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New Member
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Oct 19, 2009, 03:12 PM
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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,
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Expert
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Oct 19, 2009, 04:06 PM
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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.
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Junior Member
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Oct 19, 2009, 04:10 PM
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Originally Posted by ebaines
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.
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New Member
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Oct 19, 2009, 08:03 PM
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Any solutions?
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Junior Member
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Oct 19, 2009, 09:57 PM
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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
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Expert
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Oct 20, 2009, 05:35 AM
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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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