I have solved the following double inequality $1\leq x < x^{2}$ as follows:

$1\leq x$ is already solved

$x< x^{2}$ using a table of sings gives me $(-\infty,0) \cup (1,\infty)$

together the solution is $(1,\infty)$.

Now my question: is the following argument correct?

$1\leq x$ is already solved

For the second inequality $x< x^{2}$ I use from the first one that $1\leq x$ so I can divide both sides of the inequality (I divide by a number different from zero and a positive number which preserves the inequality) and I get $1 < x$.

Again together the solution is $(1,\infty)$.

It is valid to use the first inequality as a condition to solve the second one?

  • 1
    $\begingroup$ Yes, it is absolutely valid. $\endgroup$ – Win Vineeth Feb 17 '16 at 5:52
  • $\begingroup$ However $x=1$ is a solution. $\endgroup$ – fosho Feb 17 '16 at 6:13
  • $\begingroup$ My bad, it should be strict inequality $x < x^{2}$ $\endgroup$ – Novato Feb 17 '16 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.