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Taken from A Course of Pure Mathematics:

Irrational Numbers: If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are $1,2,3\ldots$ in succession, he will readily convince himself that he can cover the line with rational points as closely as he likes. We can state this more precisely as follows: if we take any segment $BC$ on $A$, we can find as many rational points as we please on $BC$.

Suppose, for example, that $BC$ falls within the segment $A_1A_2$. It is evident that if we choose a positive integer $k$ so that$$k.BC>1\tag1$$ And divide $A_1A_2$ into $k$ equal parts, then at least one of the points of division (say $P$) must fall inside $BC$, without coinciding with either $B$ or $C$. For if this were not so, $BC$ would be entirely included in one of the $k$ parts into which $A_1A_2$ has been divided, which contradicts the supposition $(1)$. But $P$ obviously corresponds to a rational number whose denominator is $k$. Thus at least one rational point $P$ lies between $B$ and $C$. But when we can find another such point $Q$ between $B$ and $P$, another between $B$ and $Q$, and so on indefinitely; i.e., as we asserted above, we can find as many as we please. We may express this by saying that $BC$ includes infinitely many rational points.


Question: What does the passage mean by this? I can't follow through with the proof, can you explain it to me?

I tried drawing a diagram with horizontal line $A_1A_2$ and starting off with an example, such as when $BC=1/3$ and $k=4$. However, no matter what, I just can't follow through.

The English is confusing me.

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  • $\begingroup$ Already the integers are infinitely many. Sine they are a subset of the rationals, then the rationals must also be infinite. Are you sure you don't mean the proof of irrational numbers being non-enumerable or something else? $\endgroup$ – mathreadler Apr 24 '17 at 19:15
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    $\begingroup$ The proof is about infinitely many rationals in any interval of positive length $\endgroup$ – Hagen von Eitzen Apr 24 '17 at 19:16
  • $\begingroup$ @HagenvonEitzen Yes, I started off with assuming $BC=1/3$ and $k=4$, and attempted to generalize it. $\endgroup$ – Crescendo Apr 24 '17 at 19:26
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If you divide $A_1A_2$ (which is assumed of unit length) into $k=4$ equal parts,then each part has length $\frac14$, hence is shorter than $BC$. It follows that $BC$ cannot fit between consecutive partitioning points, hence must overlap at least two (consecutive) of the $k$ parts at least partly. The partition point between these parts is in the interior of $BC$.

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