# rationalizing the denominators

• Feb 5th 2010, 05:30 PM
Vicky1997
rationalizing the denominators
Why do I have to rationalize the denominators?

In AMC( American mathematics contest ) contests I see that they never leave an answers as 2/root3 it will always be 2root3/3, but in SAT math tests they have answers such as 2/root3. What is the rule and meaning for rationalizing the denominators?(Wondering)

Thanks.

Vicky.
• Feb 5th 2010, 05:46 PM
Quacky
Times both sides by the denominator
$\displaystyle \frac {2}{\sqrt3}$

If I times that fraction by
$\displaystyle \frac{\sqrt3}{\sqrt3}$, I am actually just multiplying by one. But look at what happens:

$\displaystyle \frac {2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$

$\displaystyle =\frac{2\times\sqrt3}{\sqrt3\times\sqrt3}$

$\displaystyle =\frac{2\sqrt3}{3}$

So the rule is:

For any fraction $\displaystyle \frac{a}{\sqrt{b}}$ multiply by $\displaystyle \frac{\sqrt{b}}{\sqrt{b}}$ to give $\displaystyle \frac{a\sqrt{b}}{b}$

For the purpose of your question, this is how you rationalize the denominator., but for a harder example, you may read on.

However, it gets harder if you have something such as:
$\displaystyle \frac{3}{\sqrt3+4}$

In this scenario, you have to use the 'difference of two squares' to rationalize, by making the denominator a difference of two squares, which is where $\displaystyle (a+b)(a-b)=a^2-b^2$. So apply that to the question, where $\displaystyle a=\sqrt{3}$ and b=4:

$\displaystyle \frac{3}{\sqrt3+4}\times\frac{\sqrt3-4}{\sqrt3-4}$

$\displaystyle =\frac{3(\sqrt{3}-4)}{(\sqrt3+4)(\sqrt3-4)}$

Note the 'difference of two squares' in the denominator? Using the difference of two squares rule above, we can simplify this to:

$\displaystyle =\frac{3(\sqrt{3}-4)}{{\sqrt3}^2-(4^2)}$

$\displaystyle =\frac{3(\sqrt{3}-4)}{3-16}$

$\displaystyle =\frac{3\sqrt3-12}{-13}$
• Feb 5th 2010, 05:58 PM
General
Quote:

Originally Posted by Quacky
$\displaystyle \frac {2}{\sqrt3}$

If I times that fraction by
$\displaystyle \frac{\sqrt3}{\sqrt3}$, I am actually just multiplying by one. But look what happens to the fraction:

$\displaystyle \frac {2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$

$\displaystyle =\frac{2\times\sqrt3}{\sqrt3\times\sqrt3}$

$\displaystyle =\frac{2\sqrt3}{3}$

So the rule is:

For any fraction $\displaystyle \frac{a}{\sqrt{b}}$ multiply by $\displaystyle \frac{\sqrt{b}}{\sqrt{b}}$ to give $\displaystyle \frac{a\sqrt{b}}{b}$

for $\displaystyle b \neq 0$.
• Feb 5th 2010, 06:54 PM
Vicky1997
But I still don't understand the meaning of rationalizing the denominators.

I thought the answer was incorrect if it is not rationalized, because in math contest problems I don't think I have ever seen an answer with a root in the denominator. When I recently did a practice SAT math test, I was surprised to see an answer left as 2/root3.
This is when I realized I was rationalizing my answers without truly understanding why I was doing it.

Why do I have to rationalized the denominator?
And why do certain contests consider the answer incorrect if not rationalized while SAT math allows answers with roots in the denominator?

Vicky.(Thinking)
• Feb 6th 2010, 01:42 PM
Quacky
How would you add fractions such as

$\displaystyle \frac {5}{\sqrt2}+\frac{5}{\sqrt3}$? You'd have to find and work with a common denominator of $\displaystyle \sqrt6$. It would be much easier to solve:

$\displaystyle \frac {5\sqrt2}{2}+\frac{5\sqrt3}{3}$ because you can just look at it and see that the lowest common denominator will be 6.

When dealing with fractions, you will often want simpler terms on the denominator, so rationalization helps a lot. I'm not sure why irrational denominators aren't accepted, though. I suppose they aren't seen as being fully simplified.
• Feb 6th 2010, 02:24 PM
Soroban
Hello, Vicky!

Quote:

Why do I have to rationalize the denominator?

In the Old Days, say 100 B.C. (Before Calculators), we had only our brains and a pencil.

If we want to evaluate, for example, $\displaystyle \frac{1}{\sqrt{2}}$, we would:

. . [1] Find an approximation for $\displaystyle \sqrt{2}$ on a square-root table (1.4142)

. . [2] Perform the long division: .$\displaystyle 1 \div 1.414212$

The divison looks like this: . $\displaystyle \begin{array}{cccc}& - & - - - - - - \\ 1.4142& ) & 1.0000000 \end{array}$

. . $\displaystyle \begin{array}{ccccccccccccc} & & & & & & 0. & 7 & 0 & 7 & 1 & \hdots\\ & & -- & -- & -- & -- & -- & -- & -- & --& -- & --\\ 1\;4\;1\;4\;2 & ) & 1 & 0 & 0 & 0 & 0. & 0 & 0 & 0 & 0 & \hdots\\ &&& 9 & 8 & 9 & 9 & 4 \\ && -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 \\ &&&&&&&& 0 \\ &&&& -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&&& -- & -- & -- & -- & -- \\ \end{array}$

. - - . . . . . . . . . . . . . . . $\displaystyle \begin{array}{cccccccccccc} 1 & 6 & 0 & 6 & 0 \\ 1 & 4 & 1 & 4 & 2 \\ -- & -- & -- & -- & -- \\ & 1 & 9 & 1 & 8 & 0 & \hdots \end{array}$

Pretty tedious, dividing by a 5-digit number.
. . (Worse, if we had more decimal places.)

Rationalizing, we would have: .$\displaystyle \frac{\sqrt{2}}{2}$

And I would much prefer to do: .$\displaystyle 1.414213562... \div 2$

. . Wouldn't you?

And Quacky's view on combining fractions is excellent!

• Feb 6th 2010, 04:25 PM
Quacky
Wow, very impressive answer, did you deduce that from logic, or did you learn it somewhere? (Surprised)
• Feb 6th 2010, 06:48 PM
Vicky1997
Quote:

Originally Posted by Soroban
Hello, Vicky!

In the Old Days, say 100 B.C. (Before Calculators), we had only our brains and a pencil.

If we want to evaluate, for example, $\displaystyle \frac{1}{\sqrt{2}}$, we would:

. . [1] Find an approximation for $\displaystyle \sqrt{2}$ on a square-root table (1.4142)

. . [2] Perform the long division: .$\displaystyle 1 \div 1.414212$

The divison looks like this: . $\displaystyle \begin{array}{cccc}& - & - - - - - - \\ 1.4142& ) & 1.0000000 \end{array}$

. . $\displaystyle \begin{array}{ccccccccccccc}$$\displaystyle & & & & & & 0. & 7 & 0 & 7 & 1 & \hdots\\ & & -- & -- & -- & -- & -- & -- & -- & --& -- & --\\ 1\;4\;1\;4\;2 & ) & 1 & 0 & 0 & 0 & 0. & 0 & 0 & 0 & 0 & \hdots\\ &&& 9 & 8 & 9 & 9 & 4 \\ && -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 \\ &&&&&&&& 0 \\ &&&& -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&&& -- & -- & -- & -- & -- \\ \end{array} . - - . . . . . . . . . . . . . . . \displaystyle \begin{array}{cccccccccccc}$$\displaystyle 1 & 6 & 0 & 6 & 0 \\ 1 & 4 & 1 & 4 & 2 \\ -- & -- & -- & -- & -- \\ & 1 & 9 & 1 & 8 & 0 & \hdots \end{array}$

Pretty tedious, dividing by a 5-digit number.
. . (Worse, if we had more decimal places.)

Rationalizing, we would have: .$\displaystyle \frac{\sqrt{2}}{2}$

And I would much prefer to do: .$\displaystyle 1.414213562... \div 2$

. . Wouldn't you?

And Quacky's view on combining fractions is excellent!

Thank You!!!!(Bow)