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I came across the following equality while trying to prove this Multiplicative property of the GCD:

$$\min\{a+c, b+d\} = \min\{a, b\} + \min\{c, d\} + \min\{a - \min\{a, b\}, d - \min\{c,d\}\} + \min\{b - \min\{a, b\}, c - \min\{c,d\}\}, $$

where $a,b,c,d \in \mathbf{N}$. It shouldn't be too difficult to prove - just a matter of checking all the different cases - but I'm having some trouble pinning down exactly what the cases are.

The equation is symmetric in $a, c$ and in $b, d$, right? (Or is it in $a, b$ and in $c,d$…) So that should reduce the number of cases. Also, if $a \leq b \leq c \leq d$ or $a \leq c \leq b \leq d$, for example, then clearly $a+c \leq b+d$; but if $a \leq b \leq d \leq c$ then the cases where $a+c \leq b+d$ and $a+c > b+d$ need to be considered separately.

Any help in crystallising the above would be appreciated; I admit it's not a very appetising problem.

Community
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0-0-0
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  • I should note that I've always found that such identities arise from an unnecessary splitting of cases earlier, that if avoided would not have produced these. – user21820 Feb 15 '14 at 12:44
  • In this particular instance there wasn't any prior splitting of cases. There are only two or three more steps in the whole proof, and these are straightforward: basically just converting the gcds on the complicated side of the identity (in the link) into prime power factorisations with mins. After clumping all these products together, the equality shows up in the exponent. – 0-0-0 Feb 15 '14 at 13:00
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    Ah I see why. The version of gcd you are using implicitly splits cases (and only applies to unique factorization domains), while the more general version involving just divisibility doesn't, which is why Bill Dubuque's solution doesn't have any cases. – user21820 Feb 15 '14 at 13:45

2 Answers2

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Wlog. $a\le b$, which turns the right hand side into $$ a+\min\{c,d\}+0+\min\{b-a,c-\min\{c,d\}\}.$$ If $c\le d$ this becomes $a+c$, which is clearly $\min\{a+c,b+d\}$; and if $c>d$ it becomes $$a+d+\min\{b-a,c-d\} =\min\{a+d+b-a,a+d+c-d\}=\min\{b+d,a+c\}.$$

Hagen von Eitzen
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No it's not symmetric in any of the ways you suggested. It does have a symmetry, which is that it is invariant under the permutation $(a,b)(c,d)$ or $(a,c)(b,d)$ or $(a,d)(b,c)$ but that wouldn't really help much, except to imply that you can assume $a$ to be smallest, but have to test all $6$ permutations of $b,c,d$.

user21820
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