View Full Version : Prove between any two rational numbers, there is a rational number:
 
 klbcooldude
Oct 19, 2009, 03:12 PM
Hey,
 
I'm having some trouble solving this problem:
 
Prove or disprove the following: between any two rational numbers, there is a rational number.
 
Cheers,
 ebaines
Oct 19, 2009, 04:06 PM
Hmm. How about this:
 
Consider two rational numbers A and B, where B>A.  The difference between these number is D = B-A.  Hence A+D = B.  Note that D is a rational number.
 
Now divide D by a known irrational number that is >1, such as sqrt(2).  Call this E, where:  E = D/sqrt(2).  E is definitely irrational, and positive, and smaller than D.  So A < A+E < A+D.  Remember A+D = B.  So:  A < A+E < B, and A+E is irrational.  Hence no matter the values of rational numbers A and B, there is always an irrational number that lies between.
 Nhatkiem
Oct 19, 2009, 04:10 PM
Hmm. how about this:
 
Consider two rational numbers A and B, where B>A.  The difference between these number is D = B-A.  Hence A+D = B.  Note that D is a rational number.
 
Now divide D by a known irrational number that is >1, such as sqrt(2).  Call this E, where:  E = D/sqrt(2).  E is definitely irrational, and positive, and smaller than D.  So A < A+E < A+D.  Remember A+D = B.  So:  A < A+E < B, and A+E is irrational.  Hence no matter the values of rational numbers A and B, there is always an irrational number that lies between.
 
Sorry ebaines,  I think you misread the question. It asks for the proof of a rational (not irrational) number between two rational numbers.
 klbcooldude
Oct 19, 2009, 08:03 PM
Any solutions?
 Nhatkiem
Oct 19, 2009, 09:57 PM
Any solutions?
 
I'm going to work from ebaines's example.
 
 
Consider two rational numbers A and B, where B>A. The difference between these number is D = B-A. Hence A+D = B. Note that D is a rational number.
 
Now divide D by a known rational number that is >1, such as 2. Call this E, where: E = D/2. E is definitely rational, and positive, and smaller than D. So A < A+E < A+D. Remember A+D = B. So: A < A+E < B, and A+E is rational. Hence no matter the values of rational numbers A and B, there is always an rational number that lies between.
 
 
Basically I took ebaines explanations, and changed irrational to rational values, hope you didn't mind me doing that ebaines :o
 ebaines
Oct 20, 2009, 05:35 AM
Sorry ebaines,  I think you misread the question. it asks for the proof of a rational (not irrational) number between two rational numbers.
 
Oops - I guess I did misread it!   Proving the existence of irrational numbers between any two rational numbers just seemed more...  interesting.  But as you suggest, changing the sqrt(2) to any rational number > 1 does the trick to prove that that there's always another rational number between any two rational numbers.  It's just a short additional step to prove that there are an infinite number of rational numbers - and an infinite number of irrational numbers - between any two rational numbers, no matter how close together they are.