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Let $x$ be a nonnegative real number and denote $[x]$ as the greatest integer less than or equal to $x$. We will attempt to prove that $\big[\sqrt{x}\big] = \big[\sqrt{[x]}\big]$.

First suppose that $x$ is a perfect square. Then the equation trivially holds.

Assuming that $x$ is not a perfect square, we have $\sqrt{x} \not\in \mathbb{Z}$.

If $[x] \leq x$, then it clear that $\sqrt{[x]} \leq \sqrt{x}$ where $\sqrt{[x]}$ may or may not be an integer.

Further, it is true that there are no integers in the interval $\big(\sqrt{[x]},\sqrt{x}\big)$; because if such an integer $q$ existed, we would deduce that

$$[x] < q^2 < x$$

which is a blatant contradiction since $q^2 \in \mathbb{Z}$.

Because there are no integers in $\big(\sqrt{[x]},\sqrt{x}\big)$, it follows that

$$\big[\sqrt{x}\big] \leq \sqrt{[x]} \leq \sqrt{x} \tag{1}$$

But since $\big[\sqrt{x}\big]$ is the closest integer to $\sqrt{x}$ that satisfies $(1)$, it follows that it must also be the closest integer to $\sqrt{[x]}$, that is to say, it is the greatest integer less than or equal to $\sqrt{[x]}$.

Therefore, $\big[\sqrt{x}\big] = \big[\sqrt{[x]}\big]$.

Are there any problems with the logic of the above proof? Thank you.

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  • $\begingroup$ What is $[x] $? $\endgroup$ – YoTengoUnLCD Feb 10 '16 at 15:29
  • $\begingroup$ My apologies. It is the floor of $x$. $\endgroup$ – Jeremiah Dunivin Feb 10 '16 at 15:29
  • $\begingroup$ If $[x]\leq x$ then you can only conclude that $\sqrt{[x]}\leq \sqrt{x}$. Don't confuse $\leq$ with $<$. $\endgroup$ – Thomas Andrews Feb 10 '16 at 15:38
  • $\begingroup$ You're right. But even if I change the inequality to $\leq$, the proof still seems to hold. $\endgroup$ – Jeremiah Dunivin Feb 10 '16 at 15:45
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Your proof is fine.

(An alternate approach, which is really very close to your proof, but perhaps clearer.)

Let $n=\lfloor \sqrt{x}\rfloor$, so $$n\leq \sqrt{x}<n+1\implies \\n^2\leq x<(n+1)^2\implies\\n^2\leq \lfloor x\rfloor <(n+1)^2\implies \\ n\leq \sqrt{\lfloor x\rfloor}<n+1$$

The key trick is that $n^2\leq x$ means $n^2\leq \lfloor x\rfloor$, because $n^2$ is an integer.

To prove $\lfloor f(x)\rfloor = \lfloor g(x)\rfloor$, we merely need to show that:

$$\lfloor f(x)\rfloor \leq g(x)\text{ and }\\ \lfloor g(x)\rfloor \leq f(x)$$

(Worth figuring out why this is sufficient.) Often, one of these will be "obvious." For example, as in your case, you have $f(x)\leq g(x)$ for all $x$, so $\lfloor f(x)\rfloor \leq g(x)$.)

So the only real part you needed above was that $$n=\lfloor\sqrt{x}\rfloor\leq \sqrt{\lfloor x\rfloor}$$

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  • $\begingroup$ Nevermind, I see how $n$ is the greatest integer. I would hate to use your answer for my homework, so I was wondering if there are any other flaws besides the one that you mentioned in your comment above. $\endgroup$ – Jeremiah Dunivin Feb 10 '16 at 16:13
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$x$ lies between two perfect squares

$$n^2\le x<(n+1)^2$$ and as $n^2$ is an integer

$$n^2\le\lfloor x\rfloor\le x<(n+1)^2.$$

Then taking the square root

$$n\le\sqrt{\lfloor x\rfloor}\le\sqrt x<n+1$$

which is another way to say

$$\lfloor \sqrt{\lfloor x\rfloor}\rfloor=\lfloor \sqrt x\rfloor.$$

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