Lesson RATIONAL AND IRRATIONAL NUMBERS

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RATIONAL NUMBER
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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
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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.

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