Why is the set of rational numbers dense, and set of integers numbers not?

Dec 2018
42
2
Amsterdam
If we have two sets:
  1. Set one is the set of rational numbers with the usual less-than ordering
  2. Set two is the set of integers numbers with the usual less-than ordering

Why is the set of rational numbers dense, and set of integers numbers not?

Density is that for all choices of x and y with x < y there is a ∈ A with x < a < y.

Can somebody show me this via an example?
 
Dec 2015
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Earth
Because two rational numbers are more close to each other than two integer numbers.
Simply there are more values of \(\displaystyle a\) that can be found in \(\displaystyle (x,y)\).
 
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Dec 2018
42
2
Amsterdam
Because two rational numbers are more close to each other than two integer numbers.
Simply there are more values of \(\displaystyle a\) that can be found in \(\displaystyle (x,y)\).
What do you mean with this, can you give an example maybe?
 

topsquark

Math Team
May 2013
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What do you mean with this, can you give an example maybe?
You can always find a (countably) infinite number of rational numbers between any two rational numbers. You can't do that with the integers.

-Dan
 
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Dec 2018
42
2
Amsterdam
You can always find a (countably) infinite number of rational numbers between any two rational numbers. You can't do that with the integers.

-Dan
So for the integers an example would be that there is no number between 2 and 3. Is this correct?
 
May 2016
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USA
So for the integers an example would be that there is no number between 2 and 3. Is this correct?
Yes, but it might be more illuminating to say that there are three integers between 3 and 7, but you cannot say that there is a finite number of rational numbers between 1/7 and 1/3.
 
Aug 2012
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You can always find a rational between any two rationals.

You can't always find an integer between any two. For example there's no third integer between 2 and 3.
 
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