Dividing the square root of $x$ by $x$. Is $\frac{\sqrt{x}}{x}=\frac{1}{x}$ or $=\frac{1}{\sqrt{x}}$?

Question

So, we have:$$\frac{\sqrt{x}}{x}$$

When I worked this out, it gave me two different results: $\frac{1}{x}$ and $\frac{1}{\sqrt{x}}$ , which is illogical.

First method I used: $\frac{\sqrt{x}}{x}=\frac{x}{x^2}=\frac{x}{x\times x}=\frac{1}{x}$

I got rid of square roots, by squaring the numerator, and augmented the denominator proportionally, squaring it, so that I don't change the ratio.

Second method I used: $\frac{\sqrt{x}}{x}=\frac{\sqrt{x}}{\sqrt{x^2}}=\sqrt{\frac{x}{x^2}}=\sqrt{\frac{x}{x\times x}}=\sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}$

So, I just made it all a square root (Because: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$), and in the end, $\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$

They can't be both correct, because, as an example:

Let $x=2$

$\frac{1}{x}=\frac{1}{2}=0,5$

$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}}\approx{0,707}$

Hereupon, either $\frac{\sqrt{2}}{2}=\frac{2}{x^2}=\frac{1}{2}$ or $\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{\sqrt{2^2}}=\frac{1}{\sqrt{2}}$

I guess the second one is correct, because we can start with $\frac{1}{\sqrt{2}}$ and work it out to be $\frac{\sqrt{2}}{2}$ , but we can't start with $\frac{1}{2}$ and work it out to be $\frac{\sqrt{2}}{2}$

But why is that? I can't seem to wrap my head around it.

Please help.

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Notes:

• There's a similar question to this one (Why is the square root of $x$ divided by $x$ equal to $1$ divided by the square root of $x$), although its answer is surely connected to this question, it doesn't directly adress this issue. I wouldn't have asked if I could understand it just with the other answer.
• (Also, english isn't my native language, although I think I made a good job of explaining my issue, sorry if it wasn't clear)

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Edit 2: Wow, stackexchange is awesome. My first question, but it got solved in minutes. Thanks to the people who answered and commented.

Answer 1

In method one, squaring the numerator and denominator of a fraction does (generally) change the value. For example, starting with the fraction $\frac23$, squaring the numerator and denominator gives $\frac49$, which is not equal to the original fraction.

Answer 2

To avoid errors it's best not to use division and instead multiply reciprocals. By this I mean we have $\frac1x = x^{-1}$ and $\sqrt{x} = x^\frac12,$ so $$\frac{\sqrt{x}}{x} = x^{-1}\cdot x^{\frac12} = x^{-\frac12} = \frac1{\sqrt{x}}$$

You've almost erroneously concluded that $\frac{\sqrt{x}}{x} = \frac{x}{x^2}$. To see this substitute $y = \frac {\sqrt{x}}x$ and what you've written becomes $y=y^2$ which is false for almost all numbers, especially negative ones.