Number System

1.0Rational numbers

The rational numbers are the numbers that can be expressed in the form of $\frac{p}{q}$, where $p$ and q are integers and coprime and $\mathrm{q}\ne 0$. For e.g., $-3,0,4.33$ etc.

Coprime Numbers: Co-prime numbers or relatively prime numbers are those numbers that have their HCF (Highest Common Factor) as 1. For e.g. $(2,3);(5,6);(7,8)$ etc.

Integers: An integer is a whole number that can be positive, negative, or zero.

2.0Rational numbers between two numbers

One way to find a rational number between two rational numbers is to find their average, called mean.

To find a rational number between $x$ and $y$, we will find the mean of $x$ and y. i.e. $\frac{x+y}{2}$ which is a rational number lying between x and y . This number will be the mid-value between the given two numbers.

• Q. Find 3 rational numbers between 2 and 5 . Solution: Let $\mathrm{a}=2,\mathrm{b}=5$ A rational number between 2 and $5=\frac{2+5}{2}=\frac{7}{2}$ Second rational number between 2 and $\frac{7}{2}=\frac{2+\frac{7}{2}}{2}=\frac{11}{4}$ Third rational number between $\frac{7}{2}$ and $5=\frac{\frac{7}{2}+5}{2}=\frac{17}{4}$ Hence, three rational numbers between 2 and 5 are : $\left[\frac{7}{2},\frac{11}{4},\frac{17}{4}\right]$

$(n+1)\text{ Rule }$

• When denominators of both rational numbers are different then we need to make their denominators same. In case $n$ rational numbers are required between 2 rational numbers, multiply the numerator and denominator of both rational number by $\mathrm{n}+1$. We can write $n$ rational numbers by increasing the numerator by 1 at a time.
• When denominators of both rational numbers are same then we just have to multiply the numerator and denominator of both rational numbers ar required between two given rational numbers.
• Q. Find 4 rational numbers between 4 and 5 . Solution: Let $\mathrm{a}=4,\mathrm{b}=5$ and $\mathrm{n}=4$ $\frac{\mathrm{a}(\mathrm{n}+1)}{(\mathrm{n}+1)}=\frac{4(4+1)}{(4+1)}$ and $\frac{\mathrm{b}(\mathrm{n}+1)}{(\mathrm{n}+1)}=\frac{5(4+1)}{(4+1)}$ $=\frac{4(5)}{5}=\frac{20}{5}$ and $\frac{5(5)}{5}=\frac{25}{5}\Rightarrow \frac{20}{5},\left[\frac{21}{5},\frac{22}{5},\frac{23}{5},\frac{24}{5}\right],\frac{25}{5}$ Hence, 4 rational numbers between 4 and 5 are $\left[\frac{21}{5},\frac{22}{5},\frac{23}{5},\frac{24}{5}\right]$ or $[4.2,4.4,4.6,4.4]$
• We can directly write 4 rational numbers between 4 and 5 as 4.1, 4.2, 4.3, 4.4. But it is not advisable to use this method in subjective test format.
• While using $(\mathrm{n}+1)$ rule to find rational numbers, make sure to make denominator of unlike fractions same.

3.0Decimal expansion of rational numbers

Every rational number can be expressed as terminating decimal or non-terminating but repeating decimals.

Terminating decimal (The remainder becomes zero)

The word "terminate" means "end". A decimal that ends is a terminating decimal. OR A terminating decimal doesn't keep going. A terminating decimal will have finite number of digits after the decimal point. For e.g. $\frac{3}{4}=0.75,\frac{8}{10}=0.8,\frac{5}{4}=1.25,\frac{25}{16}=1.5625$

• Q. Express $\frac{7}{8}$ in the decimal form. Solution:
  
  We have, $\therefore \frac{7}{8}=0.875$

Non-terminating & repeating (recurring decimal)

(The remainder never becomes zero)

A decimal in which a digit or a set of finite number of digits repeats periodically is called non-terminating repeating (recurring) decimals. For e.g. $\frac{5}{3}=1.6666$......... $=1.\overline{6}$ $\frac{7}{11}=\mathrm{0.636363.}\dots \dots ...=0.\overline{63}$ $\frac{1}{999}=0.001001001\dots \dots \dots =0.\overline{001}$

• Q. Express $\frac{2}{11}$ as a decimal fraction. Solution: we have
  
  $\therefore \frac{2}{11}=0.181818\dots ...=0.\overline{18}$
• Q. If $\frac{1}{7}=0.\overline{142857}$ write the decimal expansion of $\frac{2}{7},\frac{3}{7},\frac{4}{7}$, and $\frac{5}{7}$ without actually doing the long division. Explanation: As we have, $\frac{1}{7}=0.\overline{142857}$ $\frac{2}{7}=2\times \frac{1}{7}=0.\overline{285714};\frac{3}{7}=3\times \frac{1}{7}=0.\overline{428571}$ $\frac{4}{7}=4\times \frac{1}{7}=0.\overline{571428};\frac{5}{7}=5\times \frac{1}{7}=0.\overline{714285}$.

Method to convert non-terminating decimal to the form \frac{p}{q}.

In a non-terminating repeating decimals, we have two types of decimal representations (a) Pure recurring decimal (b) Mixed recurring decimal

(a) Pure recurring decimal

Step-1:

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(b) Mixed recurring decimal

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• Q. Express each of the following pure recurring decimals in the form $\frac{p}{q}$. (i) $0.\overline{6}$ (ii) $0.\overline{585}$ (iii) 23.43 Solution: (i) Let $\mathrm{x}=0.\overline{6}\Rightarrow \mathrm{x}=0.666$ ...(1) Here, we have only one repeating digit, so we multiply both sides of eq. (1) by 10 to get $\Rightarrow 10\mathrm{x}=6.66$ ...(2) On subtracting (1) from (2), we get; $\Rightarrow 10\mathrm{x}-\mathrm{x}=(6.66\dots ...)-.(0.66\dots ....$. $\Rightarrow 9x=6\Rightarrow x=\frac{6}{9}$ $\Rightarrow \mathrm{x}=\frac{2}{3}$ Hence, $0.\overline{6}=\frac{2}{3}$ (ii) Let $\mathrm{x}=0.\overline{585}\Rightarrow \mathrm{x}=0.585585585$.. Here, we have three repeating digits, so we multiply both sides of eq. (1) by 1000 to get $\Rightarrow 1000\mathrm{x}=585.585585$ ...(1) On subtracting eq. (1) from eq. (2), we get $1000\mathrm{x}-\mathrm{x}=(\mathrm{585.585585......})-.(\mathrm{0.585585.....})$ ...(2) $\Rightarrow 999\mathrm{x}=585$ $\Rightarrow \mathrm{x}=\frac{585}{999}=\frac{195}{333}=\frac{65}{111}$ Hence, $0.\overline{585}=\frac{65}{111}$ (iii) Let $\mathrm{x}=23.\overline{43}\Rightarrow \mathrm{x}=23.434343$ ...(1) Multiplying both sides of eq. (1) by 100, we get $100\mathrm{x}=2343.4343$ ...(2) Subtracting (1) from (2) we get $100\mathrm{x}-\mathrm{x}=(\mathrm{2343.4343.......})-.(\mathrm{23.4343......}$. $\Rightarrow 99\mathrm{x}=2320\Rightarrow \mathrm{x}=\frac{2320}{99}$ Hence, $23.\overline{43}=\frac{2320}{99}$ Alternate method We have, $23.\overline{43}=23+0.\overline{43}=23+\frac{43}{99}$ Using the above rule, we have $0.\overline{43}=\frac{43}{99}$

• Q. Express the following mixed recurring decimals in the form $\frac{\mathbf{p}}{\mathbf{q}}$. (i) $0.3\overline{2}$ (ii) $0.12\overline{3}$ (iii) $15.7\overline{12}$ Solution: (i) Let $\mathrm{x}=0.3\overline{2}$ Clearly, there is just one digit on the right side of the decimal point which is without bar. So, we multiply both sides by 10 so that only the repeating digit is left on the right side of the decimal point. $\therefore 10\mathrm{x}=3.\overline{2}$ $\Rightarrow 10\mathrm{x}=3+0.\overline{2}$ $\Rightarrow 10\mathrm{x}=3+\frac{2}{9}$ $\Rightarrow 10\mathrm{x}=\frac{9\times 3+2}{9}$ $\Rightarrow 10\mathrm{x}=\frac{29}{9}$ $\Rightarrow \mathrm{x}=\frac{29}{90}$ (ii) Let $\mathrm{x}=0.12\overline{3}$ Clearly, there are two digits on the right side of the decimal point which is without bar. Now, we multiply both sides of equation by $10^2=100$ so that only the repeating digit is left on the right side of the decimal point.

(iii) Let $\mathrm{x}=15.7\overline{12}$ Clearly, there is just one digit on the right side of the decimal point which is without bar. Now, we multiply both sides by 10 so that only the repeating digit is left on the right side of the decimal point. $\therefore 10\mathrm{x}=157.\overline{12}$ $\Rightarrow 10\mathrm{x}=157+0.\overline{12}$ $\Rightarrow 10\mathrm{x}=157+\frac{12}{99}$ $\Rightarrow 10\mathrm{x}=157+\frac{4}{33}$ $\Rightarrow 10\mathrm{x}=\frac{157\times 33+4}{33}$ $\Rightarrow 10\mathrm{x}=\frac{5181+4}{33}$ $\Rightarrow 10\mathrm{x}=\frac{5185}{33}$ $\Rightarrow \mathrm{x}=\frac{5185}{330}$ $=\frac{1037}{66}$

4.0Irrational numbers

Around 400 BC , followers of the famous mathematician and philosopher, Pythagoras, were the first to discover the numbers which were not rationals such as the length of the diagonal of a square with side one unit long and the ratio of circumference to the diameter of a circle. These and other needs led to the introduction of the Irrational Numbers.

A number is called an irrational number, if it cannot be written in the form $p/q$, where $p$ & q are integers, coprime and $\mathrm{q}\ne 0$. All Non-terminating & Non-repeating decimal numbers are Irrational numbers.

For e.g. $\sqrt{2},\sqrt{3},3\sqrt{2},2+\sqrt{3},\sqrt{2+\sqrt{3}},\pi$, etc.... Decimal expansion of irrational numbers Every irrational number can be expressed as non-terminating and non-repeating decimal. For e.g. $\sqrt{2}=1.4142135$....... Note : An irrational number between two numbers $a$ and $b$ is $\sqrt{ab}$, where ab cannot be $a$ perfect square.

• Q. (i) Find two irrational numbers between 2 and 2.5 (ii) Find one rational and one irrational number between 0.101001000100001 and 0.1001000100001 ...... (iii) Find two rational numbers between $\sqrt{2}$ and $\sqrt{3}$ (iv) Find two irrational numbers between $\sqrt{2}$ and $\sqrt{3}$ Explanation: (i) Irrational number between 2 and $2.5=\sqrt{(2)(2.5)}=\sqrt{(2)\left(\frac{5}{2}\right)}=\sqrt{5}$ Again, irrational number between 2 and $\sqrt{5}=\sqrt{(2)(\sqrt{5})}=\sqrt{2\sqrt{5}}$. So, required irrational numbers between 2 and 2.5 are $\sqrt{5}$ and $\sqrt{2\sqrt{5}}$. (ii) One rational number between given numbers $=0.101$ (Terminating decimal) One irrational number between given numbers $=0.1002000100001\dots$ (Non-Terminating Non-repeating). (iii) $\sqrt{2}=1.4142\dots$ and $\sqrt{3}=1.732\dots$ Now $\sqrt{2}<1.5<1.6<\sqrt{3}$ $\Rightarrow 1.5$ and 1.6 i.e. $\frac{3}{2}$ and $\frac{8}{5}$ are two rational between $\sqrt{2}$ and $\sqrt{3}$. (iv) $2<2.1<2.2<3$ $\Rightarrow \sqrt{2}<\sqrt{2.1}<\sqrt{2.2}<\sqrt{3}$ $\Rightarrow \sqrt{2.1}$ and $\sqrt{2.2}$ are two irrationals between $\sqrt{2}$ and $\sqrt{3}$.

5.0Real numbers

Rational numbers together with irrational numbers are said to be real numbers. That is, a real number is either rational or irrational. For e.g. $2,-\frac{3}{2},0,1.5,\sqrt{2},\sqrt[3]{5},\sqrt[5]{11},\pi$ etc. are real numbers.

Representation of real numbers on the number line by means of magnifying glass

6.0Representation of rational numbers on the number line

The process of visualization of numbers on the number line through a magnifying glass is known as successive magnification. Sometimes, we are unable to check the numbers like 3.765 and $4.\overline{26}$ on the number line we seek the help of magnifying glass by dividing the part into subparts and subparts again into equal subparts to ensure the accuracy of the given number. For e.g. represent 3.765 on the number line. This number lies between 3 and 4 . The distance 3 and 4 is divided into 10 equal parts. Then the first mark to the right of 3 will represent 3.1 and second 3.2 and so on. Now, 3.765 lies between 3.7 and 3.8. We divide the distance between 3.7 and 3.8 into 10 equal parts. 3.76 will be on the right of 3.7 at the $6^{\text{th }}$ mark, and 3.77 will be on the right of 3.7 at the $7^{\text{th }}$ mark and 3.765 will lie between 3.76 and 3.77 and so on.

• Q. Represent 3.728 on the number line through successive magnification. Explanation We have to locate the point 3.728 on the number line. This number lies between 3 and 4 . First go to 3.7. You divide the portion of the number line between 3 and 4 in 10 equal parts. Now first mark from the left will give you 3.1, the $2^{\text{nd }}$ mark will give you 3.2 and so on. To get 3.7 you reach at $7^{\text{th }}$ mark. Again to get 3.72, you divide the portion of the number line between 3.7 and 3.8 in 10 equal parts, to get 3.72 , you reach $2^{\text{nd }}$ mark from the left.
  
  Again to reach 3.728 you further divide the portion of the number line between 3.72 to 3.73 in 10 equal parts.
  
  To get the point 3.728 on the number line you reach 8th point from right to 3.72 on this subdivision.
  
• Q. Visualize 4.26 on the number line, up to 4 decimal places. Explanation: We can locate the point 4.2626 on the number line. $4.\overline{26}=\mathrm{4.262626.}\dots ...=4.2626$ (up to 4 decimal places)
  

Representation of irrational numbers on the number line

Consider the number line and mark a point O on it and let it represent zero. Let A represent 1 unit on the number line. $\mathrm{So},\mathrm{OA}=1$. At A draw AB perpendicular to OA . Let $\mathrm{AB}=\mathrm{OA}=1$ unit $\therefore$ By Pythagoras Theorem, $\mathrm{OB}=\sqrt{(\mathrm{OA}{)}^2+(\mathrm{AB}{)}^2}=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{1+1}=\sqrt{2}$

• Q. Represent $\sqrt{3.2}$ units geometrically on the number line. Explanation:
  
  Let $\ell$ be the number line. Draw a line segment $\mathrm{AB}=3.2$ units and $\mathrm{BC}=1$ unit. Find the midpoint 0 of AC . Draw a semicircle with centre O and radius OA or OC. Draw $\mathrm{BD}\perp \mathrm{AC}$ intersecting the semicircle at D . Then $\mathrm{BD}=\sqrt{3.2}$ units. Now, with centre B and radius BD , draw an arc intersecting the number line $\ell$ at P . Hence, $\mathrm{BD}=\mathrm{BP}=\sqrt{3.2}$ units

Operations on real number

Following are some useful results on real numbers. (i) Negative of an irrational number is an irrational number. For e.g. $\sqrt{2}$ is an irrational number because its value i.e $1.414\dots$ is a non terminating non repeating decimal. If we consider the value of $-\sqrt{2}$ i.e $1.414\dots$, it is still a non terminating non repeating decimal so $-\sqrt{2}$ is also an irrational number. (ii) The sum or difference of a rational number and an irrational number is an irrational number. For e.g. let us consider the value of $2+\sqrt{2}$ where 2 is a rational number and $\sqrt{2}$ is an 1.414...... irrational number, their sum will be $\frac{+2.000\dots \dots .}{3.414\dots \dots .}$ which is a non terminating non repeating decimal Similarly difference of a rational & an irrational number will be an irrational number. (iii) The product of a non-zero rational number and an irrational number is an irrational number. For e.g. Let us consider the value of $2\times 1.01001000100001\dots$ which is equals to $2.020020002\dots$ which is an irrational number. (iv) The sum, difference, product and quotient of two irrational numbers need not be an irrational number. For e.g. $\sqrt{2}$ and $-\sqrt{2}$ are two irrational numbers and $\sqrt{2}+(-\sqrt{2})=\sqrt{2}-\sqrt{2}=0$ which is rational. Similarly if we multiply $\sqrt{2}$ by $2\sqrt{2},\sqrt{2}\times 2\sqrt{2}=4$ which is rational. If we divide $\sqrt{12}$ by $\sqrt{3},\frac{\sqrt{12}}{\sqrt{3}}=\sqrt{4}=2$ which is rational. Also on the other hand, $\sqrt{2}+\sqrt{3}$ is irrational. $\sqrt{5}-\sqrt{2}$ is irrational, $\sqrt{2}\times \sqrt{3}=\sqrt{6}$ is irrational and so on.

7.0Some identities using radical sign

Let $a$ and $b$ be positive real numbers, then (i) $\sqrt{\mathrm{ab}}=\sqrt{\mathrm{a}}\sqrt{\mathrm{b}}$ (ii) $\sqrt{\frac{a}{b}}=\frac{\sqrt{\mathrm{a}}}{\sqrt{\mathrm{b}}}$ (iii) $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$ (iv) $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$ (v) $(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{d})=\sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}$ (vi) $(\sqrt{a}+\sqrt{b}{)}^2=a+2\sqrt{ab}+b$ (vii) Conjugate of $a+\sqrt{b}$ is $a-\sqrt{b}$

• Q. (i) Add $7+\sqrt{5}$ and $-7+\sqrt{5}$ (ii) Multiply $7\sqrt{3}$ by $2\sqrt{3}$ (iii) Divide $6\sqrt{27}$ by $3\sqrt{12}$ Solution: (i) $(7+\sqrt{5})+(-7+\sqrt{5})=(7-7)+(\sqrt{5}+\sqrt{5})=0+2\sqrt{5}=2\sqrt{5}$ (ii) $7\sqrt{3}\times 2\sqrt{3}=7\times 2\times 3=42$ (iii) $\frac{6\sqrt{27}}{3\sqrt{12}}=\frac{\sqrt{36\times 27}}{\sqrt{9\times 12}}=\frac{\sqrt{972}}{\sqrt{108}}=\sqrt{9}=3$
• Q. Simplify: (i) $(\sqrt{5}+3)(\sqrt{5}+2)$ (ii) $(\sqrt{7}+\sqrt{2}{)}^2$ (iii) $(6+\sqrt{2})(6-\sqrt{2})$ Solution: (i) $(\sqrt{5}+3)(\sqrt{5}+2)=5+2\sqrt{5}+3\sqrt{5}+6=11+5\sqrt{5}$ (ii) $(\sqrt{7}+\sqrt{2}{)}^2=(\sqrt{7}{)}^2+(\sqrt{2}{)}^2+2\sqrt{7}\sqrt{2}=7+2+2\sqrt{14}=9+2\sqrt{14}$ (iii) $(6+\sqrt{2})(6-\sqrt{2})=(6{)}^2-(\sqrt{2}{)}^2=36-2=34$

8.0Rationalisation

The process of converting the irrational terms to rational (mostly in denominator) in a mathematical expression is known as rationalisation. Why do we need rationalisation? Let us take an example. Suppose your teacher wants you to calculate the value of $\frac{1}{\sqrt{2}}$ and value of $\sqrt{2}$ is 1.414 (approx.) To calculate this you need to divide 1 by 1.414 which will be a tedious task to do. But if we do some manipulation in the given expression as $\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{(\sqrt{2}{)}^2}=\frac{\sqrt{2}}{2}$ $\Rightarrow \frac{1.414}{2}=0.707$ As you can see multiplying $\frac{1}{\sqrt{2}}$ by $\sqrt{2}$ both in numerator and denominator converted the irrational term i.e. $\sqrt{2}$ into a rational term i.e. 2 . This has converted the calculation into an easier one.

• Rationalisation converts irrational terms to rational in the denominator which can make the calculation easier.

9.0Rationalising factor

In the above example, we have multiplied $\frac{1}{\sqrt{2}}$ by $\sqrt{2}$ in both numerator and denominator to convert $\sqrt{2}$ into 2 in the denominator. So, the factor by which we multiply numerator and denominator to rationalise the denominator is called rationalising factor.

Rationalising factors in different cases:

(i) $\frac{1}{\sqrt{\mathrm{a}}}\to$ The rationalising factor will be $\sqrt{\mathrm{a}}$ $\frac{1}{\sqrt{a}}\times \frac{\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}}{a}$ (ii) $\frac{1}{a\pm \sqrt{b}}$ The rationalising factor will be $a\mp \sqrt{b}$ $\frac{1}{a\pm \sqrt{b}}\times \frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}=\frac{a\mp \sqrt{b}}{(a{)}^2-(\sqrt{b}{)}^2}=\frac{a\mp \sqrt{b}}{a^2-b}$ (iii) $\frac{1}{\sqrt{\mathrm{a}}\pm \sqrt{\mathrm{b}}}\to$ The rationalising factor will be $\sqrt{\mathrm{a}}\mp \sqrt{\mathrm{b}}$ $\frac{1}{\sqrt{a}\pm \sqrt{b}}\times \frac{\sqrt{a}\mp \sqrt{b}}{\sqrt{a}\mp \sqrt{b}}=\frac{\sqrt{a}\mp \sqrt{b}}{(\sqrt{a}{)}^2-(\sqrt{b}{)}^2}=\frac{\sqrt{a}\mp \sqrt{b}}{a-b}$

• Q. Simplify the following by rationalising the denominator : (i) $\frac{1}{\sqrt{7}+\sqrt{6}}$ (ii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ Solution: (i) We have, $\frac{1}{\sqrt{7}+\sqrt{6}}=\frac{1}{\sqrt{7}+\sqrt{6}}\times \frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}-\sqrt{6}}$ (Multiply and divide by $\sqrt{7}-\sqrt{6}$ )

(ii) Multiplying the numerator and denominator by the conjugate of $3-\sqrt{2}$, we get

• Q. If $\frac{3+2\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$, where $a$ and $b$ are rational numbers. Find the values of $a$ and $b$. Solution: L.H.S. $=\frac{3+2\sqrt{2}}{3-\sqrt{2}}=\frac{(3+2\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$ (Use rationalization) $=\frac{9+3\sqrt{2}+6\sqrt{2}+4}{9-2}=\frac{13}{7}+\frac{9}{7}\sqrt{2}$ $\therefore \frac{13}{7}+\frac{9}{7}\sqrt{2}=a+b\sqrt{2}$ equating the rational and irrational parts; We get $a=\frac{13}{7},b=\frac{9}{7}$
• Q. If $x=3-2\sqrt{2}$, find $x^2+\frac{1}{x^2}$. Solution: We have, $x=3-2\sqrt{2}$

Thus, $\mathrm{x}+\frac{1}{\mathrm{x}}=3-2\sqrt{2}+3+2\sqrt{2}$ $x+\frac{1}{x}=6$ Squaring both sides ${\left(x+\frac{1}{x}\right)}^2=6^2$ $x^2+\frac{1}{x^2}+2=36$ ${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=34$

10.0Laws of exponents

Let $\mathrm{a}>0$ be a real number and p and q be rational numbers. Then, we have (i) $a^p\cdot a^q=a^{p+q}$ (ii) ${\left(a^p\right)}^q=a^{pq}$ (iii) $\frac{a^p}{a^q}=a^{p-q}$ (iv) $a^p{b}^p=(ab{)}^p$ (v) $a^{-p}=\frac{1}{a^p}$

• Any non zero real number raised to power zero is equal to 1 . e.g. $(x{)}^0=1,(1000{)}^0=1$
• Q. Evaluate the following: (i) $(\sqrt[3]{64}{)}^{\frac{-1}{2}}$ (ii) $(\sqrt{25}{)}^{-7}\times (\sqrt{5}{)}^{-5}$ (iii) ${\left(\frac{4}{5}\right)}^7÷{\left(\frac{5}{4}\right)}^{-5}$ (iv) ${\left(\frac{121}{169}\right)}^{-3/2}$ Solution: (i) $(\sqrt[3]{64}{)}^{\frac{-1}{2}}={\left[(64{)}^{\frac{1}{3}}\right]}^{\frac{-1}{2}}=(64{)}^{\frac{1}{3}\times \frac{-1}{2}}=(64{)}^{\frac{-1}{6}}$ $={\left(2^6\right)}^{\frac{-1}{6}}=2^{6\times \left(\frac{-1}{6}\right)}=2^{-1}=\frac{1}{2}$ (ii) $(\sqrt{25}{)}^{-7}\times (\sqrt{5}{)}^{-5}$

(iii) ${\left(\frac{4}{5}\right)}^7÷{\left(\frac{5}{4}\right)}^{-5}={\left(\frac{4}{5}\right)}^7\times {\left(\frac{4}{5}\right)}^{-5}$ $={\left(\frac{4}{5}\right)}^{7-5}={\left(\frac{4}{5}\right)}^2=\frac{16}{25}$ (iv) ${\left(\frac{121}{169}\right)}^{-3/2}={\left(\frac{11\times 11}{13\times 13}\right)}^{-3/2}$

• Q. Simplify: (i) ${\left(\frac{4x^{5}y}{16x{y}^4}\right)}^3$ (ii) $\frac{3x^2{y}^{-3}}{12x^6{y}^3}$ (iii) $5x^{2}y\left(2x^4{y}^{-3}\right)$ (iv) ${\left(2{\mathrm{fh}}^4\right)}^4(\mathrm{fg}{)}^6$ (v) ${\left(\frac{-2a^3{b}^2{x}^0}{3a^2{b}^3{\mathbf{c}}^7}\right)}^{-2}$ Solution: (i) ${\left(\frac{4x^{5}y}{16x{y}^4}\right)}^3={\left(\frac{1}{4}{x}^{5-1}\cdot y^{1-4}\right)}^3={\left(\frac{1}{4}{x}^4\cdot y^{-3}\right)}^3$ $=\frac{1}{64}{x}^{12}{y}^{-9}=\frac{x^{12}}{64y^9}$ (ii) $\frac{3x^2{y}^{-3}}{12x^6{y}^3}=\frac{1}{4}{x}^{2-6}{y}^{-3-3}$ $=\frac{1}{4}{\mathrm{x}}^{-4}{\mathrm{y}}^{-6}=\frac{1}{4{\mathrm{x}}^4{\mathrm{y}}^6}$ (iii) $5x^{2}y\left(2x^4{y}^{-3}\right)=10x^2+4\cdot y^{1-3}$ $=10x^6{y}^{-2}=\frac{10x^6}{y^2}$ (iv) ${\left(2{\mathrm{fg}}^4\right)}^4(\mathrm{fg}{)}^6$ $=2^4{\mathrm{f}}^4{\mathrm{g}}^{16}{\mathrm{f}}^6{\mathrm{g}}^6$ $=16{\mathrm{f}}^{4+6}{\mathrm{g}}^{16+6}$ $=16f^{10}{\mathrm{g}}^{22}$ (v) ${\left(\frac{-2a^3{b}^2{c}^0}{3a^2{b}^3{c}^7}\right)}^{-2}$ $={\left(\frac{-2}{3}{a}^{3-2}{b}^{2-3}{c}^{0-7}\right)}^{-2}={\left(\frac{-2}{3}{a}^1{b}^{-1}{c}^{-7}\right)}^{-2}$ $={\left(\frac{-2}{3}\right)}^{-2}{a}^{-2}{b}^2{c}^{14}={\left(-\frac{3}{2}\right)}^{-2}{a}^{-2}{b}^2{c}^{14}$ $=\frac{9}{4}\frac{b^2{c}^{14}}{a^2}$