0
$\begingroup$

I am trying to find the solution to the inequality $t^{-1/n}<1$ when $n$ is an integer and $t$ is positive, however when using

Reduce[t^(-1/n) >= 1, t]

it returned an error message stating that reduce was not able to solve this inequality (Mathematica 12.0). I also tried

Assuming[Element[n, Integers] && n > 0, Reduce[t^(-n^(-1)) <= 1, t]]

If I use

f[n_] := Reduce[t^(-1/n) <= 1, t]; 
f /@ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

I get the correct intervals of $t\geq1$. How can I show this for every $t>0$ and $n\geq1$ using Mathematica.

$\endgroup$
3
  • $\begingroup$ wolfram alpha can solve this. $\endgroup$
    – Xminer
    Commented Nov 19, 2019 at 2:47
  • 1
    $\begingroup$ Mathematica is not yet capable of reading minds, so you need to explicitly tell it your assumptions: FullSimplify[Reduce[t^(-1/n) < 1 && t > 1 && n ∈ Integers && n > 0, t], t > 1 && n ∈ Integers && n > 0] $\endgroup$ Commented Nov 19, 2019 at 2:52
  • $\begingroup$ But you are basically implementing the answer and checking whether it is true. What if the answer was not available or not immediate? That is why I posed the question. $\endgroup$
    – DMH16
    Commented Nov 19, 2019 at 2:58

1 Answer 1

3
$\begingroup$

You state that n is an integer and that t is positive. Include both of these constraints in Reduce

Reduce[t^(-1/n) <= 1 && Element[n, Integers] && t > 0, t] // Simplify

(* t^(-1/n) ∈ Reals && 
 n ∈ Integers && ((n <= -1 && 0 < t <= 1) || (n >= 1 && t >= 1)) *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.