What is the best way to handle near duplicates? Here are two examples of what I mean.

This question asks about zero divisors in commutative rings. However, this older question asks a slightly more restrictive question, although the answers contain the general solution.

This question asks a question about tensor products for which this question asks to prove even more. However, the new question allows a solution method which the old question did not.

In both circumstances, the new question is very nearly a duplicate of the old question, yet in both cases the new question has some merit the old one does not. What to do?

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It would be great if there was an option to merge questions instead of closing them as abstract duplicates. This has been discussed here but no decision on this has been taken. –  user17762 Dec 21 '12 at 21:56
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If you don't, or can't, close as an exact duplicate, you can still leave notes on each problem linking to the other, with an explanation of what you're doing. –  Gerry Myerson Dec 22 '12 at 6:36
    
Related. Arguably, this post is a near-duplicate of the linked post, in fact. ;) –  Cameron Buie Jan 1 '13 at 19:02
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2 Answers

Maybe the simplest way to go about it (which errs on the side of inclusivity) is to interpret "exact duplicate" as "exact" and not "almost the same."

Of course, what constitutes "the same" is subject to subjectivity.

One thing I've learned is that in practice it usually doesn't matter if question A and question B are similar and question A's solutions contains the answer for question B: people will tend to block closure for the most minor difference in the two questions. That's the main reason I didn't bother flagging the post you were referring to as duplicate.

I totally agree it leaves the door open to a flood of slightly mutated duplicates, and it is not an optimal solution. but it seems easier to apply than trying to sort out what exactly constitutes a duplicate.

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If we do face a flood of slightly mutated duplicates, there is a solution for that. –  Rahul Dec 31 '12 at 6:36
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Here is a similar example:

how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$

This question was closed as an exact duplicate. But $7 \neq 2$ and this makes a difference. Because the new question has then a neater solution, which doesn't work for $7$:

Let $x_n=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ where there are $n$ square roots. [or $x_1=\sqrt{2}, x_n=\sqrt{2+\sqrt{x_{n-1}}}$].

Then it is an easy induction (double angle formula) exercise that

$$x_n =2 \cos(\frac{\pi}{2^{n+1}})\,.$$

The convergence and the limit of $x_n$ follow now much easier from here.

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