Lesson RATIONAL AND IRRATIONAL NUMBERS
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RATIONAL NUMBER __________________ Any number which {{{red(can)}}} be expressed in the form {{{p/q}}} where 'p' and 'q' (q not equal to 1) are integers mutually prime to each other (this means 'p' and 'q' have no common factors; in other words H.C.F. of 'p' and 'q' is 1) is called a rational number. e.g. 56, -235.6, 5/7, {{{sqrt(16)}}}, etc Note: {{{-235.6 = -2356/10 = -1178/5}}}. Thus -235.6 can be expressed as a ratio of two integers -1178 and 5 and -1178 and 5 have no factors common between them. IRRATIONAL NUMBER ____________________ Any number which {{{red(cannot)}}} be expressed in the form {{{p/q}}} where 'p' and 'q' ('q' not equal to 1) are integers mutually prime to each other (this means 'p' and 'q' have no common factors; in other words H.C.F. of 'p' and 'q' is 1) is called an irrational number. e.g. {{{sqrt(5)}}}, {{{pi}}}, {{{-sqrt(8)}}}, etc Note: Let us prove that {{{sqrt(5)}}} is an irrational number. Let us assume that {{{sqrt(5)}}} is a rational number. Then it can be expressed as {{{sqrt(5) = p/q}}} where 'p' and 'q' are mutually prime integers and 'q' unequal to 1. Squaring both sides {{{5 = (p/q)^2}}} or {{{5*q = p^2/q}}} ______(1) Now, as 'q' is an integer so '5q' is also an integer. But as 'p' and 'q' has no common factors and 'q' is not equal to 1, so {{{p^2/q}}} cannot be an integer. So, there is a contradiction! Left side of eqn.(1) is an integer but the right side is not. This cannot be true. So our very assumption that {{{sqrt(5)}}} is a rational number must be wrong. Hence, {{{sqrt(5)}}} is an irrational number. If you have any query regarding this lesson you may send your query to me. Email: partha.s.bhowmick@gmail.com