Timeline for Is the set $ AA+A $ always at least as large as $ A+A $?
Current License: CC BY-SA 3.0
5 events
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| Apr 28, 2015 at 13:51 | comment | added | Oliver Roche-Newton | Another more obvious example for which $|AA+A| \ll |A|^2$ is the set $A=\{1,2,\dots,N\}$. Then $AA+A \subset \{1,N^2+N\}$. I think part of the difficulty in showing that $AA+A$ is always "large" comes from the fact that there are very different sets (i.e. both arithmetic and geometric progressions) for which $A+AA$ is relatively "small". | |
| Apr 27, 2015 at 19:22 | comment | added | Yakk |
@TomDeMedts adding 0 adds 1 to element to both A and AA. Adding all negations doubles the number of elements of A and AA if A was strictly positive. First add negations, then add 0, for n elements in A', and 2n-3 elements in AA'.
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| Apr 27, 2015 at 15:53 | comment | added | Tom De Medts | I'm not sure whether $|AA|=2n-1$ is minimum possible. For instance, if $A = \{ -1,0,1 \}$, then $AA = A$. Perhaps if you replace your set $A$ by $A \cup -A$ you also get interesting examples; I haven't verified this. | |
| Apr 27, 2015 at 14:05 | history | edited | Neil Strickland | CC BY-SA 3.0 |
Corrected 2n+1 to 2n-1
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| Apr 27, 2015 at 12:43 | history | answered | Neil Strickland | CC BY-SA 3.0 |