klbcooldude Posts: 6, Reputation: 1 New Member #1 Oct 19, 2009, 03:12 PM
Prove between any two rational numbers, there is a rational number:
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 Posts: 12,132, Reputation: 1307 Expert #2 Oct 19, 2009, 04:06 PM

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 Posts: 120, Reputation: 9 Junior Member #3 Oct 19, 2009, 04:10 PM
Originally Posted by ebaines

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 Posts: 6, Reputation: 1 New Member #4 Oct 19, 2009, 08:03 PM
Any solutions?
 Nhatkiem Posts: 120, Reputation: 9 Junior Member #5 Oct 19, 2009, 09:57 PM
Originally Posted by klbcooldude
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 Posts: 12,132, Reputation: 1307 Expert #6 Oct 20, 2009, 05:35 AM
Originally Posted by Nhatkiem
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.

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