Category : 8th Class
RATIONAL NUMBERS
FUNDAMENTALS
Rational Number:-
e.g., \[\frac{1}{2},\frac{2}{2},\frac{-1}{2},0,\frac{3}{-\,2}\] etc.
e.g., \[\frac{1}{2},\frac{2}{5},\frac{-3}{-2},-\left( -\frac{1}{2} \right)\]etc.
Negative Rational Numbers:-
e.g., \[\frac{-1}{2}.\frac{3}{-2},\frac{-7}{11}......\]etc.
Standard form of Rational Number:-
e.g., \[\frac{-1}{2},\frac{5}{6},\frac{8}{11}\]……etc.
Note:- There are infinite rational numbers between any two rational numbers.
Property of Rational Number
e.g., find 2 rational number between \[\frac{1}{3}\]and \[\frac{1}{2}\]
Solution:- Let \[x=\frac{1}{3}\] and \[y=\frac{1}{3}\] and y > x.
Then, Rational no. between\[\frac{1}{3}\]and\[\frac{1}{2}\]is
\[\frac{1}{2}\left( \frac{1}{3}+\frac{1}{2} \right)=\frac{1}{2}\left( \frac{2+3}{6} \right)=\frac{5}{12}\]
Again Let \[x=\frac{5}{12}\] and \[y=\frac{1}{2}\] and y > x. then
Rational no. between \[\frac{5}{12}\] and \[\frac{1}{2}\] is
\[\frac{1}{2}\left( \frac{5}{12}+\frac{1}{2} \right)=\frac{1}{2}\left( \frac{5+6}{12} \right)=\frac{1}{2}\times \frac{11}{12}=\frac{11}{24}\]
Hence the Rational Numbers between \[\frac{1}{3}\] and \[\frac{1}{2}\] are \[\frac{5}{12}\] and \[\frac{11}{24}\].
Then 'n' rational number lying between x and y are \[\left( x+d \right),\left( x+2d \right),\left( x+3d \right),\_\_\_\left( x+nd \right).\]
Example:- Find 9 rational number between 2 and 3.
Solution:- Let x = 2 and y = 3 then y > x
Now \[\mathbf{d}=\frac{y-x}{n+1}=\frac{3-2}{9+1}=\frac{1}{10}\]
Then, rational number are, 2 + 0.1, 2 + 0.2, 2 + 0.3, 2 + 0.4, 2 + 0.5, 2 + 0.6, 2 + 0.7, 2 + 0.8, 2 + 0.9 = 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8 and 2.9.
Representation of Rational Number on the Number line
Let O represent 0 (zero) and A represent 1. So divide OA into 4 equal parts, each point in the middle representing P, Q and R. Point R represent\[\frac{3}{4}\].
Operations on Rational Numbers
Example: Find the sum of the rational numbers \[\frac{-4}{9},\frac{15}{12}\] and \[\frac{-7}{18}\].
Solution: \[\frac{-4}{9}+\frac{15}{12}+\frac{-7}{18}=\frac{-16+45-14}{36}=\frac{15}{36}=\frac{5}{12}\]
Properties of Addition of Rational Number
E.g., Let \[a=3\], \[b=-2,\] then \[a+b=3+\left( -2 \right)-1\]
E.g., Let \[a=\frac{1}{2}\]and \[b=\frac{1}{3}\] then
To check whether, a + b = b + a
\[\Rightarrow \frac{1}{2}+=\frac{1}{3}+\frac{1}{2}\]
\[\Rightarrow \frac{5}{6}=\frac{5}{6}\]
E.g., \[a=1,\]\[b=-\,2\] and \[c=3\] then,
\[1+\left( -2+3 \right)=\left( 1-2 \right)+3\]
\[1+1=-1+3\]
2 = 2
Existence of additive identity (property of zero):-
E.g., \[2+0=2,\,-2+0=-2,\,\,3+0=3,\,\,\frac{-1}{2}+0=\frac{-1}{2}\]
Existence of additive inverse;-
For\[\frac{a}{b}\], it is\[-\frac{a}{b}\]
e.g., For\[\frac{1}{2}\], it is\[-\frac{1}{2}\]
(\[-\frac{1}{2}\]is a additive inverse of\[\frac{1}{2}\])
\[-\frac{3}{2}\Rightarrow \frac{3}{2}\] (\[\frac{3}{2}\] is a additive inverse of \[-\frac{3}{2}\])
Note:- Additive inverse of the rational number ‘0’ is 0 itself.
Subtraction of Rational Number:-
If \[\frac{p}{q}\]and \[\frac{r}{s}\] be two rational number it follows \[\frac{r}{s}-\frac{p}{q}=\frac{r}{s}+\left( -\frac{p}{q} \right)\]
e.g., subtract \[\frac{-2}{7}\] from \[\frac{3}{4}\].
Solution:- \[\frac{3}{4}-\left( -\frac{2}{7} \right)=\frac{3}{4}+\frac{2}{7}=\frac{21+8}{28}=\frac{29}{28}\]
Multiplication of Rational Number;
If \[\frac{a}{b}\] and \[\frac{c}{d}\] are two rational numbers, then \[\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}\]
Example:- Multiply \[\frac{-17}{30}\] by \[\frac{15}{-34}\]
Solution:- \[\frac{-17}{30}\times \frac{15}{34}=\frac{-17\times 15}{30\times -34}=\frac{1}{4}\]
Properties of multiplication of Rational Numbers:
e.g., \[\frac{2}{3}\times \frac{3}{4}=\frac{2\times 3}{3\times 4}=\frac{1}{2}\]
e.g., \[\frac{2}{3}\times \frac{3}{4}=\frac{3}{4}\times \frac{2}{3}\Rightarrow \frac{6}{12}=\frac{6}{12}\]
\[\frac{1}{2}=\frac{1}{2}\]
e.g., \[\frac{1}{2}\times \left( \frac{2}{3}\times \frac{3}{4} \right)=\left( \frac{1}{2}\times \frac{2}{3} \right)\times \frac{3}{4}\]
\[\frac{6}{24}=\frac{6}{24}\]
e.g., \[\left( \frac{p}{q}\times 1 \right)=\frac{p}{q}\], \[\left( \frac{3}{4}\times 1 \right)=\frac{3}{4}\], \[\left( \frac{-5}{2}\times 1 \right)=\frac{-5}{2}\]
Hence, we say that \[\frac{a}{b},\frac{b}{a}\] are multiplicative inverse of each other
e.g., (i) \[\frac{2}{3}\times \frac{3}{2}=1\]
(ii) \[\left( \frac{-3}{4}\times \frac{-4}{3} \right)=\frac{12}{12}=1\]
\[\frac{a}{b}\times \left( \frac{c}{d}\times \frac{e}{f} \right)=\left( \frac{a}{b}\times \frac{c}{d} \right)+\left( \frac{a}{b}\times \frac{e}{f} \right).\] This property is called distributive property for multiplication over addition.
e.g., \[\frac{1}{2}\times \left( \frac{2}{3}+\frac{3}{4} \right)=\frac{1}{2}\left( \frac{8+9}{12} \right)=\frac{1}{2}\times \frac{17}{12}=\frac{17}{24}\]
Example:- \[\frac{14}{57}\div \frac{42}{19}=\frac{14}{57}\times \frac{19}{42}=\frac{14\times 19}{57\times 42}=\frac{1}{9}\]
For example:-
are non-terminating repeating decimals.
For Example:
Method of expressing recurring decimals as rational number:
Example:
We can express non-terminating recurring decimals in the form of rational numbers.
Example-1:- Let us write \[0.2\overline{45}\] in the form of rational number.
Solution:- Let x =\[0.2\overline{45}\] (i)
Then \[10x=2.4545\].............. (ii)
Also, \[1000x=245.4545\]........... (iii)
On subtracting (ii) from (iii), we get: \[990x=245\Leftrightarrow x=\frac{245}{990}=\frac{49}{198}\].
Hence, \[0.2\overline{45}=\frac{40}{198}.\]
Example-2:- Let us find the rational form of\[0.\overline{428571}\].
Solution:- The periodicity of the recurring decimal is 6. So multiply the decimal fraction by 106, \[0.\overline{428571}\] = x (say)
\[{{10}^{6}}=1000000\,x=428571.\overline{428571}\]
\[x=0.\overline{428571}\]
\[99999x=428571\]
\[\therefore x=\frac{428571}{999999}=\frac{3}{7}\]
Example-3:- Express \[15.0\overline{2}\] as a rational Number
Solution:- Here, the whole number obtained by writing digits in there order =1502. The whole number made by the non-recurring digits in order = 150,
The number of digits after the decimal point = 2 (two)
The number of digits after the decimals point do not recur = one
\[\therefore 15.0\overline{2}=\frac{1502-150}{{{10}^{2}}-{{10}^{1}}}=\frac{1352}{90}=\frac{676}{45}\]
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