Recall that a rational number is one that can be written as a fraction. As a decimal, it either terminates (so it can be written with a finite number of digits) or it repeats infinitely. Some examples of rational numbers are:
$$
5,\;\;7.5,\;\;\frac{{625}}
{{1000}} = 0.625,\;\;\frac{1}
{3} = 0.33333...
$$
Irrational numbers, on the other hand, are decimals that neither terminate nor repeat. For instance $$\sqrt 2 = {\text{1}}{\text{.41421}}...$$ In fact, the square root of any number that is not a perfect square is irrational.
One task that we sometimes have to do in mathematics problems is to rationalize the denominator or numerator of a fractional expression. This means that we want either the denominator or the numerator of a fraction to be a rational number instead of an irrational one.
WHY? Why do we want to rationalize the denominator or the numerator? It is the same answer, after all, just in a different form. Well, for one thing, this gives us a standard form, which makes it easier to compare answers and combine numbers. More importantly, knowing how to rationalize either the numerator or the denominator will become useful when trying to simplify more complicated expressions. In later math classes, you may not be asked to make sure your answer has a rational denominator, but the method will be useful when solving more difficult problems. As with many symbolic operations, the more ways that you know how to manipulate expressions, the better you will be able to face and solve harder and more involved problems. |
So, let’s look at an example:
Rationalize the denominator:
$$
\frac{5}
{{2\sqrt 7 }}
$$
So, our denominator, right now, contains an irrational number:
$$\sqrt 7 $$. How do we make it rational?
Well, first of all, we can’t just multiply, divide, add, and subtract whatever we want. That would change the value of the expression. All we can really do is multiply by 1 or add 0. It doesn’t seem like either of these are very useful. However, if we multiply by 1 in another form, this method will, in fact, be very useful.
The number, 1, can be written in many different ways. For instance, $$
\frac{3}{3} = 1,\;\;\;\frac{{7.5}}{{7.5}} = 1,\;\;\frac{{ - 107}}{{ - 107}} = 1$$.
So, as long as a fraction we are multiplying by has the same numerator as the denominator, then it equals 1, and we are just fine. Now, let’s look back to our example.
$$
\frac{5}
{{2\sqrt 7 }}
$$
What number can we multiply $$
\sqrt 7 $$ by to make it rational? Certainly multiplying by a rational number will not help. How about multiplying it by itself? If you refer to Simplifying Radicals, you will see that:
$$
\sqrt 7 \cdot \sqrt 7 = \sqrt {7 \cdot 7} = \sqrt {7^2 } = 7
$$
And, 7 is indeed a rational number. So, now we know that if we multiply the bottom by $$
\sqrt 7 $$, it will become rational. However, we need to remember that we have to multiply by 1, otherwise we will change the value of the expression. So, let’s multiply by 1 in the form of:
$$
\frac{{\sqrt 7 }}
{{\sqrt 7 }}
$$
So, we get:
$$
\frac{5}
{{2\sqrt 7 }} \cdot \frac{{\sqrt 7 }}
{{\sqrt 7 }} = \frac{{5\sqrt 7 }}
{{2 \cdot 7}} = \frac{{5\sqrt 7 }}
{{14}}
$$
And, we are done. Our denominator is rationalized (although, now, our numerator is not).
Rationalize the denominator:
$$
\frac{6}
{{3 - \sqrt 5 }}
$$
When we first look at this problem, we may think that we can just multiply it by $$
\frac{{\sqrt 5 }}{{\sqrt 5 }}$$, since there is a $$\sqrt 5 $$
in the denominator. However, what happens when we do this? Multiplying the denominator by $$
\sqrt 5 $$, gives us:
$$
\sqrt 5 \left( {3 - \sqrt 5 } \right) = 3\sqrt 5 - 5
$$
So, that didn’t work! We still have an irrational number in the denominator. So, when we have a square root added to or subtracted from a rational number, it just won’t cut it to multiply it by the same square root. This is where the conjugate comes in.
The conjugate of $$a + b\sqrt m $$ is $$a - b\sqrt m$$. In other words, the conjugate is the same as the original expression, except the sign of the radical changes. The conjugate is very useful to us, because when we multiply an expression by its conjugate, we get a rational expression. Let’s see why this happens with our example:
The conjugate of $$3 - \sqrt 5 $$ is $$3 + \sqrt 5 $$. Multiplying them together, we get:
$$
\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right) = 9 + 3\sqrt 5 - 3\sqrt 5 - 5 = 4
$$
Thus, we got rid of the radicals, because using the distributive property over the polynomials gave us a positive and negative of the same radical – therefore cancelling it out. So, now we know what we need to multiply the top and bottom by (multiplying by 1 in the form of a fraction with the same denominator and numerator) to rationalize the denominator – the conjugate of the denominator:
So, we have:
$$\frac{6}{{3 - \sqrt 5 }} \cdot \frac{{3 + \sqrt 5 }}{{3 + \sqrt 5 }} = \frac{{6\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}} = \frac{{18 + 6\sqrt 5 }}{4}$$
Now, we have rationalized the denominator, but to finish simplifying this expression, we need to divide any common factors in the top and bottom. Notice that all of the terms in the numerator as well as the denominator are even, so they are all divisible by 2. So, we can divide (factor) 2 out of each of the top terms, and out of the bottom to get:
$$\frac{{18 + 6\sqrt 5 }}{4} = \frac{{2\left( {9 + 3\sqrt 5 } \right)}}{{2 \cdot 2}}$$
So, we can rewrite the right hand side as the product:
$$\frac{2}{2} \cdot \frac{{\left( {9 + 3\sqrt 5 } \right)}}{2}$$
Thus, since $$\frac{2}{2} = 1$$, our final answer is:
$$\frac{{\left( {9 + 3\sqrt 5 } \right)}}{2}$$
Now, we are done.
Rationalize the numerator:
$$\frac{{2 + 7\sqrt 3 }}{{1 - 2\sqrt 5 }}$$
So, here, we just need to rationalize the numerator – not the denominator. [In fact, we can’t rationalize both the numerator and the denominator unless the expression can be reduced to a rational number – in other words, a fraction. But if it is irrational, either the top or the bottom must remain an irrational number.]
We use the same technique to rationalize the numerator as we did with the denominator. We take the conjugate (of the numerator this time!) and multiply the whole expression by the conjugate over itself (which is another way of multiplying by 1).
The conjugate of $$2 + 7\sqrt 3 $$ is $$2 - 7\sqrt 3 $$. So, we multiply by 1 in the form of $$
\frac{{2 - 7\sqrt 3 }}{{2 - 7\sqrt 3 }}$$:
$$
\frac{{2 + 7\sqrt 3 }}
{{1 - 2\sqrt 5 }} \cdot \frac{{2 - 7\sqrt 3 }}
{{2 - 7\sqrt 3 }} = \frac{{\left( {4 - 14\sqrt 3 + 14\sqrt 3 + 49 \cdot 3} \right)}}
{{\left( {2 - 7\sqrt 3 - 4\sqrt 5 + 14\sqrt {5 \cdot 3} } \right)}}
$$
Simplifying, we get:
$$
\frac{{\left( {4 - 14\sqrt 3 + 14\sqrt 3 + 49 \cdot 3} \right)}}
{{\left( {2 - 7\sqrt 3 - 4\sqrt 5 + 14\sqrt {5 \cdot 3} } \right)}} = \frac{{151}}
{{2 - 7\sqrt 3 - 4\sqrt 5 + 14\sqrt {15} }}
$$
The denominator doesn’t look very nice now, but there is nothing more we can combine or cancel, and our numerator is rationalized, so we are done.
Some practice problems to check your skills:
1. Rationalize the denominator and simplify your result:
$$
\frac{6}
{{3\sqrt {17} }}
$$
Solution: $$
\frac{{2\sqrt {17} }}
{{17}}
$$
2. Rationalize the denominator and simplify your result:
$$
\frac{{16}}
{{1 + \sqrt 5 }}
$$
Solution: $$
\frac{{4\left( {1 - \sqrt 5 } \right)}}
{{ - 1}} = - 4\left( {1 - \sqrt 5 } \right) = - 4 + 4\sqrt 5
$$
3. Rationalize the numerator and simplify your result:
$$
\frac{{\sqrt {19} }}
{{4\sqrt 2 }}
$$
Solution: $$
\frac{{19}}
{{4\sqrt {38} }}
$$
4. Rationalize the numerator and simplify your result:
$$
\frac{{4 - 7\sqrt 2 }}
{6}
$$
Solution: $$
\frac{{ - 41}}
{{3\left( {4 + 7\sqrt 2 } \right)}} = \frac{{ - 41}}
{{12 + 21\sqrt 2 }}
$$
5. Rationalize the denominator and simplify your result:
$$
\frac{{2 - \sqrt 7 }}
{{4 + \sqrt 7 }}
$$
Solution: $$
\frac{{16 - 6\sqrt 7 }}
{9}
$$