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?
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.
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.
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 $\displaystyle \frac{2}{3}$ 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?