1

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.

Tzina Chris
  • 193
ben
  • 21
  • 4
    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
  • 1
    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
  • 1
    So since $-1 < 0 < 1$, we have $(-1)^2 < 0 < 1$? – anomaly Sep 11 '16 at 02:56
  • Can you please tell us what $n$ is? Are they natural numbers or real numbers? – makansij Feb 17 '19 at 19:27
  • Also I think @ben got the definition of upper bound wrong. upper bound just means $\le$ not $<$ – makansij Feb 17 '19 at 19:36
  • To fix your error, you only need $0\le \sqrt{n}\le K \implies 0\le n=(\sqrt{n})^2 \le K^2$. – David P Sep 26 '23 at 18:45

2 Answers2

7

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.

Did
  • 279,727
Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
  • 1
    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 $<$ – makansij Feb 17 '19 at 19:35
0

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.

Lucas Gallindo
  • 161