I can't understand how to solve this question . I'll be thankful if you help me. Please solve it and post your complete solution & a little description of what you did.

Between which integers is $-3 + \sqrt 5$ situated?

Sorry for my bad English


closed as off-topic by John Joy, MCT, qaphla, user negative one over twelve, user228113 May 20 '16 at 2:22

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Another take.

Notice that $$-3+\sqrt4\lt-3+\sqrt5\lt-3+\sqrt9\Rightarrow\\-1\lt-3+\sqrt5\lt0$$

Since $$|0-(-1)|=1$$ is the smallest distance between two (different) integers, this is the required interval.


$$2^2<\sqrt{5}^2<3^2$$ so $2<\sqrt{5}<3$. Thus your number is between $-1$ and $0$

  • $\begingroup$ By far the simplest solution. +1! $\endgroup$ – zz20s May 20 '16 at 1:51

Assuming negative roots are permitted...


$\implies 2<\lvert\sqrt{5}\rvert<3$.

$\sqrt{5}$ is positive or negative so your number is between $-3-3$ and $-3+3$


  • $\begingroup$ Typically, in real analysis $\sqrt x$ is the positive square root of $x$. $\endgroup$ – user228113 May 20 '16 at 2:21
  • $\begingroup$ @G.Sassatelli Thanks for the heads up that's good to know. Analysis is almost certainly the wrong domain in this case, it's tagged algebra - precalculus and numerical-methods and the user has low reputation. $\endgroup$ – user334732 May 20 '16 at 7:50
  • 1
    $\begingroup$ [numerical-methods] is an inappropriated tag. By "in real analysis" I meant "when you do not deal with complex numbers". I don't think anyone really sees $-3+\sqrt 5$ as a multi-valued function indicating either of the roots of $t^2+6t+4$. You would typically write $-3\pm\sqrt5$. $\endgroup$ – user228113 May 20 '16 at 12:03
  • $\begingroup$ @G.Sassatelli complex numbers aren't involved in $\sqrt{x}=-y$ for $x\geq0$. You're confusing with $\sqrt{-x}=iy$ which isn't the case here. $\endgroup$ – user334732 Sep 22 '16 at 17:53

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