# A course of Pure mathematics incorrect

I thought I'd go through 'A Course Of Pure Mathematics' by G H Hardy And I came to this part:

If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are 1, 2, 3, . . . 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 Λ, we can find as many rational points as we please on BC. Suppose, for example, that BC falls within the segment A1A2. It is evident that if we choose a positive integer k so that k · BC > 1, ∗ (1) and divide A1A2 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 A1A2 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 then 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.

But what is he talking about?? This is obviously not true. I drew a picture but for some reason I can't upload it so I'll just describe it.

Let A1A2 be 6 units long and BC 1 unit long. B will exist 3 away from A1 and C 4. Then let k = 2. This satisfies 2*BC > 1 because 2*1 > 1.

Now we divide A1A2 into 2 equal parts. 6 / 2 = 3 So P will exist 3 units away from A1. But this corresponds with B.

I know he's right. Am I reading this wrong? This doesn't make sense at all.

It looks like $$A_1A_2$$ has length $$1$$ by definition (which is OK, since aif $$BC$$ does not lie within such an interval, it contains an integer). If you prefer, replace $$k\cdot BC > 1$$ by $$k\cdot BC > |A_1A_2|$$, and the proof works without that requirement.
Then $$\dfrac{A1\ A2}{k} \lt BC$$.