > 1.

Let r_0 be the first remainder from the long division process such that:



So we can clearly see that a_1, a_2 , ... , a_q are the decimal expansion digits of the division of p/q.

So we know that the remainders r_0, r_1, r_2, ..., r_q exist in [0,q-1]

Using the Pigeonhole Principle, there exists some pair r_m and r_n such that:



And so we can conclude that: