Number Systems

1. 1. Write the following in decimal form and say what kind of decimal expansion each has:
   
   
   1. \(\frac{36}{100}\)
      Solution:
      We divide \(36\) by \(100\). \[ \frac{36}{100}=0.36 \] Since the digits stop after two decimal places, the decimal expansion is Terminating.
   
   
   2. \(\frac{1}{11}\)
      Solution:
      Divide \(1\) by \(11\): \[\require{enclose} \begin{array}{l|l} \phantom{100}0.090909\ldots \\ 11\enclose{longdiv}{1.000000} \\ \underline{\phantom{1000}99\phantom{10000}}\\ \phantom{10000}100\\ \underline{\phantom{100000}99\phantom{100}}\\ \phantom{1000000}100\\ \underline{\phantom{10000000}99\phantom{1}}\\ \phantom{100000000}1\ldots \end{array}\] Therefore, \[ \frac{1}{11}=0.\overline{09} \] The block \(09\) repeats continuously. Hence, the decimal expansion is Non-terminating recurring.
   
   
   3. \(4\frac{1}{8}\)
      Solution:
      First convert the mixed fraction into an improper fraction. \[ 4\frac{1}{8} = \frac{(4\times8)+1}{8} = \frac{32+1}{8} = \frac{33}{8} \] Now divide \(33\) by \(8\): \[ \require{enclose} \begin{array}{l|l} \phantom{100}4.125 \\\ 8\enclose{longdiv}{33.000} \\ \underline{\phantom{10}32\phantom{1000}} \\ \phantom{100}10 \\ \underline{\phantom{1000} 8\phantom{100}}\\ \phantom{1000}20 \\ \underline{\phantom{10000}16\phantom{1}} \\ \phantom{100000}40 \\ \underline{\phantom{100000}40} \\ \phantom{1000000}0 \end{array} \] Therefore, \[ 4\frac{1}{8}=4.125 \] Since the decimal digits stop, it is a Terminating decimal expansion.
   
   
   4. \(\frac{3}{13}\)
      Solution:
      Divide \(3\) by \(13\): \[ \require{enclose} \begin{array}{l|l} \phantom{10}0.2307692307692\ldots \\ 13\enclose{longdiv}{3.000000000} \\ \underline{\phantom{100}26\phantom{100000000}} \\ \phantom{1000}40 \\ \underline{\phantom{1000}39\phantom{10000000}} \\ \phantom{10000}100 \\ \underline{\phantom{100000}91\phantom{100000}} \\ \phantom{1000000}90 \\ \underline{\phantom{1000000}78\phantom{10000}} \\ \phantom{1000000}120 \\ \underline{\phantom{1000000}117\phantom{1000}} \\ \phantom{100000000}30\\ \underline{\phantom{100000000}26\phantom{100}}\\ \phantom{1000000000}4\ldots \end{array} \] Therefore, \[ \frac{3}{13}=0.\overline{230769} \] The digits \(230769\) repeat continuously. Hence, it is Non-terminating recurring.
   
   
   5. \(\frac{2}{11}\)
      Solution:
      Divide \(2\) by \(11\): \[ \require{enclose} \begin{array}{l|l} \phantom{100}0.181818\ldots \\ 11\enclose{longdiv}{2.00000} \\ \underline{\phantom{100}11\phantom{10000}} \\ \phantom{1000}90 \\ \underline{\phantom{1000}88\phantom{1000}} \\ \phantom{10000}20 \\ \underline{\phantom{10000}11\phantom{100}} \\ \phantom{100000}90 \\ \underline{\phantom{100000}88\phantom{10}} \\ \phantom{1000000}2 \end{array} \] Therefore, \[ \frac{2}{11}=0.\overline{18} \] The digits \(18\) repeat continuously. Hence, the decimal expansion is Non-terminating recurring.
   
   
   6. \(\frac{329}{400}\)
      Solution:
      Divide \(329\) by \(400\): \[ \require{enclose} \begin{array}{l|l} \phantom{1000}0.8225 \\ 400\enclose{longdiv}{329.0000} \\ \underline{\phantom{1000}3200\phantom{1000}} \\ \phantom{100000}900 \\ \underline{\phantom{100000}800\phantom{100}} \\ \phantom{100000}1000 \\ \underline{\phantom{1000000}800\phantom{10}} \\ \phantom{1000000}2000 \\ \underline{\phantom{1000000}2000\phantom{1}}\\ \phantom{1000000000}0 \end{array} \] Therefore, \[ \frac{329}{400}=0.8225 \] Since the decimal expansion stops after four decimal places, it is a Terminating decimal expansion.

It is given that: \[ \frac{1}{7}=0.\overline{142857} \] Now multiply both sides successively.

For \(\frac{2}{7}\): \[ \begin{aligned} \frac{2}{7} &=2\times\frac{1}{7}\\ &=2\times0.\overline{142857}\\ &=0.\overline{285714} \end{aligned} \]
For \(\frac{3}{7}\): \[ \begin{aligned} \frac{3}{7} &=3\times\frac{1}{7}\\ &=3\times0.\overline{142857}\\ &=0.\overline{428571} \end{aligned} \]
For \(\frac{4}{7}\): \[ \begin{aligned} \frac{4}{7} &=4\times\frac{1}{7}\\ &=4\times0.\overline{142857}\\ &=0.\overline{571428} \end{aligned} \]
For \(\frac{5}{7}\): \[ \begin{aligned} \frac{5}{7} &=5\times\frac{1}{7}\\ &=5\times0.\overline{142857}\\ &=0.\overline{714285} \end{aligned} \]
For \(\frac{6}{7}\): \[ \begin{aligned} \frac{6}{7} &=6\times\frac{1}{7}\\ &=6\times0.\overline{142857}\\ &=0.\overline{857142} \end{aligned} \]
Hence, \[ \boxed{\bbox[5pt]{ \begin{aligned} \frac{2}{7}&=0.\overline{285714}\\ \frac{3}{7}&=0.\overline{428571}\\ \frac{4}{7}&=0.\overline{571428}\\ \frac{5}{7}&=0.\overline{714285}\\ \frac{6}{7}&=0.\overline{857142} \end{aligned} }} \]

1. Let \[ x=0.\overline{6} \] Therefore, \[ x=0.6666\ldots \tag{1} \] Multiply both sides by \(10\): \[ 10x=6.6666\ldots \tag{2} \] Subtract equation (1) from equation (2): \[ \require{cancel} \begin{aligned} 10x &= 6.6666\ldots \\ x &= 0.6666\ldots \\ \hline 9x &= 6 \end{aligned} \] Divide both sides by \(9\): \[ x=\frac{6}{9} \] Simplifying, \[ x=\frac{\cancelto{2}{6}}{\cancelto{3}{9}} \] Hence, \[ x=\frac{2}{3} \] Therefore, \[ 0.\overline{6}=\frac{2}{3} \]



2. \(0.4\overline{7}\)
   
   Solution:
   Let \[ x=0.4\overline{7} \] Therefore, \[ x=0.47777\ldots \tag{1} \] Since one non-repeating digit exists before the repeating digit, multiply by \(10\): \[ 10x=4.77777\ldots \tag{2} \] Now multiply again by \(10\): \[ 100x=47.77777\ldots \tag{3} \] Subtract equation (2) from equation (3): \[ \begin{aligned} 100x &= 47.77777\ldots \\ 10x &= 4.77777\ldots \\ \hline 90x &= 43 \end{aligned} \] Divide both sides by \(90\): \[ x=\frac{43}{90} \] Therefore, \[ 0.4\overline{7}=\frac{43}{90} \]



3. \(0.\overline{001}\)
   
   Solution:
   Let \[ x=0.\overline{001} \] Therefore, \[ x=0.001001001\ldots \tag{1} \] Here the repeating block contains \(3\) digits. Multiply both sides by \(1000\): \[ 1000x=1.001001001\ldots \tag{2} \] Subtract equation (1) from equation (2): \[ \begin{aligned} 1000x &= 1.001001001\ldots \\ x &= 0.001001001\ldots \\ \hline 999x &= 1 \end{aligned} \] Divide both sides by \(999\): \[ x=\frac{1}{999} \] Hence, \[ 0.\overline{001}=\frac{1}{999} \]

Let \[ x=0.99999\ldots \tag{1} \] Multiply both sides of equation (1) by \(10\): \[ 10x=9.99999\ldots \tag{2} \] Subtract equation (1) from equation (2): \[ \require{cancel} \begin{aligned} 10x &= 9.99999\ldots \\ x &= 0.99999\ldots \\ \hline 9x &= 9 \end{aligned} \] Divide both sides by \(9\): \[ x=\frac{9}{9} \] Therefore, \[ x=1 \] Hence, \[ 0.99999\ldots = 1 \].

In the decimal expansion of \(\frac{1}{17}\), the remainder obtained during division determines the repeating digits. While dividing by \(17\), the possible non-zero remainders are: \[ 1,2,3,4,\ldots,16 \] Therefore, the maximum possible number of digits in the repeating block is: \[ 16 \] Now let us verify this by performing long division.

\[ \require{enclose} \begin{array}{l} \phantom{100}0.0588235294117647\ldots \\ 17\enclose{longdiv}{1.0000000000000000} \\ \underline{\phantom{10000}85\phantom{10000000000000}} \\ \phantom{10000}150 \\ \underline{\phantom{10000}136\phantom{1000000000000}} \\ \phantom{10000}140 \\ \underline{\phantom{10000}136\phantom{1000000000000}} \\ \phantom{1000000}40 \\ \underline{\phantom{1000000}34\phantom{100000000000}} \\ \phantom{10000000}60 \\ \underline{\phantom{10000000}51\phantom{10000000000}} \\ \phantom{100000000}90 \\ \underline{\phantom{100000000}85\phantom{1000000000}} \\ \phantom{1000000000}50 \\ \underline{\phantom{1000000000}34\phantom{100000000}} \\ \phantom{1000000000}160 \\ \underline{\phantom{1000000000}153\phantom{10000000}} \\ \phantom{100000000000}70 \\ \underline{\phantom{100000000000}68\phantom{1000000}} \\ \phantom{1000000000000}20 \\ \underline{\phantom{1000000000000}17\phantom{100000}} \\ \phantom{10000000000000}30 \\ \underline{\phantom{10000000000000}17\phantom{10000}} \\ \phantom{10000000000000}130 \\ \underline{\phantom{10000000000000}119\phantom{1000}} \\ \phantom{100000000000000}110 \\ \underline{\phantom{100000000000000}102\phantom{100}} \\ \phantom{1000000000000000}80 \\ \underline{\phantom{1000000000000000}68\phantom{100}} \\ \phantom{1000000000000000}120 \\ \underline{\phantom{1000000000000000}119\phantom{10}} \\ \phantom{100000000000000000}10 \end{array} \] From the division, we get: \[ \frac{1}{17} = 0.\overline{0588235294117647} \] The repeating block is: \[ 0588235294117647 \] Counting the digits: \[ 16 \text{ digits} \] Hence, the maximum number of digits in the repeating block is: \[ \boxed{16} \]

Let us observe some rational numbers having terminating decimal expansions.

Rational Number	Decimal Expansion	Prime Factors of Denominator
\(\frac{1}{2}\)	\(0.5\)	\(2\)
\(\frac{3}{4}\)	\(0.75\)	\(2^2\)
\(\frac{7}{5}\)	\(1.4\)	\(5\)
\(\frac{9}{20}\)	\(0.45\)	\(2^2\times5\)
\(\frac{11}{125}\)	\(0.088\)	\(5^3\)

From the above examples, we observe that:

• The denominator contains only the prime factor \(2\),
• or only the prime factor \(5\),
• or both \(2\) and \(5\).

Therefore, we conclude: \[ \boxed{ \text{A rational number has a terminating decimal expansion} } \] \[ \boxed{ \text{if the denominator } q \text{ has only prime factors } 2 \text{ and/or } 5. } \] In other words, \[ q=2^m\times5^n \] where \(m\) and \(n\) are non-negative integers.

Three numbers whose decimal expansions are non-terminating and non-recurring are:

1. \[ \pi \approx 3.1415926535\ldots \] The digits continue infinitely without repeating in a fixed pattern.


2. \[ e \approx 2.7182818284\ldots \] This is another irrational number with a non-terminating non-recurring decimal expansion.


3. \[ \sqrt{2}\approx 1.4142135623\ldots \] Since \(\sqrt{2}\) cannot be expressed as a rational number, its decimal expansion is non-terminating and non-recurring.

First convert the given rational numbers into decimal form. \[ \frac{5}{7}\approx0.714285\ldots \] \[ \frac{9}{11}\approx0.818181\ldots \] Therefore, we need irrational numbers lying between: \[ 0.714285\ldots \quad \text{and} \quad 0.818181\ldots \] We know the following irrational numbers: \[ \pi \approx 3.141592653\ldots \] \[ e \approx 2.718281828\ldots \] \[ \sqrt{2}\approx1.414213562\ldots \] Now choose suitable rational numbers to subtract so that the results lie within the required interval.

First irrational number: \[ \pi-2.4 \] \[ \approx3.141592653-2.4 \] \[ \approx0.741592653\ldots \] Since \[ 0.714285\ldots < 0.741592653\ldots < 0.818181\ldots \] therefore, \[ \pi-2.4 \] is an irrational number between the given rational numbers.

Second irrational number: \[ \sqrt{2}-0.6 \] \[ \approx1.414213562-0.6 \] \[ \approx0.814213562\ldots \] Since \[ 0.714285\ldots < 0.814213562\ldots < 0.818181\ldots \] therefore, \[ \sqrt{2}-0.6 \] is an irrational number between the given rational numbers.

Third irrational number: \[ e-2 \] \[ \approx2.718281828-2 \] \[ \approx0.718281828\ldots \] Since \[ 0.714285\ldots < 0.718281828\ldots < 0.818181\ldots \] therefore, \[ e-2 \] is also an irrational number between the given rational numbers.

Hence, three irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) are: \[ \boxed{ \pi-2.4,\; \sqrt{2}-0.6,\; e-2 } \]

1. \[ \sqrt{23} \]
   Solution:
   We know that: \[ 4^2=16 \] and \[ 5^2=25 \] Since \[ 16<23<25 \] therefore, \[ 4<\sqrt{23}<5 \] The decimal expansion of \(\sqrt{23}\) is: \[ \sqrt{23}\approx4.7958315\ldots \] The decimal expansion is non-terminating and non-recurring. Hence, \[ \boxed{\sqrt{23}\text{ is an Irrational Number}} \]



2. \[ \sqrt{225} \]
   Solution:
   Prime factorisation of \(225\): \[ 225=5\times5\times3\times3 \] Therefore, \[ \sqrt{225} = \sqrt{5^2\times3^2} \] \[ \sqrt{225}=5\times3 \] \[ \sqrt{225}=15 \] Since \(15\) is an integer and can be written as: \[ \frac{15}{1} \] therefore, \[ \boxed{\sqrt{225}\text{ is a Rational Number}} \]



3. \[ 0.3796 \]
   Solution:
   The decimal expansion: \[ 0.3796 \] terminates after four decimal places. Therefore, it is a terminating decimal. Every terminating decimal is rational. Hence, \[ \boxed{0.3796\text{ is a Rational Number}} \]



4. \[ 7.478478478\ldots \]
   Solution:
   Observe that the block: \[ 478 \] repeats continuously. Therefore, \[ 7.478478478\ldots = 7.\overline{478} \] This is a non-terminating recurring decimal. Every recurring decimal is rational. Hence, \[ \boxed{ 7.478478478\ldots \text{ is a Rational Number} } \]



5. \[ 1.101001000100001\ldots \]
   Solution:
   Observe the decimal expansion carefully:
   • The decimal expansion never terminates.
   • The digits do not repeat in any fixed pattern.
   • The number of zeros between \(1\)'s keeps increasing.
   Therefore, the decimal expansion is: \[ \text{Non-terminating and Non-recurring} \] Hence, \[ \boxed{ 1.101001000100001\ldots \text{ is an Irrational Number} } \]