R Under development (unstable) (2023-12-13 r85679 ucrt) -- "Unsuffered Consequences"
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> 
>  library(rcdd)
If you want correct answers, use rational arithmetic.
See the Warnings sections in help pages for
    functions that do computational geometry.
> 
>  A <- rbind(
+     c(0,  1.000,   -1,    0,    0,    0),
+     c(0,  1.000,   -1,    0,    0,    0),
+     c(0, -0.001,    1,    0,    0,    0),
+     c(0, -0.001,    0,    1,    0,    0),
+     c(0, -0.001,    0,    0,    1,    0),
+     c(0, -0.001,    0,    0,    0,    1),
+     c(1,  1.000,   -1,   -1,   -1,   -1),
+     c(1,  0.300,   -1,   -1,   -1,    0),
+     c(1,  7.990,   -1,   -3,   -5,   -8))
> 
>  b <- c(0, 0, 0, 0)
> 
>  out <- lpcdd(A, b)
>  out
$solution.type
[1] "Inconsistent"

$dual.direction
[1]  0  0  4  2  0  0  8 -3 -1

> 
>  all(A[out$dual.direction < 0, 1] == 1)
[1] TRUE
> 
>  fred <- rbind(out$dual.direction) %*% A
>  fred <- as.numeric(fred)
>  sally <- fred[2]
>  fred <- fred[- c(1, 2)]
>  all.equal(max(abs(fred)), 0)
[1] TRUE
>  
>  sally >= 0
[1] FALSE
>  ### if FALSE proves the constraints cannot be satisfied
> 
> 
>  Arat <- 1000 * A
>  Arat <- round(Arat)
>  Arat <- z2q(Arat, 0 * Arat + 1000)
>  Arat
      [,1] [,2]      [,3] [,4] [,5] [,6]
 [1,] "0"  "1"       "-1" "0"  "0"  "0" 
 [2,] "0"  "1"       "-1" "0"  "0"  "0" 
 [3,] "0"  "-1/1000" "1"  "0"  "0"  "0" 
 [4,] "0"  "-1/1000" "0"  "1"  "0"  "0" 
 [5,] "0"  "-1/1000" "0"  "0"  "1"  "0" 
 [6,] "0"  "-1/1000" "0"  "0"  "0"  "1" 
 [7,] "1"  "1"       "-1" "-1" "-1" "-1"
 [8,] "1"  "3/10"    "-1" "-1" "-1" "0" 
 [9,] "1"  "799/100" "-1" "-3" "-5" "-8"
> 
>  brat <- d2q(b)
>  brat
[1] "0" "0" "0" "0"
> 
>  out.rat <- lpcdd(Arat, brat)
>  out.rat
$solution.type
[1] "Inconsistent"

$dual.direction
[1] "0"  "0"  "4"  "2"  "0"  "0"  "8"  "-3" "-1"

> 
>  out$solution.type == out.rat$solution.type
[1] TRUE
>  all.equal(out$dual.direction, q2d(out.rat$dual.direction))
[1] TRUE
> 
>  fred <- qmatmult(rbind(out.rat$dual.direction), Arat)
>  sally <- fred[2]
>  fred <- fred[- c(1, 2)]
>  all(fred == "0")
[1] TRUE
>  
>  qsign(sally) >= 0
[1] FALSE
>  ### if FALSE proves the constraints cannot be satisfied
> 
> 
> proc.time()
   user  system elapsed 
   0.17    0.04    0.18