Timeline for For a finite set A of positive reals, prove that the set A + A - A contains at least as many positive as negative elements
Current License: CC BY-SA 4.0
9 events
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| May 24, 2023 at 22:44 | comment | added | Timothy Chow | @NickS If you delete your answer then we would lose Timo Reichert's useful comment. Maybe Timo should add this example to the question. | |
| May 24, 2023 at 19:31 | comment | added | Nick S | @TimoReichert I will delete the answer then. Note though that we also know that $0 \in M$ which may help. | |
| May 24, 2023 at 16:57 | comment | added | Timo Reichert | I agree, I have already shown that we cannot prove the statement for $A + M$, where $M$ is an arbitrary symmetric set around 0. We would have to somehow use the fact that $M$ gets created by $A - A$. The counterexample I used for this was $M = \{-125, -121, -117, -113, -109, -106, 106, 109, 113, 117, 121, 125\}$ and $A = \{1, 4, 8, 12, 16, 20\}$, which produces the elements $\{-124, -121, -120, -117, -116, -113, -112, -109, -108, -105, -102, -101, -98, -97, -94, -93, -90, -89, -86, 107, 110, 113, 114, 117, 118, 121, 122, 125, 126, 129, 133, 137, 141, 145\}.$ 19 are negative and 15 positive. | |
| May 24, 2023 at 15:37 | comment | added | Emil Jeřábek | $|A-D|\le|A+D|$ is certainly not true for arbitrary sets of positive numbers $A$ and $D$: e.g., let $A=D=\{1,2,4\}$; then $A+D=\{2,3,4,5,6,8\}$ has $6$ elements, but $A-D=\{-3,-2,-1,0,1,2,3\}$ has $7$. More generally, if $A=D$ is a Sidon set with $n$ elements, then $|A+D|=(n+1)n/2$, but $|A-D|=n(n-1)+1$, which is almost twice as big. | |
| May 24, 2023 at 15:27 | comment | added | Emil Jeřábek | Is $C=A-A$ here? | |
| May 24, 2023 at 15:18 | comment | added | Nick S | @TimoReichert Ty, fixed. | |
| May 24, 2023 at 15:17 | history | edited | Nick S | CC BY-SA 4.0 |
edited body
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| May 24, 2023 at 12:50 | comment | added | Timo Reichert | I think the last line should be $F(a,b)=F(c,d) \Leftrightarrow G(a,d)=G(c,b)$, right? Also, you meant to use the set $D$ in the functions instead of $C$, correct? But yes, this is what we have to do and it is written in a very nice and compact way! | |
| May 23, 2023 at 21:22 | history | answered | Nick S | CC BY-SA 4.0 |