CBSE Class 9 Maths Advanced Number System Notes Set 01

Number System

A. Introduction to Number System and Rational Numbers

Classification of Numbers

1. Natural numbers. The numbers we count with form the set N = { 1, 2, 3, 4, ... }.

2. Whole numbers. Put 0 together with the natural numbers and you get the whole numbers, W = { 0, 1, 2, 3, 4, ... }.

3. Integers. Whole numbers together with the negatives of the natural numbers make up the integers, Z (or I) = { ..., −3, −2, −1, 0, 1, 2, 3, ... }.

4. Rational numbers. Any number that can be written as \( \frac{p}{q} \) with p and q integers and q ≠ 0 is rational. Examples: \( \frac{2}{3} \) and \( \frac{-37}{15} \). Every natural number, whole number and integer is also a rational number.

5. Real numbers. Numbers that can stand for real, measurable quantities are real numbers, and each of them has a place on the number line. They include all rational and all irrational numbers.

6. Prime numbers. A natural number greater than 1 whose only divisors are 1 and itself is prime, e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23.

7. Composite numbers. Natural numbers greater than 1 that are not prime. The number 1 is neither prime nor composite.

8. Co-prime numbers. Two numbers (not necessarily prime) are co-prime when their H.C.F. is 1, e.g. 4 and 9. Any two consecutive numbers are always co-prime.

9. Even and odd numbers. Integers divisible by 2 are even: { ..., −2, 0, 2, ... }. Integers that are not divisible by 2 are odd: { ..., −3, −1, 1, 3, ... }.

Rational Numbers in Decimal Form

Terminating decimals. Take a rational number \( \frac{p}{q} \) in which p and q are co-prime. Its decimal expansion stops after a few digits exactly when the prime factors of q are only 2s and 5s, i.e. q = \( 2^m \times 5^n \) with m, n non-negative integers. Examples: \( \frac{1}{2} \) = 0.5, \( \frac{11}{16} \) = 0.6875, \( \frac{3}{20} \) = 0.15.

Non-terminating repeating (recurring) decimals. If q has any prime factor other than 2 and 5, the decimal never stops, but a digit or block of digits repeats again and again. Examples: \( \frac{2}{3} \) = 0.6666... = \( 0.\overline{6} \) and \( \frac{5}{11} \) = 0.454545... = \( 0.\overline{45} \).

Rational Numbers on the Real Number Line

Successive magnification. Looking at a number line through an imaginary magnifying glass, again and again, is called successive magnification. It is needed for numbers such as 3.765 or \( 4.\overline{26} \), which cannot be located on an ordinary unit-scale line.

Method for such numbers.

1. Find the two consecutive integers between which the number lies.
2. Split that gap into ten equal parts and find the two consecutive decimal marks that enclose the number.
3. Magnify that small part, split it into ten again, and repeat until the number is located.

Converting a Recurring Decimal into a Fraction

Long method.

Step 1: put the given mixed recurring decimal equal to x.
Step 2: count the non-repeating digits after the decimal point; call this n.
Step 3: multiply both sides by \( 10^n \) so that only repeating digits follow the decimal point.
Step 4: multiply the result of step 3 by \( 10^m \), where m is the number of repeating digits.
Step 5: subtract the equation of step 3 from that of step 4, which removes the repeating part.
Step 6: divide both sides by the coefficient of x.
Step 7: reduce the fraction \( \frac{p}{q} \) to lowest terms.

Direct method. Numerator: write all the digits (non-repeating digits followed by one set of the repeating digits) as a whole number and subtract the number formed by the non-repeating digits alone. Denominator: as many 9s as there are repeating digits, followed by as many 0s as there are non-repeating digits after the decimal point.

Finding Rational Numbers between Two Numbers

Method I. To insert n rational numbers between a and b (a < b), multiply the top and bottom of both a and b by \( \frac{n+1}{n+1} \). The numerators then differ enough to list n fractions with the same denominator between them, by increasing the numerator step by step.

Method II (repeated averages). Take a < b. Adding a to both sides of a < b gives 2a < a + b, so a < \( \frac{a+b}{2} \). Adding b to both sides gives a + b < 2b, so \( \frac{a+b}{2} \) < b. Hence the mean \( \frac{a+b}{2} \) lies between a and b; this is the first rational number.

For the next numbers, average again: the mean of a and \( \frac{a+b}{2} \) is \( \frac{3a+b}{4} \), and the mean of \( \frac{a+b}{2} \) and b is \( \frac{a+3b}{4} \). So a < \( \frac{3a+b}{4} \) < \( \frac{a+b}{2} \) < \( \frac{a+3b}{4} \) < b, and the process can go on without end.

Solved Examples

Example 1. Is (39, 93) a co-prime pair ?
Answer:
The H.C.F. of 39 and 93 is 3, which is not 1.
So (39, 93) is not co-prime.

Example 2. Represent \( \frac{3}{7} \) on a real number line.
Answer:
(i) Draw a line XY that extends endlessly in both directions.
(ii) Mark a point O on it for 0.
(iii) Fix a unit length and mark OA = 1 unit.
(iv) Divide OA into 7 equal parts. The third division point P from O represents \( \frac{3}{7} \).

Example 3. Represent \( \frac{7}{5} \) on a real number line.
Answer:
Write \( \frac{7}{5} \) = \( 1\frac{2}{5} \).
(i) Draw the line XY and mark O for 0.
(ii) Mark OA = 1 unit and OB = 2 units.
(iii) Divide each of OA and AB into 5 equal parts. The point P, two parts beyond A, represents \( \frac{7}{5} \).

Example 4. Represent \( \frac{-13}{4} \) on a real number line.
Answer:
Write \( \frac{-13}{4} \) = \( -3\frac{1}{4} \).
(i) Draw the line XY and mark O for 0.
(ii) Towards the left of O mark OA = 1, OB = 2, OC = 3 and OD = 4 units.
(iii) Divide each of OA, AB, BC and CD into 4 equal parts. The point P one part to the left of C (that is, at −3 and one quarter) represents \( \frac{-13}{4} \).

Example 5. Represent 2.5 on a real number line.
Answer:
Mark 0, 1, 2, 3 and divide each unit into 10 equal parts. The fifth mark after 2 is 2.5.

Example 6. Represent 2.65 on a real number line by the process of magnification.
Answer:
2.65 lies between 2 and 3. Divide that gap into 10 parts and magnify it: 2.6 and 2.7 enclose the number.
Magnify the part between 2.6 and 2.7 and divide it into 10 parts again (2.61, 2.62, ..., 2.69). The middle mark is 2.65.

Example 7. Visualise the representation of \( 5.3\overline{7} \) on the number line up to 5 decimal places.
Answer:
5.3777... lies between 5 and 6; magnify to see 5.3 and 5.4; magnify again to see 5.37 and 5.38; again to see 5.377 and 5.378; the last magnification shows the number between 5.3777 and 5.3778 (the red dots show where 5.37777 falls on each magnified line).

Example 8. Express \( 0.\overline{6} \) in \( \frac{p}{q} \) form.
Answer:
Let x = 0.6666... (i). Multiply by 10: 10x = 6.666... (ii).
Subtract (i) from (ii): 9x = 6, so x = \( \frac{6}{9} \) = \( \frac{2}{3} \).

Example 9. Express \( 0.\overline{47} \) in \( \frac{p}{q} \) form.
Answer:
Let x = 0.474747... (i). Two digits repeat, so multiply by 100: 100x = 47.4747... (ii).
Subtract (i) from (ii): 99x = 47, so x = \( \frac{47}{99} \).

Example 10. Express \( 0.12\overline{3} \) in \( \frac{p}{q} \) form.
Answer:
Let x = 0.12333... (i). Multiply by 100: 100x = 12.333... (ii). Multiply (ii) by 10: 1000x = 123.333... (iii).
Subtract (ii) from (iii): 900x = 111, so x = \( \frac{111}{900} \) = \( \frac{37}{300} \).

Example 11. Express the following in \( \frac{p}{q} \) form using the direct method : (i) \( 0.\overline{45} \) (ii) \( 0.7\overline{37} \) (iii) \( 0.46\overline{573} \).
Answer:
(i) \( 0.\overline{45} = \frac{45-0}{99} = \frac{45}{99} = \frac{5}{11} \).
(ii) \( 0.7\overline{37} = \frac{737-7}{990} = \frac{730}{990} = \frac{73}{99} \).
(iii) \( 0.46\overline{573} = \frac{46573-46}{99900} = \frac{46527}{99900} \).

Example 12. Find 4 rational numbers between 2 and 3.
Answer:
Multiply the top and bottom of 2 and 3 by (4 + 1) = 5: 2 = \( \frac{10}{5} \) and 3 = \( \frac{15}{5} \).
Four fractions between them are \( \frac{11}{5} \), \( \frac{12}{5} \), \( \frac{13}{5} \), \( \frac{14}{5} \).

Example 13. Find 3 rational numbers between \( \frac{1}{3} \) and \( \frac{1}{2} \).
Answer:
Mean of the two: \( \frac{\frac13+\frac12}{2} = \frac{5/6}{2} = \frac{5}{12} \), so \( \frac{1}{3} \) < \( \frac{5}{12} \) < \( \frac{1}{2} \).
Mean of \( \frac{1}{3} \) and \( \frac{5}{12} \): \( \frac{\frac{4}{12}+\frac{5}{12}}{2} = \frac{9}{24} \).
Mean of \( \frac{5}{12} \) and \( \frac{1}{2} \): \( \frac{\frac{5}{12}+\frac{6}{12}}{2} = \frac{11}{24} \).
The three numbers are \( \frac{9}{24} \), \( \frac{5}{12} \), \( \frac{11}{24} \), in order \( \frac{1}{3} \) < \( \frac{9}{24} \) < \( \frac{5}{12} \) < \( \frac{11}{24} \) < \( \frac{1}{2} \).

B. Irrational Numbers

A real number that is not rational is irrational. Its decimal expansion neither ends nor repeats. Examples: \( \sqrt{2} \), \( \sqrt[3]{4} \), \( 2+\sqrt{3} \), \( \sqrt{2+\sqrt{3}} \), \( \sqrt[4]{\sqrt[7]{3}} \), \( \pi \).

Proving that a Number is Irrational

Irrationality is proved by contradiction (an indirect proof): assume the opposite statement is true, and show that this assumption leads to something impossible, so the original claim must hold.

Inserting Irrational Numbers between Two Real Numbers

If a and b are two positive real numbers, then \( \sqrt{ab} \) lies between them, and it is irrational as long as ab is not a perfect square.

Irrational Numbers: Decimal Form and Properties

\( \sqrt{2} \) = 1.414213... and \( \sqrt{3} \) = 1.732050807... go on for ever without any repeating block.

Properties.

1. The negative of an irrational number is irrational, e.g. \( -\sqrt{3} \) and \( -\sqrt[4]{5} \).
2. The sum or difference of a rational and an irrational number is always irrational.
3. The sum, difference or product of two irrational numbers may be rational or irrational.
4. The product of a rational number and an irrational number may be rational or irrational (it is irrational when the rational number is not zero).

Geometrical Representation of Real Numbers

To show \( \sqrt{x} \) on the number line (x a positive real number):

Step I: take the positive number x.
Step II: draw a line and mark a point A on it.
Step III: mark B on the line with AB = x units.
Step IV: from B mark C on the line, 1 unit beyond B, so that ABC is a straight line.
Step V: find the mid-point O of AC by drawing its perpendicular bisector.
Step VI: draw a semicircle with centre O and radius OC.
Step VII: draw a perpendicular to AC through B, meeting the semicircle at D. Then BD = \( \sqrt{x} \).
Step VIII: with B as centre and BD as radius draw an arc that cuts the line (OC produced) at E. The distance BE = \( \sqrt{x} \).

Why it works. AB = x and BC = 1, so AC = x + 1 and OA = OC = OD = \( \frac{x+1}{2} \). Then OB = AB − OA = \( \frac{x-1}{2} \). In the right triangle OBD, BD² = OD² − OB² = \( \left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2 = \frac{4x}{4} = x \), so BD = \( \sqrt{x} \). This shows that \( \sqrt{x} \) exists for every real x > 0.

Solved Examples

Example 14. Prove that \( \sqrt{2} \) is an irrational number.
Answer:
Suppose, on the contrary, that \( \sqrt{2} \) is rational. Then \( \sqrt{2} \) = \( \frac{a}{b} \) with a, b positive integers having H.C.F. 1.
Squaring: 2 = \( \frac{a^2}{b^2} \), so \( a^2 = 2b^2 \). Thus \( a^2 \) is even, so a is a multiple of 2; write a = 2c.
Then 4c² = 2b², so b² = 2c², hence b² is even and b is also a multiple of 2.
Now a and b share the factor 2, contradicting H.C.F. = 1. Therefore \( \sqrt{2} \) is irrational.

Example 15. Prove that \( 3-\sqrt{5} \) is an irrational number.
Answer:
Assume \( 3-\sqrt{5} \) is rational, equal to \( \frac{a}{b} \) with a, b co-prime positive integers.
Then \( \sqrt{5} = 3-\frac{a}{b} = \frac{3b-a}{b} \). Since a and b are integers, \( \frac{3b-a}{b} \) is rational, so \( \sqrt{5} \) would be rational.
That contradicts the fact that \( \sqrt{5} \) is irrational. Hence \( 3-\sqrt{5} \) is irrational.

Example 16. Insert an irrational number between 2 and 3.
Answer:
The number \( \sqrt{2\times 3} = \sqrt{6} \) lies between them (about 2.449) and is irrational.

Example 17. Find two irrational numbers between 2 and 2.5.
Answer:
First method. \( \sqrt{2\times 2.5} = \sqrt{5} \) is irrational (no rational number has square 5). Also \( \sqrt{2\times\sqrt{5}} \) lies between 2 and \( \sqrt{5} \) and is irrational.
Second method. Write non-terminating, non-repeating decimals between 2 and 2.5, such as 2.101001000100001... and 2.201001000100001...

Example 18. Plot \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \) on a number line.
Answer:
Take a horizontal line X'OX with O as origin (0). Take OA = 1 unit and draw AB perpendicular to OA with AB = 1 unit. Join OB.
OB = \( \sqrt{OA^2+AB^2} = \sqrt{1^2+1^2} = \sqrt{2} \) units. With centre O and radius OB draw an arc cutting the line at P. Then OP = \( \sqrt{2} \).
Draw BC perpendicular to OB with BC = 1 and join OC: OC = \( \sqrt{(\sqrt2)^2+1^2} = \sqrt{3} \). An arc with centre O and radius OC meets the line at Q, so OQ = \( \sqrt{3} \).
Draw CD perpendicular to OC with CD = 1 and join OD: OD = \( \sqrt{(\sqrt3)^2+1^2} = \sqrt{4} = 2 \) units. Draw DE perpendicular to OD with DE = 1 and join OE: OE = \( \sqrt{2^2+1^2} = \sqrt{5} \). An arc with centre O and radius OE meets the line at R, so OR = \( \sqrt{5} \).

Another method for \( \sqrt{2} \) and \( \sqrt{3} \). Mark O (zero) and A (one) on a number line and draw the right triangle OAB with AB = OA = 1. Then OB = \( \sqrt{2} \). An arc with centre O and radius OB cuts the line at C, so OC = \( \sqrt{2} \). Next draw CE perpendicular to the line with CE = 1; then OE = \( \sqrt{(\sqrt2)^2+1^2} = \sqrt3 \), and an arc with centre O and radius OE cuts the line at D, so OD = \( \sqrt{3} \).

C. Surds and their Applications

Surd. An irrational number of the form \( \sqrt[n]{a} \) (a rational and positive) is called a surd. Here a is the radicand, the sign \( \sqrt[n]{\ } \) is the radical sign and the number n is the order of the surd. \( \sqrt[n]{a} \) is read as the nth root of a and equals \( a^{1/n} \).

Laws of Surds

1. \( (\sqrt[n]{a})^n = \sqrt[n]{a^n} = a \)
2. \( \sqrt[n]{a}\times\sqrt[n]{b} = \sqrt[n]{ab} \)
3. \( \sqrt[n]{a}\div\sqrt[n]{b} = \sqrt[n]{\frac{a}{b}} \)
4. \( \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} = \sqrt[n]{\sqrt[m]{a}} \)
5. \( \sqrt[n]{a} = \sqrt[np]{a^p} \) and \( \sqrt[n]{a^m} = \sqrt[np]{a^{mp}} \)

Operations on Surds

Addition and subtraction. Only like surds (same order and same radicand) can be added or subtracted, using the distributive rule: \( a\sqrt[n]{p}+b\sqrt[n]{p} = (a+b)\sqrt[n]{p} \) and \( a\sqrt[n]{p}-b\sqrt[n]{p} = (a-b)\sqrt[n]{p} \).

Multiplication and division. First bring the surds to the same order, using the L.C.M. of their orders. Then \( a\sqrt[n]{p}\times b\sqrt[n]{q} = ab\sqrt[n]{pq} \) and \( \frac{a\sqrt[n]{p}}{b\sqrt[n]{q}} = \frac{a}{b}\sqrt[n]{\frac{p}{q}} \).

Comparison. If x > y > 0 and n > 1 is an integer, then \( \sqrt[n]{x} > \sqrt[n]{y} \). To compare surds of different orders, convert them to the same order first.

Rationalisation

If the product of two surds is rational, each is called a rationalising factor (R.F.) of the other. Turning a surd into a rational number by multiplying with a suitable factor is rationalisation. When a denominator contains a root, the process of changing the fraction into an equivalent one with a rational denominator is called rationalising the denominator.

The rationalising factor of \( a^{1/n} \) is \( a^{1-\frac{1}{n}} \) (a a real number). The R.F. of \( a+b\sqrt{c} \) is \( a-b\sqrt{c} \) (a, b, c rational).

Solved Examples

Example 19. Simplify \( \sqrt{75}-\sqrt{45}+\sqrt{50}-\sqrt{32} \).
Answer:
√75 = 5√3, √45 = 3√5, √50 = 5√2, √32 = 4√2.
The expression becomes \( 5\sqrt3-3\sqrt5+5\sqrt2-4\sqrt2 = 5\sqrt3-3\sqrt5+\sqrt2 \).

Example 20. Simplify : \( 5\sqrt[3]{250}+7\sqrt[3]{16}-14\sqrt[3]{54} \).
Answer:
∛250 = 5∛2, ∛16 = 2∛2 and ∛54 = 3∛2.
So the value is \( 25\sqrt[3]{2}+14\sqrt[3]{2}-42\sqrt[3]{2} = (25+14-42)\sqrt[3]{2} = -3\sqrt[3]{2} \).

Example 21. Simplify : \( \sqrt[3]{2}\times\sqrt[4]{3} \).
Answer:
L.C.M. of 3 and 4 is 12, so make the order 12: \( \sqrt[12]{2^4}\times\sqrt[12]{3^3} = \sqrt[12]{16\times 27} = \sqrt[12]{432} \).

Example 22. Simplify : \( \sqrt{8a^5b}\times\sqrt[3]{4a^2b^2} \).
Answer:
Make the order 6: \( \sqrt[6]{(8a^5b)^3}\times\sqrt[6]{(4a^2b^2)^2} = \sqrt[6]{2^{13}a^{19}b^7} \).
Take out sixth powers: \( = 2^2a^3b\sqrt[6]{2ab} = 4a^3b\sqrt[6]{2ab} \).

Example 23. Divide : \( \sqrt{24}\div\sqrt[3]{200} \).
Answer:
Use order 6: \( \frac{\sqrt[6]{24^3}}{\sqrt[6]{200^2}} = \sqrt[6]{\frac{13824}{40000}} = \sqrt[6]{\frac{216}{625}} \).

Example 24. Which is greater : (i) \( \sqrt[3]{6} \) and \( \sqrt[5]{8} \) (ii) \( \sqrt{\frac12} \) and \( \sqrt[3]{\frac13} \) ?
Answer:
(i) L.C.M. of 3 and 5 is 15. \( \sqrt[3]{6} = \sqrt[15]{6^5} = \sqrt[15]{7776} \) and \( \sqrt[5]{8} = \sqrt[15]{8^3} = \sqrt[15]{512} \). Since 7776 > 512, \( \sqrt[3]{6} \) is greater.
(ii) L.C.M. of 2 and 3 is 6. \( \sqrt{\frac12} = \sqrt[6]{\frac18} \) and \( \sqrt[3]{\frac13} = \sqrt[6]{\frac19} \). As 8 < 9, \( \frac{1}{8} \) > \( \frac{1}{9} \), so \( \sqrt{\frac12} \) is greater.

Example 25. Arrange \( \sqrt{2} \), \( \sqrt[3]{3} \) and \( \sqrt[4]{5} \) in ascending order.
Answer:
L.C.M. of 2, 3, 4 is 12. \( \sqrt2 = \sqrt[12]{64},\ \sqrt[3]{3} = \sqrt[12]{81},\ \sqrt[4]{5} = \sqrt[12]{125} \).
Since 64 < 81 < 125, the order is \( \sqrt2 < \sqrt[3]{3} < \sqrt[4]{5} \).

Example 26. Rationalise the denominator of \( \frac{1}{\sqrt{162}} \).
Answer:
\( \sqrt{162} = 9\sqrt2 \), so \( \frac{1}{9\sqrt2}\times\frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{18} \).

Example 27. Rationalise the denominator of \( \frac{1}{7+5\sqrt3} \).
Answer:
Multiply by \( \frac{7-5\sqrt3}{7-5\sqrt3} \): the denominator becomes 49 − 75 = −26.
Result: \( \frac{7-5\sqrt3}{-26} = \frac{5\sqrt3-7}{26} \).

Example 28. Rationalise the denominator of \( \frac{a^2}{\sqrt{a^2+b^2}+b} \).
Answer:
Multiply by \( \frac{\sqrt{a^2+b^2}-b}{\sqrt{a^2+b^2}-b} \): the denominator is (a² + b²) − b² = a².
So the expression equals \( \frac{a^2(\sqrt{a^2+b^2}-b)}{a^2} = \sqrt{a^2+b^2}-b \).

Example 29. If \( \frac{3+2\sqrt2}{3-\sqrt2} = a+b\sqrt2 \) with a, b rational, find a and b.
Answer:
Multiply top and bottom by 3 + √2: numerator (3 + 2√2)(3 + √2) = 9 + 3√2 + 6√2 + 4 = 13 + 9√2; denominator 9 − 2 = 7.
So the left side is \( \frac{13}{7}+\frac{9}{7}\sqrt2 \). Comparing the rational and irrational parts: a = \( \frac{13}{7} \), b = \( \frac{9}{7} \).

Example 30. If \( x=\frac{1}{2+\sqrt3} \), find the value of \( x^3-x^2-11x+3 \).
Answer:
x = 2 − √3 after rationalising, so x − 2 = −√3. Squaring: x² − 4x + 4 = 3, i.e. x² − 4x + 1 = 0.
Rewrite: x³ − x² − 11x + 3 = x(x² − 4x + 1) + 3(x² − 4x + 1) = 0 + 0 = 0.

Example 31. If \( x=3-\sqrt8 \), find the value of \( x^3+\frac{1}{x^3} \).
Answer:
\( \frac1x = \frac{1}{3-\sqrt8} = 3+\sqrt8 \), so \( x+\frac1x = 6 \).
Use \( x^3+\frac{1}{x^3} = \left(x+\frac1x\right)^3-3\left(x+\frac1x\right) = 6^3-3(6) = 216-18 = 198 \).
The value is 198.

Example 32. If \( \sqrt{5} \) = 2.236 and \( \sqrt{2} \) = 1.414, evaluate \( \frac{3}{\sqrt5+\sqrt2}+\frac{4}{\sqrt5-\sqrt2} \).
Answer:
Combine: \( \frac{3(\sqrt5-\sqrt2)+4(\sqrt5+\sqrt2)}{5-2} = \frac{7\sqrt5+\sqrt2}{3} \).
Substitute: \( \frac{7\times2.236+1.414}{3} = \frac{17.066}{3} \) = 5.689 (approximately).

D. Exponents

Exponents of Real Numbers

Positive integral power. For a real number a and a natural number n, \( a^n \) = a × a × ... × a (n factors). Here a is the base, n the exponent and \( a^n \) is the nth power of a. Example: \( 2^3 \) = 8. For any non-zero a, \( a^0 \) = 1; for instance \( 3^0 \) = 1 and \( \left(\frac34\right)^0 \) = 1.

Negative integral power. For non-zero a and positive integer n, \( a^{-n} = \frac{1}{a^n} \). So \( a^n \) is defined for every integer n.

Rational exponents. For a positive real number a and a positive integer n, the principal nth root of a is the unique positive real number x with \( x^n = a \); it is written \( a^{1/n} \) or \( \sqrt[n]{a} \).

Remark. If a is negative and n is even, the principal nth root is not defined among real numbers, because an even power of a real number is never negative. So \( (-9)^{1/2} \) has no meaning in real numbers.

Laws of Rational Exponents

For positive reals a, b and rational m, n:

1. \( a^m\times a^n = a^{m+n} \)
2. \( a^m\div a^n = a^{m-n} \)
3. \( (a^m)^n = a^{mn} \)
4. \( a^{-n} = \frac{1}{a^n} \)
5. \( a^{m/n} = (a^m)^{1/n} = (a^{1/n})^m = \sqrt[n]{a^m} = (\sqrt[n]{a})^m \)
6. \( (ab)^m = a^mb^m \)
7. \( \left(\frac ab\right)^m = \frac{a^m}{b^m} \)
8. \( a^{b^n} = a^{b\cdot b\cdots b\ (n\ \text{times})} \) (the tower of powers is evaluated from the top).

Solved Examples

Example 33. Evaluate : (i) \( 5^8\div5^3 \) (ii) \( \left(\frac34\right)^{-3} \).
Answer:
(i) \( 5^{8-3} = 5^5 = 3125 \).
(ii) \( \left(\frac34\right)^{-3} = \left(\frac43\right)^{3} = \frac{64}{27} \).

Example 34. Evaluate : (i) \( \left(\frac12\right)^5\times\left(\frac{-2}{3}\right)^4\times\left(\frac35\right)^{-1} \) (ii) \( \left(\frac23\right)^3\times\left(\frac25\right)^{-3}\times\left(\frac35\right)^2 \).
Answer:
(i) \( \frac{1}{32}\times\frac{16}{81}\times\frac53 = \frac{80}{7776} = \frac{5}{486} \).
(ii) \( \frac{8}{27}\times\frac{125}{8}\times\frac{9}{25} = \frac53 \).

Example 35. Simplify : (i) \( \frac{(25)^{3/2}\times(243)^{3/5}}{(16)^{5/4}\times(8)^{4/3}} \) (ii) \( \frac{16\times2^{n+1}-4\times2^n}{16\times2^{n+2}-2\times2^{n+2}} \).
Answer:
(i) 25^{3/2} = 125, 243^{3/5} = 27, 16^{5/4} = 32, 8^{4/3} = 16. So \( \frac{125\times27}{32\times16} = \frac{3375}{512} \).
(ii) Numerator: \( 32\cdot2^n-4\cdot2^n = 2^n(32-4) \); denominator: \( 64\cdot2^n-8\cdot2^n = 2^n(64-8) \). The ratio is \( \frac{28}{56} \) = \( \frac{1}{2} \).

Example 36. Simplify \( \left(\frac{81}{16}\right)^{-3/4}\times\left[\left(\frac{25}{9}\right)^{-3/2}\div\left(\frac52\right)^{-3}\right] \).
Answer:
(81/16)^{-3/4} = (3/2)^{-3} = 8/27.
(25/9)^{-3/2} = (5/3)^{-3} = 27/125 and (5/2)^{-3} = 8/125, so the bracket is \( \frac{27}{125}\div\frac{8}{125} = \frac{27}{8} \).
Product: \( \frac{8}{27}\times\frac{27}{8} = 1 \).

Example 37. Prove that : \( \frac{x^{-1}}{x^{-1}+y^{-1}}+\frac{x^{-1}}{x^{-1}-y^{-1}} = \frac{2y^2}{y^2-x^2} \).
Answer:
Write negative powers as reciprocals: \( \frac{x^{-1}}{x^{-1}+y^{-1}} = \frac{1/x}{(x+y)/(xy)} = \frac{y}{x+y} \) and \( \frac{x^{-1}}{x^{-1}-y^{-1}} = \frac{1/x}{(y-x)/(xy)} = \frac{y}{y-x} \).
Add: \( \frac{y(y-x)+y(y+x)}{(y+x)(y-x)} = \frac{2y^2}{y^2-x^2} \). Hence proved.

Example 38. Find the value of x : \( \left(\frac35\right)^x\left(\frac53\right)^{2x} = \frac{125}{27} \).
Answer:
Rewrite (3/5)^x = (5/3)^{−x}. Then \( \left(\frac53\right)^{-x+2x} = \left(\frac53\right)^x = \frac{125}{27} = \left(\frac53\right)^3 \).
The bases are equal, so x = 3.

Example 39. If \( 25^{x-1} = 5^{2x-1}-100 \), find x.
Answer:
25^{x−1} = 5^{2x−2}. Put it in the equation: \( 5^{2x-2}-5^{2x-1} = -100 \).
Since 5^{2x−1} = 5^{2x−2} × 5: \( 5^{2x-2}(1-5) = -100 \), so 5^{2x−2} = 25 = 5².
Hence 2x − 2 = 2 and x = 2.

Example 40. Assuming that x is a positive real number and a, b, c are rational numbers, show that \( \left(\frac{x^b}{x^c}\right)^a\left(\frac{x^c}{x^a}\right)^b\left(\frac{x^a}{x^b}\right)^c = 1 \).
Answer:
Each bracket simplifies by the law of division: \( (x^{b-c})^a(x^{c-a})^b(x^{a-b})^c \).
Multiplying exponents: \( x^{ab-ac}\cdot x^{bc-ab}\cdot x^{ac-bc} \). Add the exponents: ab − ac + bc − ab + ac − bc = 0.
So the product is \( x^0 = 1 \). Hence proved.