# prove that the $\sqrt{n}$ is unbounded

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.

• A more direct proof would be to note that for any $K\ge 0$, $\sqrt{K^2} = K$. – copper.hat Sep 10 '16 at 20:56
• 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 $-K<0$. – Winther Sep 10 '16 at 22:05
• So since $-1 < 0 < 1$, we have $(-1)^2 < 0 < 1$? – anomaly Sep 11 '16 at 2:56
• Can you please tell us what $n$ is? Are they natural numbers or real numbers? – guimption Feb 17 at 19:27
• Also I think @ben got the definition of upper bound wrong. upper bound just means $\le$ not $<$ – 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.

• 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. – Wojowu Sep 10 '16 at 21:15
• @Wojowu I think copper.hat's solution in the comments is good enough. – Jam Sep 10 '16 at 21:17
• furthermore, the OP got the definition of upper bound wrong. upper bound just means $\le$ not $<$ – 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.