Chapter 6.6, Problem 36WE

(a)

To determine

To Explain:

Why the statement is incorrect.A student who used long division to find the repeating decimal for $\frac{1}{17}$ claimed that there were 20 digits in the repeating block of digits.

Answer to Problem 36WE

Since it has 16 digits in the repeating block of digits.

Explanation of Solution

Calculation:

Given

  $\frac{1}{17}$

Since the denominator is prime and does not have any factor of 2 or 5,

and any number of the form $\frac{1}{x}$ will have atmost (x-1) possible remainders as you divide 1 by x .You will only stop if you reach 0.

But if you reach same number twice, then it will start to repeat.

So, any number of the form $\frac{1}{p}$ , where p is prime will have maximum (p-1) number of digits in the repeating block of the decimal expansion.

So, $\frac{1}{17}$ will have maximum (17-1) = 16 digits in the repeating block of the decimal expansion.

But the student in the given statement states that there are 20 digits in the repeating block , which cannot be true since maximum possible numbers in the repeating block are 16.

  $\frac{1}{17}=0.\overline{0588235294117647}$

So, the number of digits in the repeating block of the decimal expansion is 16.

(b)

To determine

To Find:

The repeating decimal for $\frac{1}{17}$ .

Answer to Problem 36WE

  $\frac{1}{17}=0.\overline{0588235294117647}$

Explanation of Solution

Calculation:

Given

  $\frac{1}{17}$

By long division method:

So, $\frac{1}{17}=0.\overline{0588235294117647}$