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When you add two fractions using the above method, the result might not be in lowest terms As a final step in the addition of two fractions, you can reduce the result to lowest terms But you don t always have to By now you must wonder, Does it matter whether or not a fraction is in lowest terms The answer is, It depends If you want to express a fraction in the simplest or most elegant possible way, you should reduce it But you will sometimes come across situations where it s better to put a fraction, ratio, or proportion in a form other than lowest terms

Therefore, the dynamic equations for the error v v , b = b are b p(t) 0 1 0 x(t) = v(t) = 0 0 1 x(t) + 0 0 0 b(t) F and using the de nition of Q in eqn 0 0 0 2 0 = Q = 0 va 2 0 0 b

047 05 056 068 1 2 22 33 47 56 1 2

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Suppose you want to describe how many people in Country X have green eyes, per 100,000 population You might discover that one out of every 10 people has green eyes You could say that 1/10 of the people have green eyes, or the ratio of people with green eyes to all the people in Country X is 1:10 But if you use 100,000 population as a basis, you ll have to say that the proportion of people with green eyes is 10,000 per 100,000 population in Country X That s not even close to the lowest form, but it conveys the intended meaning better In pure theory, it doesn t matter if a fraction, ratio, or proportion is in lowest terms or not These days, nearly everyone uses computers in complicated calculations, and the machines don t care about lowest terms You can input the numbers as they are, and the computer will output the data in any form you want

va (t) b (t)

Here s a challenge!

Start with the general equation for adding two fractions: a /b + c /d = (ad + bc)/(bd ) Based on this, and on the rules you already know, prove that a /b c /d = (ad bc)/(bd ) as long as b 0 and d 0

For any time t, due to the fact that in this example F and are time invariant, the state error covariance can be found in closed form as

1 2 3 4 8 7 6 5

Table 6-2 shows how the subtraction formula is derived from the addition formula Near the end of this proof parentheses and brackets aren t enough for grouping of an expression, so braces must be used! They look exactly the same as they braces you use to enclose lists of symbols representing set elements, but the purpose is different

(4144)

Table 6-2 Derivation of a general formula for the subtraction of one fraction from another, based on the formula for addition of fractions As you read down the left-hand column, each statement is equal to all the statements above it

transmitter is ready, with your music or audio source connected, we can move on to testing the induction loop receiver

Veri cation that eqn (4144) is the solution in Exercise 418 In this example, 1 ( ) = exp (F ) = 0 0

Statements a /b c /d a /b + [ (c /d )] a /b + [ 1(c /d )] a /b + ( 1/1)(c /d ) a /b + ( 1)c /1d a /b + ( 1)c /d [ad + b( 1)c]/bd [ad + ( 1)bc]/bd [ad + ( 1)(bc)]/bd {ad + [ (bc)]}/bd (ad bc)/bd QED Reasons Begin here Convert subtraction to addition of a negative Principle of the sign-changing element Divisive identity element : substitute 1/1 for 1 Multiplication of fractions to right of plus sign Multiplicative identity element: substitite d for 1d Formula for addition of two fractions, considering ( 1)c as a single quantity Commutative law for multiplication Group elements b and c with parentheses Principle of sign-changing element (the other way around) Convert addition of a negative to subtraction Mission accomplished

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