Exercise 40 Page 687

We want to write the repeating decimal 0.4336 as a fraction. To do so, we will start by writing this number as a sum of decimals and as a sum of fractions.

Number	0.4336
Sum of Decimals	0.43+0.0036+0.000036+0.00000036+...
Sum of Fractions	43/100+ 36/10 000+36/1 000 000+36/100 000 000+...

Consider the above sum of fractions. Note that from the 2^(nd) term on, we can think of it as a geometric series that has a first term of a_1= 3610 000. To find its common ratio r, we can divide any term of the sequence by its previous term. For simplicity, we will divide a_2 by a_1.

a_2÷ a_1

SubstituteII

a_2= 36/1 000 000, a_1= 36/10 000

36/1 000 000 ÷ 36/10 000

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Evaluate

DivFracByFracDiv

a/b÷c/d=a/b*d/c

36/1 000 000 * 10 000/36

SplitIntoFactors

Split into factors

36/10 000(100) * 10 000/36

CancelCommonFac

Cancel out common factors

36/10 000(100) * 10 000/36

SimpQuotProd

Simplify quotient and product

1/100

Each term of the sequence can be obtained by multiplying the previous term by the common ratio 1100.

Let's substitute r= 1100 and a_1= 3610 000 in the formula for the sum of an infinite geometric series.

S=a_1/1-r

SubstituteII

r= 1/100, a_1= 36/10 000

S=3610 000/1- 1100

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Simplify right-hand side

OneToFrac

Rewrite 1 as 100/100

S=3610 000/100100- 1100

SubFrac

Subtract fractions

S=3610 000/99100

FracToDiv

a/b=a÷ b

S=36/10 000 ÷ 99/100

DivFracByFracDiv

a/b÷c/d=a/b*d/c

S=36/10 000 (100/99)

MultFrac

Multiply fractions

S=3600/990 000

ReduceFrac

a/b=.a /3600./.b /3600.

S=1/275

We found that the series formed by the sum of the terms, starting from the 2^(nd), is 1275. With this information we can express the given number as a sum of two fractions. 0.4336= 43/100+1/275 Finally, we will add the fractions to obtain the value of 0.4336 expressed as a single fraction.

0.4336=43/100+1/275

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Simplify right-hand side

ExpandFrac

a/b=a * 11/b * 11

0.4336=473/1100+1/275

ExpandFrac

a/b=a * 4/b * 4

0.4336=473/1100+4/1100

AddFrac

Add fractions

0.4336=477/1100