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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.

therefore

$|\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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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    $\begingroup$ 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
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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.

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