1. ## Rational 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?

2. Originally Posted by webguy
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?
$\displaystyle \frac{\sqrt{3}}{3}$ is not a rational number ... are you sure that is what the book says? a direct quote would be helpful.

3. Book says: To rationalise the fraction $\displaystyle 1/(\surd3)$, multiply both the numerator and denominator by $\displaystyle \surd3$ giving: $\displaystyle (\surd3)/3$.

4. Originally Posted by webguy
Book says: To rationalise the fraction $\displaystyle 1/(\surd3)$, multiply both the numerator and denominator by $\displaystyle \surd3$ giving: $\displaystyle (\surd3)/3$.
There is a failure to communicate there. At least a failure to translate correctly.
The word rationalize does not mean strictly to make rational in mathematics-speak.
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.

5. Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?

6. Originally Posted by webguy
Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?
goes back to the days before calculators ... making decimal approximations was easier with a whole number in the denominator.

7. Originally Posted by webguy
Understood, thanks . However, why is it better to have an irrational number as the numerator instead of the denominator?
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.

8. Hello, webguy!

Edit: Plato beat me to it . . . *sigh*
. . . .But I refuse to delete all this typing . . . so there!

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

Probably the same reason we prefer positive denominators.

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 has three signs: .$\displaystyle \begin{Bmatrix}\text{one in the numerator} \\ \text{one in the denominator} \\ \text{one "out in front"} \end{Bmatrix}$
So that $\displaystyle \dfrac{2}{3}$ is actually: $\displaystyle +\dfrac{+2}{+3}$

We can change any two signs without changing the value of the fraction.

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

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 a lot of work to do!

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

Then we must divide: .$\displaystyle 1 \div 1.4142$ . . . a very unplesant task!

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?

9. Thank you for all of your answers! I think I can declare this thread as solved