I wanted to check if my answer to proving that the $\sqrt{n}$ is unbounded works.

If the $\sqrt{n}$ is bounded then there exists a $K$ s.t. $|\sqrt{n}|< K$, for all n.


$|\sqrt{n}| < K \Rightarrow -K < \sqrt{n} < K \Rightarrow (-K)^2 < (\sqrt{n})^2 < (K)^2 \Rightarrow K^2 < n < K^2 $

and since this is impossible, the sequence $\sqrt{n}$ is unbounded.

  • 4
    $\begingroup$ A more direct proof would be to note that for any $K\ge 0$, $\sqrt{K^2} = K$. $\endgroup$ – copper.hat Sep 10 '16 at 20:56
  • 1
    $\begingroup$ If we multiply an inequality by a negative number then we need to reverse the inequality, e.g. $1<2$ multiplied by $-1$ gives $-1 > -2$ not $-1 < -2$. This is the error in your argument as squaring the inequality $-K < \sqrt{n}$ is equivalent to multiplying by $-K<0$. $\endgroup$ – Winther Sep 10 '16 at 22:05
  • $\begingroup$ So since $-1 < 0 < 1$, we have $(-1)^2 < 0 < 1$? $\endgroup$ – anomaly Sep 11 '16 at 2:56
  • $\begingroup$ Can you please tell us what $n$ is? Are they natural numbers or real numbers? $\endgroup$ – guimption Feb 17 at 19:27
  • $\begingroup$ Also I think @ben got the definition of upper bound wrong. upper bound just means $\le$ not $<$ $\endgroup$ – guimption Feb 17 at 19:36

Not a valid proof.

From $$-K < \sqrt{n} < K$$

one cannot conclude that

$$K^2 < n < K^2$$

Example: $-3<1<3$ is true but $9<1<9$ is not true because $9<1$ is not true.

  • 1
    $\begingroup$ Usually, once pointing out an error, you can also point out a suggested fix. Not necessary of course, but in many cases OP will appreciate that. $\endgroup$ – Wojowu Sep 10 '16 at 21:15
  • $\begingroup$ @Wojowu I think copper.hat's solution in the comments is good enough. $\endgroup$ – Jam Sep 10 '16 at 21:17
  • $\begingroup$ furthermore, the OP got the definition of upper bound wrong. upper bound just means $\le$ not $<$ $\endgroup$ – guimption Feb 17 at 19:35

Let $K\in I\!R$ be an bound to $\sqrt{n}$. So $|\sqrt{n}|< K \forall n$.

But take $n'=ceil(K^2)+1$, where $n'$ is the smallest integer greater than $K^2$ ($K\in I\!R$, so $K^2$ is not necessarily is an integer) plus one. Then $\sqrt{n'}>\sqrt{K^2}=K$, so $K$ can't be a bound to $\sqrt{n}$. We showed this without specifying $K$, so it is valid for any bound, and therefore no bound exists.


Your Answer

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

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