How do I rationalise the denominator of a fraction which consists of surds?
[Recall: the numerator of a fraction is the top number; the denominator refers to the bottom number. A surd is an irrational number, e.g. √3, √5, etc.]Given the following fraction:(a+√b)/(c-√b), where a,b and c are non-negative integers and b is not a square number.We can see that the denominator (c-√b) is irrational. To rationalise the denominator we take the following steps:1. Multiply BOTH the numerator and the denominator of our fraction by (c+√b) in order to eliminate the irrational surd in the denominator. Note: we perform this multiplication to both the numerator and denominator in order to preserve the value of the original fraction .2. We now have for our numerator: (a+√b)(c+√b), and for our denominator: (c-√b)(c+√b). Expand these brackets, thus we obtain the following fraction:(ac+(a+c)√b+b) / (c2 - b)Clearly we have succeeded in rationalising our denominator (whilst still maintaining the value of our original fraction) since (c2-b) is clearly a rational number, as required.Example:Write (5+7√3)/(5-√3) in the form a+b√3, where a and b are rational.Soln: We carry out the steps stated above;Multiply numerator and denominator by (5+√3), in doing so eliminating the irrational surd from our denominator. We thus obtain:{(5+7√3)(5+√3)} / {(5-√3)(5+√3)} (expand brackets)=(46+12√3) / (25+5√3-5√3 - 3)=(46+12√3) / (22)=23/11+(6/11)√3.Clearly, from our orginal hypothesis, a=23/11, b=6/11 are both rational numbers, thus we are done.
