Are There Infinitely Many Rational Numbers Between Two Irrational Numbers?

[Dustinsfl]

Show that there are infinitely many rational numbers between two different irrational numbers and vice versa.

So I started as such:
WLOG let $a,b$ be irrational numbers such that $a<b$. By theorem (not sure if there is a name for it), we know that there exist a rational number $x$ such that $a<x<b$.

If I can show there is another irrational between $a$ and $b$, I could then use the fact that between every two rational numbers there is a rational number and repeated applications would show that there are infinitely many.

 

[Sudharaka]

Show that there are infinitely many rational numbers between two different irrational numbers and vice versa.

So I started as such:
WLOG let $a,b$ be irrational numbers such that $a<b$. By theorem (not sure if there is a name for it), we know that there exist a rational number $x$ such that $a<x<b$.

If I can show there is another irrational between $a$ and $b$, I could then use the fact that between every two rational numbers there is a rational number and repeated applications would show that there are infinitely many.

Hi dwsmith, :)

I would argue it like this. Let \(a\) and \(b\) be two different irrational numbers and suppose that there are only a finite number of rational numbers in-between \(a\) and \(b\). So we have,

\[a<x_{0}<x_{1}<\cdots<x_{n}<b\]

where \(x_{0}<x_{1}<\cdots<x_{n}\) are rational numbers. Now there is no rational number between, \(a\) and \(x_{0}\) which leads to a contradiction since http://www.proofwiki.org/wiki/Between_Every_Two_Reals_Exists_a_Rational that between any two real numbers there exist a rational number.

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Let $a$ and $b$ be two different irrational numbers.
Suppose there are only a finite number of rational numbers between $a$ and $b$.
$$
a < x_1 < x_2 < x_3 < \cdots < x_n < b\quad\text{where} \ x_i\in\mathbb{Q}
$$
We know that between any two rational numbers there exists another rational number.
So
$$
a < x_1 < y_1 < x_2 < \cdots < y_{n - 1} < x_n < b
$$
where $y_i = \frac{x_i + x_{i + 1}}{2}\in\mathbb{Q}$.
We can continue this process indefinitely; furthermore, the rational numbers in $(a,b)\sim\mathbb{Z}^+$.
Hence, they cannot be finite.

 

[Sudharaka]

Let $a$ and $b$ be two different irrational numbers.
Suppose there are only a finite number of rational numbers between $a$ and $b$.
$$
a < x_1 < x_2 < x_3 < \cdots < x_n < b\quad\text{where} \ x_i\in\mathbb{Q}
$$
We know that between any two rational numbers there exists another rational number.
So
$$
a < x_1 < y_1 < x_2 < \cdots < y_{n - 1} < x_n < b
$$
where $y_i = \frac{x_i + x_{i + 1}}{2}\in\mathbb{Q}$.
We can continue this process indefinitely; furthermore, the rational numbers in $(a,b)\sim\mathbb{Z}^+$.
Hence, they cannot be finite.

Hi dwsmith, :)

When you assume "there are only a finite number of rational numbers between $a$ and $b$" there is also the possibility that you have only one rational number in between $a$ and $b$. I think this case is not covered in your proof.

Kind Regards,
Sudharaka.

 

[Dustinsfl]

What about saying that the rationals are dense in the reals so between (a,b) there are infinitely many rationals?

 

[Plato2]

Show that there are infinitely many rational numbers between two different irrational numbers and vice versa.

This is a ridiculous question.
The fundamental theorem is: Between any two numbers there is a rational number.
If $a<b$ then $\exists x_1\in\mathbb{Q}$ such that $a<x_1<b$.
Thus if $n>1$ then $\exists x_{n+1}\in\mathbb{Q}$ such that $x_n<x_{n+1}<b$

 

[Dustinsfl]

Dont shoot the messenger. I didn't create the question.