For a real number x, let f(x) be a function such that f(x)=x−⌊x⌋; and let A(x) be a set such that A(x)={f(xn):n∈N}.
Prove the following:
(1) x is rational if and only if A(x) is finite;
(2) x is irrational if and only if A(x) is dense in the interval [0,1).
Have fun :)
#Algebra
#Sets
#IrrationalNumbers
#Decimals
#FloorFunction
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Suppose A(x) is finite. Then by the Pigeonhole Principle, there are two distinct integers m and n such that f(xm)=f(xn). So xm−xn=k is an integer. Then x=m−nk is rational.
Suppose A(x) is dense in [0,1). Then it is certainly infinite. So x can't be rational, by (1).
What remains is to show that if x is irrational, then A(x) is dense in [0,1).
Log in to reply
You can do this by the Pigeonhole Principle as well:
http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset
We see that f(x)={x}, i.e. it is the fractional part of x.
(1) ⟶ In this direction we assume that x is rational and try to prove A(x) is finite. Since x is rational, we can write it as x=qp where p and q are coprime integers and q is not zero.
Claim: A(x) will have exactly q elements.
The elements in A(x) will be {qp},{q2p},{q3p},{q4p},….
Since p and q are integers, these terms can be written as qp mod q,q2p mod q,q3p mod q,q4p mod q,…. There can be only q such terms, viz q0,q1,q2,…qq−1. Since q is a finite number, A(x) also has a finite number of elements.