rationalizing

rationalizing - Rationalizing denominators and numerators...

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Rationalizing denominators and numerators Introductory example . Suppose that you are given the problem of obtaining an approximate decimal value for the expression 1 - 2 1 given that 2 ~ 1.41421 rounded to 6 figures. Also suppose that you are not allowed to use a calculator . How can you tackle this problem? The straightforward approach is to divide = - 1.41421 1 0.41421 into 1 by performing an "old fashioned" long division. Note that = 1 0 .41421 10 4.1421 , so we can divide 4.1421 into 10 to obtain the required decimal approximation. 2 . 4 1 4 2 3 ___________________ 4 . 1 4 2 1 | 1 0 . 0 0 0 0 0 0 0 0 0 8 . 2 8 4 2 ________ 1 . 7 1 5 8 0 1 . 6 5 6 8 4 ___________ 0 . 0 5 8 9 6 0 0 . 0 4 1 4 2 1 _____________ 0 . 0 1 7 5 3 9 0 0 . 0 1 6 5 6 8 4 _____________ 0 . 0 0 0 9 7 0 6 0 0 . 0 0 0 8 2 8 4 2 ________________ 0 . 0 0 0 1 4 2 1 8 0 0 . 0 0 0 1 2 4 2 6 3 ________________ 0 . 0 0 0 0 1 7 9 1 7 This calculation shows that the expression 1 - 2 1 ~ 2.4142, rounded to 5 digits. An alternative way of evaluating the expression 1 - 2 1 is to use the technique of " rationalizing the denominator ". This technique makes use of the difference of two squares formula: = ( ) - a b ( ) + a b - a 2 b 2 ------- (i). __________ The denominator - 2 1 in the expression 1 - 2 1 is an irrational number. However, the product ( ) - 2 1 ( ) + 2 1 is a rational number. In fact it is the integer 1, since, using (i) with = a 2 and = b 1 gives: = ( ) - 2 1 ( ) + 2 1 - ( ) 2 2 1 2 = - 2 1 = 1. Multiplying both numerator and denominator of 1 - 2 1 by + 2 1 gives:
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1 - 2 1 = + 2 1 ( ) - 2 1 ( ) + 2 1 = + 2 1 1 = + 2 1 ~ 2.41421. ________ This last value agrees with the value 2.4142 obtained previously, if it is rounded to 5 digits. Example 1 : Rationalize the denominator for each of the following. Check your answer by using a calculator to verify that the value obtained by the final expression agrees approximately with the value of the original expression. (a) = 2 3 2 3 (b) 4 - 5 1 (c) 6 + 7 2 (d) 2 - 3 6 (e) 1 - 3 2 (f) + 1 2 + 2 2 Solution : (a) 2 3 = 2 3 = 2 3 . 3
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This note was uploaded on 11/12/2011 for the course MATH 111 taught by Professor Uri during the Spring '08 term at Rutgers.

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rationalizing - Rationalizing denominators and numerators...

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