Thm 140: If $\xi > \eta$, then $\eta + \nu = \xi$ has exactly one solution $\nu$. (All lowercase Greek letters represent Dedekind cuts.)

Summary of proof and the alleged error:

Consider the set of all rational numbers of the form $X - Y$ (with $X > Y$) where $X$ is a lower number for $\xi$ and $Y$ is a lower number for $\eta$.

Landau then proves that this set defines a cut and claims that it is the solution $\nu$ to the equation in the statement of the theorem. (He has also already shown that this solution is unique.)

To prove this, he must show that $\eta + \nu$ and $\xi$ are equal, which simply consists of showing that every lower number of the former is a lower number of the latter and vice versa.

Now here is where I believe he makes an error: he writes:

"Every lower number for $\nu + \eta$ is of the form $$(X - Y) + Y_1$$ where X is a lower number for $\xi$, ** $Y$ is an upper number for $\eta$ **, $Y_1$ a lower number for $\eta$, and $X > Y$.

Now we have ** $Y > Y_1$ **, [so] $$((X - Y) + Y_1) + (Y - Y_1) = \cdots$$

The issue here is that he originally says that $Y$ is a lower number of $\eta$, but then makes it an upper number so that he can subtract $Y_1$ from $Y$ (he doesn't define negative numbers until the section on real numbers, so only subtraction that yields a positive number is 'legal' at this point).

Am I right? If so, how would you repair this proof?

If I'm wrong, what am I missing here?

Thanks in advance.


1 Answer 1


The lower numbers for $-\eta$ should be the negatives of the upper numbers for $\eta$. So in setting up the lower numbers for $\xi-\eta$ it should be numbers $X-Y$ where $X$ is a lower number for $\xi$ and $Y$ is an upper number for $\eta$. This would make the steps in the proof work, with $Y_1$ a lower number for $\eta$, but would mean the wording of the proof should change to "consider rationals of the form $X-Y$ where $X$ is a lower number for $\xi$ and $Y$ is an upper number for $\eta$," at the start of the proof.

  • $\begingroup$ Good suggestion. I'm going to have to think about this some more. Before I choose an answer, I'm going to see what others have to say. $\endgroup$ Jul 18, 2013 at 19:06
  • 1
    $\begingroup$ If we take the set of $X-Y$ where $X,Y$ are respectively lower numbers for $\xi,\eta$ then it would seem this set is not bounded above. For we could take a fixed $X$ and some $Y$ very large negative, and obtain a very large positive number as $X-Y$. If as it seems Landau is working with left cuts as defining things, this looks wrong and doesn't produce a left cut, except the left cut for $+\infty.$ $\endgroup$
    – coffeemath
    Jul 18, 2013 at 19:21
  • $\begingroup$ Ah, I see. It's probably just a typo in the theorem statement then. $\endgroup$ Jul 18, 2013 at 19:23

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