MHB Are There Infinitely Many Rational Numbers Between Two Irrational Numbers?

Dustinsfl
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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.
 
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dwsmith said:
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.
 
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.
 
dwsmith said:
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.
 
What about saying that the rationals are dense in the reals so between (a,b) there are infinitely many rationals?
 
dwsmith said:
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 $x_n<x_{n+1}<b$
 
Dont shoot the messenger. I didn't create the question.
 
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