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Given a collection of numbers, one may wish to find the "closest pair": two numbers in the collection that are not the same, but whose difference is as small as possible. For instance, if we have values 4, 11, 6 and 13, then the closest pair is either (4, 6) or (11.13): in both cases the difference is 2, and there are no pairs whose difference is 1.

Let Q denote the set of all rational numbers x such that 0 < x < 1. Using a proof by contradiction, prove the following theorem:

There does not exist a positive real number $d$ such that the closest pair of elements of Q has a difference exactly equal to $d$.

Can someone explain to me what this questions is about? I seem to have difficulty even understanding the question.

BrianO
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yociyoci
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  • Assume such a number $d$ exists. So other differences cannot be smaller. How many numbers from $Q$ set (at most) can you put between 0 and 1? – z100 Nov 11 '15 at 22:47
  • You are trying to prove that rational numbers between $0$ and $1$ get arbitrarily close together by proving that there can be no number $d$ which serves as that minimum distance. – John Douma Nov 11 '15 at 22:47

1 Answers1

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Hint: The exercise asks for an argument (by contradiction) for the statement:

"There does not exist a positive real number $d$ such that the closest pair of elements of $\mathbb Q$ has a difference exactly equal to $d$."

First think about what that statement means and convince yourself that it is true. Then you can try to prove it.

It might be helpful to think of sets for which this statement is false. If you look at $\mathbb Z$, for example, then there is such a positive number $d$ such that the closest pair of elements has a difference exactly equal to $d$: The smallest distance between two members of $\mathbb Z$ is $1$.

Now the statement says that there is no positive minimum distance $d$ of pairs in $\mathbb Q$. In other words, you can find pairs in $\mathbb Q$ that are arbitrarily close together. And that seems clearly true! Now you just have to prove it by contradiction..

lodrik
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