Knowing how to order ${a,b,c,0}$ implies knowing how to order ${a,b,c,0,-a,-b,-c}$?
Suppose I have $4$ real numbers ${a,b,c,0}$ and I know
that they are all different
how to order them from smallest to largest, e.g., I know that $b<a<0<c$
Does this imply that I know how to order from smallest to largest ${a,b,c,0,-a,-b,-c}$?
linear-algebra combinatorics inequality linear-programming
asked Nov 18 at 16:36
STF
751420
add a comment |
Suppose I have $4$ real numbers ${a,b,c,0}$ and I know
that they are all different
how to order them from smallest to largest, e.g., I know that $b<a<0<c$
Does this imply that I know how to order from smallest to largest ${a,b,c,0,-a,-b,-c}$?
linear-algebra combinatorics inequality linear-programming
asked Nov 18 at 16:36
STF
751420
add a comment |
Suppose I have $4$ real numbers ${a,b,c,0}$ and I know
that they are all different
how to order them from smallest to largest, e.g., I know that $b<a<0<c$
Does this imply that I know how to order from smallest to largest ${a,b,c,0,-a,-b,-c}$?
linear-algebra combinatorics inequality linear-programming
asked Nov 18 at 16:36
STF
751420
Suppose I have $4$ real numbers ${a,b,c,0}$ and I know
that they are all different
how to order them from smallest to largest, e.g., I know that $b<a<0<c$
Does this imply that I know how to order from smallest to largest ${a,b,c,0,-a,-b,-c}$?
linear-algebra combinatorics inequality linear-programming
linear-algebra combinatorics inequality linear-programming
asked Nov 18 at 16:36
STF
751420
asked Nov 18 at 16:36
STF
751420
asked Nov 18 at 16:36
STF
751420
asked Nov 18 at 16:36
STF
751420
asked Nov 18 at 16:36
STF
751420
751420
add a comment |
add a comment |
2 Answers
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oldest
votes
No, because you don't know how to compare $c$ to $-a$ and $-b$.
It's only possible if $0$ is either the smallest or the largest of the four.
answered Nov 18 at 16:40
Arthur
110k7105186
add a comment |
You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...
answered Nov 18 at 16:41
b00n heT
10.2k12134
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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No, because you don't know how to compare $c$ to $-a$ and $-b$.
It's only possible if $0$ is either the smallest or the largest of the four.
answered Nov 18 at 16:40
Arthur
110k7105186
add a comment |
No, because you don't know how to compare $c$ to $-a$ and $-b$.
It's only possible if $0$ is either the smallest or the largest of the four.
answered Nov 18 at 16:40
Arthur
110k7105186
add a comment |
No, because you don't know how to compare $c$ to $-a$ and $-b$.
It's only possible if $0$ is either the smallest or the largest of the four.
answered Nov 18 at 16:40
Arthur
110k7105186
No, because you don't know how to compare $c$ to $-a$ and $-b$.
It's only possible if $0$ is either the smallest or the largest of the four.
answered Nov 18 at 16:40
Arthur
110k7105186
answered Nov 18 at 16:40
Arthur
110k7105186
answered Nov 18 at 16:40
Arthur
110k7105186
answered Nov 18 at 16:40
Arthur
110k7105186
110k7105186
add a comment |
add a comment |
You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...
answered Nov 18 at 16:41
b00n heT
10.2k12134
add a comment |
You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...
answered Nov 18 at 16:41
b00n heT
10.2k12134
add a comment |
You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...
answered Nov 18 at 16:41
b00n heT
10.2k12134
You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...
answered Nov 18 at 16:41
b00n heT
10.2k12134
answered Nov 18 at 16:41
b00n heT
10.2k12134
answered Nov 18 at 16:41
b00n heT
10.2k12134
answered Nov 18 at 16:41
b00n heT
10.2k12134
10.2k12134
add a comment |
add a comment |
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