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Disclosure: This is homework, but not part of the homework. This is just something that I do not understand.

$$ x = \sqrt{\frac{5}{3}} $$

$$ x = \frac{\sqrt{15}}{3} $$

Could anyone please explain this to me?

Thanks in advance.

Community
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JFW
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    See also this post: http://math.stackexchange.com/questions/47748/fractions-with-radicals-in-the-denominator/47855#47855 – Martin Sleziak Nov 13 '11 at 16:23
  • BTw, you can use LaTeX makeup by starting with an $ and ending with another $, for instance $\sqrt{\frac{5}{3}}$ becomes $\sqrt{\frac{5}{3}}$. – FUZxxl Nov 13 '11 at 16:26
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    If you only want to verify that the two are the same, (note that both are positive and) square them: $x^2$ in the former is $\displaystyle \frac{5}{3}$ and in the latter is $\displaystyle \frac{15}{9}$ and you probably know how to verify that the two are the same. – ShreevatsaR Nov 13 '11 at 17:46

5 Answers5

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$$ x= \sqrt{\frac{5}{3}} = \frac{\sqrt{5}}{\sqrt{3}} =\frac{\sqrt{5} \times\sqrt{3} }{\sqrt{3}\times \sqrt{3}} = \frac{\sqrt{15}}{3}$$

This is called rationalizing the denominator, you can practice more here.

Quixotic
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Well lets think about it. All we can do to change this so that it is exactly equal is to multiply the top and bottom by the same value. We can also easily say that $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

Putting this into terms of this question... $\frac{\sqrt{5}}{\sqrt{3}} = \sqrt{\frac{5}{3}}$

We can also that that $\sqrt{n} \times \sqrt{n} = n$. In other words, the square root of any positive number $n$ times itself is equal to $n$. Since the bottom number is equal to 3 in the converted from, we can simply multiply the numerator and denominator by $\sqrt{3}$

$\frac{\sqrt{5}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{15}}{\sqrt{9}} = \frac{\sqrt{15}}{3}$

I hope that clears things up!

Cygnini - ShadowRi5ing
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If you have the root $\sqrt{5/3}$, you can simply extend by three, yielding $\sqrt{15/9}$. Then you can proceed by the laws for roots and get $$\sqrt{15\over9} = \frac{\sqrt{15}}{\sqrt9} = \frac{\sqrt{15}}3$$

FUZxxl
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Multiply top and bottom of the fraction by $\sqrt{3}$ and you get $\sqrt{3} \cdot \sqrt{5}=\sqrt{15}$ on top and $\sqrt{3} \cdot \sqrt{3}=3$ on the bottom. The trick is to multiply the fraction by the bottom square root, thus getting rid of the square root in the bottom of the fraction. Mathematicians don't like square roots on the bottom of fraction :)

Catherine
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    "Mathematicians don't like square roots on the bottom of fraction :)" - I am arguably not a mathematician, but I prefer $\sqrt{5/3}$ to $\sqrt{15}/3$; in fact, writing $\sin 45^{\circ} = \sqrt{2}/2$ instead of $1/\sqrt{2}$ looks funny to me. I think it is high school math teachers who have some (unnecessary, IMO) aversion to denominators containing radical signs. – Srivatsan Nov 13 '11 at 20:41
  • I was only joking if you notice the smile ":)" – Catherine Nov 13 '11 at 22:26
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    Yes, I did recognize that you were joking. But considering that this site will be visited by new users as well, it's good to point that out explicitly as well, I feel. – Srivatsan Nov 13 '11 at 22:28
  • fair enough , but I am a new user too, so I think you can let me off this time :) – Catherine Nov 13 '11 at 22:41
  • Sorry if I came across as picky. I didn't intend to fault you when I wrote that comment; just clarifying my stand. =) – Srivatsan Nov 13 '11 at 23:11
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$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$

Since $\frac{\sqrt{b}}{\sqrt{b}}=1$, then we can multiply the right hand side of the above expression by $\frac{\sqrt{b}}{\sqrt{b}}$, we get:

$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}$$

Note that $\sqrt{a} \times \sqrt{b}=\sqrt{ab}$ , and note that $\sqrt{b} \times \sqrt{b}=\sqrt{b}^2=b$.

Therefore,

$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}$$

In this case,

$$\sqrt{\frac{5}{3}}=\frac{\sqrt{5}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{5 \cdot 3}}{3}=\frac{\sqrt{15}}{3}$$

This question was asked long time before, but I answer it in his way because I think someone will see his answer and will get use of it.

Hussain-Alqatari
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