Number Theory

Rational and Irrational Numbers

Rational numbers:

A rational number is a number which can be expressed as a ratio of two integers. Non–integer rational numbers (commonly called fractions) are usually written as the fraction \(\frac{a}{b}\), where \(b\) is not zero, \(a\) is called the numerator, and \(b\) the denominator.

Any repeating number with a constant cycle or numbers with terminating digits are also known as rational numbers.

Some of the examples of rational numbers are:
\(\frac{2}{3}\), \(0.3333......\), \(2.25\) etc.

The decimal representation of a real number is called a repeating decimal (or recurring decimal) if at some point it becomes periodic. For example, the decimal representation of \(\frac{1}{3} = 0.3333333..\) becomes periodic just after the decimal point, repeating the single–digit sequence "3" indefinitely. Similarly the fraction \(\frac{1}{7} = 0.142857142857.......\) is repeating with a period of 6 digits. All the repeating numbers are rational numbers. Repeating decimal can be represented using bar lines on the digits. \(\frac{1}{7} = 0.142857142857.... = 0.\overline {142857} \)

Some interesting examples of repeating decimal:

\(\cfrac{1}{7} = 0.\overline {142857} \), \(\cfrac{2}{7} = 0.\overline {285714} \)

\(\cfrac{3}{7} = 0.\overline {428571} \), \(\cfrac{4}{7} = 0.\overline {571428} \)

\(\cfrac{5}{7} = 0.\overline {714285} \), \(\cfrac{6}{7} = 0.\overline {857142} \)

We see that all the six repeating digits are same in all the fractions given above.

Conversion from Repeating Numbers to Fraction:

Note that every repeating decimal can be converted to fraction (\(p/q\) form). Suppose the number is \(0.12\overline {345} \), this number is equal to 0.12345345345345……up to infinite. In this number 12 is non repeating number, 345 is repeating number, number of repeating digits is 3 and number of non repeating digits is 2.  The \(p/q\) form is given by: \[\left[ {\frac{{{\rm{Entire}}\,\,\,\,{\rm{Number}}\; - \;{\rm{Non}}\,\,\,{\rm{repeating}}\,\,\,{\rm{number}}}}{{\mathop {99999..}\limits_{{\rm{(up}}\,{\rm{to}}\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{rep}}{\rm{.}}\,\,{\rm{digits)}}} \mathop {000000000}\limits_{\,\,\,\,\,\,\,\,\,\,{\rm{(upto}}\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{non}}\,\,{\rm{rep}}\,\,{\rm{digits)}}} }}} \right]\]

For example the number \(0.2\overline {34} \), 34 is repeating and 2 is non repeating. The equivalent fraction of this number is \(\frac{{234 - 2}}{{990}} = \frac{{232}}{{990}} = \frac{{116}}{{495}}\)

Example 1: Convert the following numbers in p/q form.
(a) \(0.94\overline {56} \)
(b) \(0.1234\overline {568} \)

(a) \(\frac{{87}}{{92}}\), (b) \(\frac{{10}}{{81}}\)
(a) \(0.94\overline {56} \) =\(\frac{{9456 - 94}}{{9900}}\)= \(\frac{{9362}}{{9900}}\,\, = \,\,\frac{{87}}{{92}}\)
(b) \(0.1234\overline {568} \) =\(\frac{{1234568 - 1234}}{{9990000}}\)= \(\frac{{10}}{{81}}\,\)
Example 2: A number is represented by \(0.{a_1}{a_2}{a_1}{a_2}{a_1}{a_2}..........\), where \({a_1},\,{a_2}\) are the single digits. Find the minimum number greater than 100, that should be multiplied with this number so that the product is always an integer.

198
The number can be written as \(\frac{{{a_1}{a_2}}}{{99}}\), hence it should be multiplied with 99 or multiple of 99. So the minimum number greater than 100 is 2×99 = 198.
Example 3: Find the sum of the numbers \(0.\overline {43} \) and \(0.5\overline {87} \)

\(1.0202...... = 1.\overline {02} \)
Method 1: The numbers are \(\frac{{43}}{{99}} + \frac{{587 - 5}}{{990}} = \frac{{43}}{{99}} + \frac{{582}}{{990}} = \frac{{1012}}{{990}}\)
The sum is more than 1, hence after decimal some part of the number is repeating and some part is non-repeating. As the denominator is 990, hence the numerator must have only two digits repeated and one non-repeating digit after decimal.
\( \Rightarrow \frac{{x - 10}}{{990}} = \frac{{1012}}{{990}}\) or \(x = 1022\)
Hence the number is \(1.0\overline {22} = 1.0\overline 2 \)
Method 2: Write the numbers column wise as
\(\begin{array}{l}0.43434343.......\\0.58787878.......\end{array}\)
Adding the two numbers, we get the sum as, \(1.0202...... = 1.\overline {02} \).
Irrational numbers:

An irrational number is any real number that is not a rational number — that is, it is a number which cannot be expressed as a fraction \(m/n\), where \(m\) and \(n\) are integers, with \(n\) non–zero. The most well known irrational numbers are \(\pi \) and\(\sqrt 2 \).  

  • Sum or difference of two irrational numbers may not be irrational.
  • Product or division of two irrational numbers may not be irrational.
  • Product of one rational and one irrational number may or may not be irrational. For example \(3 \times \sqrt 2 \)is irrational but \(0 \times \sqrt 3 \) is rational. Note that 0 is rational.

The constant \(\pi \) is irrational. 

Example 4: If \(A{\rm{ }} = {\rm{ }}0.{a_1}{a_2}{a_1}{a_2} \ldots  \ldots  \ldots  ..\infty \) and \(B{\rm{ }} = {\rm{ }}0.{b_1}{b_2}{b_3}{b_1}{b_2}{b_3} \ldots  \ldots  \ldots  \ldots ..\infty \), find the minimum positive integer  that should be multiplied to \((A + B)\) so that product is always an integer.

Solution: \(A\) and \(B\) both are repeating decimals

\(A =\frac{{{a_1}{a_2}}}{{99}}\)

\(B = \frac{{{b_1}{b_2}{b_3}}}{{999}}\)

Now the minimum positive integer that should be multiplies to \((A + B)\) to make it an integer will be LCM of 99 and 999 = 10989.