How can you take a square root of a negative number?

Specifically, I have this problem where I need to select a positive integer $q$ s.t. for $b \in ]2,5[$, $$q > \sqrt{\frac{1}{b-5}}$$ and I want to be able to conclude that $$b < 5-\frac{1}{q^2}$$

And the reason that $q$ has to be selected this way is of course:

$$b < 5-\frac{1}{q^2}$$ $$bq^2 < 5q^2-1$$ $$q^2(b-5)< -1$$

and since $b-5 < 0$ ($b \in ]2,5[$)

$$q^2 > \frac{1}{b-5}$$

But how can I select such $q$?

  • $\begingroup$ Your question is quite unclear. Do you want a $q$ in terms of $b$ such that $q^2>\frac{1}{b-5}$ $\forall\, b\in\mathbb R$? $\endgroup$ – Luke Collins Nov 8 '16 at 18:58
  • $\begingroup$ @LukeCollins $b \in ]2,5[$? $\endgroup$ – mavavilj Nov 8 '16 at 18:58
  • $\begingroup$ Where did that condition come from? $\endgroup$ – Luke Collins Nov 8 '16 at 18:59
  • $\begingroup$ How did the problem start? I ask because $q>\sqrt {1/(b-5) } $ does not actually make sense for $b <5$. However $q^2 >1/(b-5) $ does (and will hold for all real $q $). $\endgroup$ – Ian Nov 8 '16 at 19:00
  • $\begingroup$ @LukeCollins it's related to the problem where I have this problem of selecting the $q$. Read the question? $\endgroup$ – mavavilj Nov 8 '16 at 19:00

You made an error on taking the reciprocals: the smaller becomes the larger.

Explicitly, \begin{align} q>\frac1{\sqrt{b-5}}&\iff q^2>\frac1{b-5}\quad\textbf{and}\quad b-5>0\\ &\iff b-5>\frac1{q^2}\quad\textbf{and}\quad b-5>0\\ &\iff 5<b<5+\frac1{q^2}. \end{align}

  • $\begingroup$ How can $b-5 > 0$ if $b \in ]2,5[$? $\endgroup$ – mavavilj Nov 8 '16 at 19:05
  • $\begingroup$ I didn't even see this hypothesis. It means you can't select such an integer: $b$ is not in the domain of the inequation. Are you sure of the exact form of the inequation? $\endgroup$ – Bernard Nov 8 '16 at 19:12
  • $\begingroup$ I know that it has to be that $b \in ]2,5[$. I also know that the interval $]2,5[$ is the limit of $[2+i^{-2}, 5- i^{-2}]$, thus why I need the inequalities, such as $b < 5-i^{-2}$. $\endgroup$ – mavavilj Nov 8 '16 at 19:16
  • $\begingroup$ Isn't it $q>\dfrac1{\sqrt{\lvert b-5\rvert}}$? $\endgroup$ – Bernard Nov 8 '16 at 19:17
  • 1
    $\begingroup$ Then replace $b-5$ with $5-b$, and you'll obtain $5-\dfrac1{q^2}<b<5$. $\endgroup$ – Bernard Nov 8 '16 at 19:20

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