Rational Number

**webguy** · August 26th 2010, 02:20 PM

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 $(\surd3)/3$ is a rational number?

**skeeter** · August 26th 2010, 02:23 PM

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 $(\surd3)/3$ is a rational number?

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

**webguy** · August 27th 2010, 01:18 PM

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

**Plato** · August 27th 2010, 01:45 PM

Originally Posted by webguy

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

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 $\dfrac{2}{\sqrt{3}}$ we rationalize that number writing it as $\dfrac{2\sqrt{3}}{3}$ by multiplying by $\dfrac{\sqrt{3}}{\sqrt{3}}$ .

In other words we make the denominator rational.

**webguy** · August 27th 2010, 04:13 PM

Understood, thanks

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

**skeeter** · August 27th 2010, 05:20 PM

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.

**Plato** · August 27th 2010, 05:23 PM

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 $\dfrac{1}{\sqrt3}$ over against $\dfrac{\sqrt3}{3}$ which is easier to do by ‘hand’?
They are both equal. But the second is easier to work with.

**Soroban** · August 27th 2010, 06:30 PM

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
instead of the denominator?

Probably the same reason we prefer positive denominators.

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

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

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

But how do we read $\frac{2}{\text{-}3}$ ? . . . 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: . $\begin{Bmatrix}\text{one in the numerator} \\ \text{one in the denominator} \\ \text{one "out in front"} \end{Bmatrix}$
So that $\dfrac{2}{3}$ is actually: $+\dfrac{+2}{+3}$

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 $\frac{1}{\sqrt{2}}$ , we have a lot of work to do!

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

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

It starts like this:

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

We "move the decimal points" and divide:

. . $\begin{array}{cccccccccccccc} &&&&&&0 & .7 & 0 & 7 & 1 & 1 \\ && --&--&--&--&--&--&--&--&--&-- \\ 1.4142 & | & 1 & 0 & 0 & 0 & 0 & .0 & 0 & 0 & 0 & 0 \\ &&& 9 & 8 & 9 & 9 & 4 \\ && --&--&--&--&--&-- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&& --&--&--&--&--&-- \\ &&&&&& 1 & 6 & 0 & 6 & 0 \\ &&&&&& 1 & 4 & 1 & 4 & 2 \\ &&&&&&--&--&--&--&-- \\ &&&&&&& 1 & 9 & 1 & 8 & 0 \\ &&&&&&& 1 & 4 & 1 & 4 & 2 \\ &&&&&&& --&--&--&--&-- \\ &&&&&&&&5&0&3&8 \end{array}$

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

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

And the division looks like this:

. . $\begin{array}{ccccccc} && 0 & .7 & 0 & 7 & 1 \\ &&--&--&--&--&-- \\ 2 & | & 1 & .4 & 1 & 4 & 2 \\ && 1 & 4 \\ && --&-- \\ &&&& 1 & 4 \\ &&&& 1 & 4 \\ &&&& --&-- \\ &&&&&&2 \\ &&&&&& 2 \\ &&&&&& -- \end{array}$

Which way do you prefer?

**webguy** · August 28th 2010, 04:33 AM

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

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