# Math Help - Rational and Irrational numbers

1. ## Rational and Irrational numbers

Hi, I am learning about rational and irrational numbers and I came across a question that I am trying to get my head around,

m is a irrational number
n is a rational number which is greater than m

Is it possible to always find a rational number between m and n?

Thanks

2. Originally Posted by studentme
Hi, I am learning about rational and irrational numbers and I came across a question that I am trying to get my head around,

m is a irrational number
n is a rational number which is greater than m

Is it possible to always find a rational number between m and n?

Thanks
That is harder than one would think. Are you sure it didn't say "find an irrational number between the two"?

3. Hi, thanks for the quick reply. I re-read the question and it says:

"Is it possible to always find a rational number between m and n?"

4. Originally Posted by studentme
Hi, thanks for the quick reply. I re-read the question and it says:

"Is it possible to always find a rational number between m and n?"
I think this problem falls under Real Analysis rather than Pre-Algebra and Algebra, but I think the answer to the question is yes.

My reasoning: from Wikipedia's article on continued fractions:

"The sequence of rational numbers given by any infinite continued fraction converges to an irrational number, which limit is taken to be the value of the continued fraction. Moreover, every irrational number α is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values α and 1."

Also, further down on the page there is discussion of best rational approximations using either a truncated continued fraction, or else a related continued fraction involving a half-rule (see here).

Any rate, it follows that rational approximations can get arbitrarily close to irrational numbers, and that there is always a better approximation than any given one, hence it would be between m and n.

5. Originally Posted by undefined
I think this problem falls under Real Analysis rather than Pre-Algebra and Algebra, but I think the answer to the question is yes.

My reasoning: from Wikipedia's article on continued fractions:

"The sequence of rational numbers given by any infinite continued fraction converges to an irrational number, which limit is taken to be the value of the continued fraction. Moreover, every irrational number α is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values α and 1."

Also, further down on the page there is discussion of best rational approximations using either a truncated continued fraction, or else a related continued fraction involving a half-rule (see here).

Any rate, it follows that rational approximations can get arbitrarily close to irrational numbers, and that there is always a better approximation than any given one, hence it would be between m and n.
Look up "rationals dense in the reals"

6. Originally Posted by Drexel28
Look up "rationals dense in the reals"
Thanks, I see that the OP's question is a special case of the more general result from Real Analysis that you referred to. Thus, the restrictions that m is irrational and n is rational can be removed and the result is still valid.

7. Originally Posted by undefined
Thanks, I see that the OP's question is a special case of the more general result from Real Analysis that you referred to. Thus, the restrictions that m is irrational and n is rational can be removed and the result is still valid.
While covered in real analysis first (usually) it is a topological concept.

8. Originally Posted by studentme
Hi, I am learning about rational and irrational numbers and I came across a question that I am trying to get my head around,

m is a irrational number
n is a rational number which is greater than m

Is it possible to always find a rational number between m and n?

Thanks
Let $m < n$ (dropping the restrictions on rationality). It's not all that hard to show that there is a rational number r such that $m < r < n$.

To show this, assume the contrary; there is no such rational number. Let N be a positive integer and consider the numbers $i / N$ where i ranges over the integers. The distance between two consecutive members of this set is $1 / N$. If no member of the set lies between m and n, then we must have, for some i,

$i / N < m < n < (i+1) / N$,

so $n-m < 1/N$.

Since our choice of N was arbitrary, this inequality must hold for all N. An equivalent inequality is

$N < 1 / (n-m)$.

But we can always find an N such that

$N > 1 / (n-m)$,

by Archimede's principle.
This contradiction shows our assumption that no rational lies between m and n is false.

9. Originally Posted by awkward
Let $m < n$ (dropping the restrictions on rationality). It's not all that hard to show that there is a rational number r such that $m < r < n$.

To show this, assume the contrary; there is no such rational number. Let N be a positive integer and consider the numbers $i / N$ where i ranges over the integers. The distance between two consecutive members of this set is $1 / N$. If no member of the set lies between m and n, then we must have, for some i,

$i / N < m < n < (i+1) / N$,

so $n-m < 1/N$.

Since our choice of N was arbitrary, this inequality must hold for all N. An equivalent inequality is

$N < 1 / (n-m)$.

But we can always find an N such that

$N > 1 / (n-m)$,

by Archimede's principle.
This contradiction shows our assumption that no rational lies between m and n is false.
Yes, but the point is that the Archimedean principle is not fundamental.

10. Originally Posted by studentme
m is a irrational number
n is a rational number which is greater than m
Is it possible to always find a rational number between m and n?
Then there is a positive integer $K$ such $K>\frac{1}{n-m}$.
From which we get $n-\frac{1}{K}>m$.
Then note that $n-\frac{1}{K}$ must be rational and $n>n-\frac{1}{K}$.