Reduce is not giving a correct output [closed]

Question

For some reason, the output of the code below is False.

Reduce[(1/(
 2 a (-2 r + c^2 (1 + r)) (-1 + x)) (r - a c (2 + r) - 
   3 a r (-1 + x) + c^2 (-r + a (2 + r) x) - 
   Sqrt[-1 + c] Sqrt[r]
     Sqrt[(-1 + c) (1 + c)^2 r + 
     2 a (r + c (2 - (2 + c) r) + (-1 + c) (1 + c)^2 r x) + 
     a^2 (-8 + 4 c - r + 5 c r + 
        2 (r + c (2 - (2 + c) r)) x + (-1 + c) (1 + 
           c)^2 r x^2)])) > 1 && (c - a (1 - c))/c > r > a/(1 + a - c) && 1 > c > a > 0]

The way I understand the result is that Mathematica suggests that they can not all be true at the same time.

But, for example {c -> .5, a -> .1, r -> .8, x -> .15} will satisfy that.

Am I understanding it wrong? Or, is the Reduce function giving me a wrong output here?

Answer 1

This is not a bug. Your inequality contains Sqrt[-1 + c]. In general, this is a complex number, which you cannot compare unless $c \ge 1$. But this contradicts your other condition $1>c$.

Answer 2

@yarchik's answer is correct that it is not a bug. I assumed it to be a bug too quickly without really looking what expressions contained and what documentation says about possible issues.

But anyway Reduce can in general solve even such systems but we need to add that variables are real and we are solving over Complexes.

The same is with FindInstance.

This returns {} because no solutions could be found.

FindInstance[(1/(2 a (-2 r + c^2 (1 + r)) (-1 + x)) (r - 
       a c (2 + r) - 3 a r (-1 + x) + c^2 (-r + a (2 + r) x) - 
       Sqrt[-1 + c] Sqrt[
         r] Sqrt[(-1 + c) (1 + c)^2 r + 
          2 a (r + c (2 - (2 + c) r) + (-1 + c) (1 + c)^2 r x) + 
          a^2 (-8 + 4 c - r + 5 c r + 
             2 (r + c (2 - (2 + c) r)) x + (-1 + 
                c) (1 + c)^2 r x^2)])) > 1 && (c - a (1 - c))/c > r > 
   a/(1 + a - c) && 1 > c > a > 0, {a, c, r, x}]

---

{}

But if we add ...&& {a, c, r, x} \[Element] Reals, {a, c, r, x}, Complexes] a solution is found.

FindInstance[(1/(2 a (-2 r + c^2 (1 + r)) (-1 + x)) (r - 
       a c (2 + r) - 3 a r (-1 + x) + c^2 (-r + a (2 + r) x) - 
       Sqrt[-1 + c] Sqrt[
         r] Sqrt[(-1 + c) (1 + c)^2 r + 
          2 a (r + c (2 - (2 + c) r) + (-1 + c) (1 + c)^2 r x) + 
          a^2 (-8 + 4 c - r + 5 c r + 
             2 (r + c (2 - (2 + c) r)) x + (-1 + 
                c) (1 + c)^2 r x^2)])) > 1 && (c - a (1 - c))/c > r > 
   a/(1 + a - c) && 
  1 > c > a > 0 && {a, c, r, x} \[Element] Reals, {a, c, r, 
  x}, Complexes]

---

{{a -> 1/15, c -> 1/15 (1 + Sqrt[31]), r -> 46/125, x -> -17}}

So same should be possible with Reduce. (it took some time as the system is complicated but solution was found).

Reduce[(1/(2 a (-2 r + c^2 (1 + r)) (-1 + x)) (r - a c (2 + r) - 
       3 a r (-1 + x) + c^2 (-r + a (2 + r) x) - 
       Sqrt[-1 + c] Sqrt[
         r] Sqrt[(-1 + c) (1 + c)^2 r + 
          2 a (r + c (2 - (2 + c) r) + (-1 + c) (1 + c)^2 r x) + 
          a^2 (-8 + 4 c - r + 5 c r + 
             2 (r + c (2 - (2 + c) r)) x + (-1 + 
                c) (1 + c)^2 r x^2)])) > 1 && (c - a (1 - c))/c > r > 
   a/(1 + a - c) && 
  1 > c > a > 0 && {a, c, r, x} \[Element] Reals, {a, c, r, 
  x}, Complexes]

---

(0 < a < 1/
    4 && ((a < c <= a + Sqrt[2 a + a^2] && 
       a/(1 + a - c) < r < (-a + c + a c)/
        c && (-a + a c - c r + a c r)/(a c r) < x < 
        1) || (a + Sqrt[2 a + a^2] < c < 
        1 && ((a/(1 + a - c) < 
            r < -(c^2/(-2 + c^2)) && ((-a + a c - c r + a c r)/(
              a c r) < 
              x <= (-2 a c + r - a r + c r + 2 a c r - c^2 r + 
                a c^2 r - c^3 r)/(a (-1 + c) (1 + c)^2 r) - 
               2 Sqrt[-((-c^2 + 2 r - c^2 r + c^4 r - 2 c^2 r^2 + 
                  c^4 r^2)/((-1 + c)^2 (1 + c)^4 r^2))] || 
             x > 1)) || (-(c^2/(-2 + c^2)) < r < (-a + c + a c)/
            c && (-a + a c - c r + a c r)/(a c r) < x < 1))))) || (1/
    4 <= a < 1 && a < c < 1 && 
   a/(1 + a - c) < r < (-a + c + a c)/c && (-a + a c - c r + a c r)/(
    a c r) < x < 1)