Hi,

A rational number is described as being a fraction where both the numerator and denominator are integers. If this is so, why does my book say that $\displaystyle (\surd3)/3$ is a rational number?

Printable View

- Aug 26th 2010, 02:20 PMwebguyRational Number
Hi,

A rational number is described as being a fraction where both the numerator and denominator are integers. If this is so, why does my book say that $\displaystyle (\surd3)/3$ is a rational number? - Aug 26th 2010, 02:23 PMskeeter
- Aug 27th 2010, 01:18 PMwebguy
Book says: To rationalise the fraction $\displaystyle 1/(\surd3)$, multiply both the numerator and denominator by $\displaystyle \surd3$ giving: $\displaystyle (\surd3)/3$.

- Aug 27th 2010, 01:45 PMPlato
There is a failure to communicate there. At least a failure to translate correctly.

**The word***rationalize*does not mean strictlyin mathematics-speak.__to make rational__

It is considered poor form to have a radical in the denominator.

Therefore instead of $\displaystyle \dfrac{2}{\sqrt{3}}$ we*rationalize*that number writing it as $\displaystyle \dfrac{2\sqrt{3}}{3}$ by multiplying by $\displaystyle \dfrac{\sqrt{3}}{\sqrt{3}}$.

In other words we make the denominator rational. - Aug 27th 2010, 04:13 PMwebguy
Understood, thanks :). However, why is it better to have an irrational number as the numerator instead of the denominator?

- Aug 27th 2010, 05:20 PMskeeter
- Aug 27th 2010, 05:23 PMPlato
Yours is actually a very good question.

Before good calculators and/or CAS it was very difficult to divide by an irrational number (actually it still is in practice).

Consider $\displaystyle \dfrac{1}{\sqrt3}$ over against $\displaystyle \dfrac{\sqrt3}{3}$ which is easier to do by ‘hand’?

They are both equal. But the second is easier to work with. - Aug 27th 2010, 06:30 PMSoroban
Hello, webguy!

Edit: Plato beat me to it . . . *sigh*

. . . .But I refuse to delete all this typing . . . so there!

Quote:

Why is it better to have an irrational number in the numerator

instead of the denominator?

Probably the same reason we preferdenominators.*positive*

We know that $\displaystyle \frac{2}{3}$ is read "two-thirds".

And $\displaystyle \frac{\text{-}2}{3}$ is read "negative-two thirds".

And $\displaystyle \text{-}\frac{2}{3}$ is read "negative two-thirds".

But how do we read $\displaystyle \frac{2}{\text{-}3}$ ? . . . Maybe "two negative-thirds" ?

Interesting . . . This problem never came up before. .Why?

Because "they" always use positive denominators at the start.

(And didn't bother to tell us that it's a Rule.)

A bit of trivia . . .

A fraction hassigns: .$\displaystyle \begin{Bmatrix}\text{one in the numerator} \\ \text{one in the denominator} \\ \text{one "out in front"} \end{Bmatrix}$*three*

So that $\displaystyle \dfrac{2}{3}$ is actually: $\displaystyle +\dfrac{+2}{+3}$

We can change*any*__two____signs__

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the question: Why do we "rationalize" denominators?

The main reason arose back in 1000 B.C. (before calculators).

If we need the value of $\displaystyle \frac{1}{\sqrt{2}}$, we have aof work to do!*lot*

First, look up $\displaystyle \sqrt{2}$ on a square-root table and get, say, $\displaystyle 1.4142$

Then we must divide: .$\displaystyle 1 \div 1.4142$ . . . aunplesant task!*very*

It starts like this:

. . $\displaystyle \begin{array}{cccccccc}

&& --&--&--&--&-- \\ 1\;.4\;1\;4\;2 & | & 1 & .0 & 0 & 0 & 0 \end{array}$

We "move the decimal points" and divide:

. . $\displaystyle \begin{array}{cccccccccccccc}

&&&&&&0 & .7 & 0 & 7 & 1 & 1 \\ && --&--&--&--&--&--&--&--&--&-- \\

1.4142 & | & 1 & 0 & 0 & 0 & 0 & .0 & 0 & 0 & 0 & 0 \\

&&& 9 & 8 & 9 & 9 & 4 \\

&& --&--&--&--&--&-- \\

&&&& 1 & 0 & 0 & 6 & 0 & 0 \\

&&&&& 9 & 8 & 9 & 9 & 4 \\

&&&& --&--&--&--&--&-- \\

&&&&&& 1 & 6 & 0 & 6 & 0 \\

&&&&&& 1 & 4 & 1 & 4 & 2 \\

&&&&&&--&--&--&--&-- \\

&&&&&&& 1 & 9 & 1 & 8 & 0 \\

&&&&&&& 1 & 4 & 1 & 4 & 2 \\

&&&&&&& --&--&--&--&-- \\

&&&&&&&&5&0&3&8

\end{array}$

So, to four decimal places, we have: .$\displaystyle \dfrac{1}{\sqrt{2}} \:=\:0.7071$

If rationalize first, we have: .$\displaystyle \dfrac{1}{\sqrt{2}} \:=\:\dfrac{\sqrt{2}}{2}$

And the division looks like this:

. . $\displaystyle \begin{array}{ccccccc}

&& 0 & .7 & 0 & 7 & 1 \\

&&--&--&--&--&-- \\

2 & | & 1 & .4 & 1 & 4 & 2 \\

&& 1 & 4 \\

&& --&-- \\

&&&& 1 & 4 \\

&&&& 1 & 4 \\

&&&& --&-- \\

&&&&&&2 \\

&&&&&& 2 \\

&&&&&& --

\end{array}$

Which way do you prefer?

- Aug 28th 2010, 04:33 AMwebguy
Thank you for all of your answers! I think I can declare this thread as solved :)