Rational and Irrational numbers

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April 19th 2010, 09:38 PM
studentme

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
April 19th 2010, 09:40 PM
Drexel28

Quote:

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"?
April 19th 2010, 10:04 PM
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?"
April 19th 2010, 10:18 PM
undefined

Quote:

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.
April 19th 2010, 10:19 PM
Drexel28

Quote:

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"
April 19th 2010, 10:29 PM
undefined

Quote:

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.
April 19th 2010, 10:30 PM
Drexel28

Quote:

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.
April 20th 2010, 02:34 PM
awkward

Quote:

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.
April 20th 2010, 02:50 PM
Drexel28

Quote:

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.
April 20th 2010, 03:00 PM
Plato

Quote:

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?

Having read through the other replies, here is a minimal reply.
To follow this we need to agree to just basic facts.
The sum of two rationals is rational. And the positive integers are not bounded above.

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 $n>n-\frac{1}{K}$ .