How do you rationalise the denominator of 1/(sqrt(a) + sqrt(b))?

Multiply both numerator and denominator by the conjugate (sqrt(a) - sqrt(b)). This uses the identity (sqrt(a) + sqrt(b))(sqrt(a) - sqrt(b)) = a - b, which eliminates the surds from the denominator. For example, 1/(sqrt(3) + 1) times (sqrt(3) - 1)/(sqrt(3) - 1) = (sqrt(3) - 1)/(3 - 1) = (sqrt(3) - 1)/2.

What are the laws of exponents for real numbers?

a^m times a^n = a^(m+n); a^m / a^n = a^(m-n); (a^m)^n = a^(mn); (ab)^m = a^m times b^m; (a/b)^m = a^m / b^m; a^0 = 1 (a is not 0); a^(-n) = 1/a^n; a^(1/n) = nth root of a.

Samacheer Kalvi 9th Maths Real Numbers Exercise Solutions | KalviGavani

Q: What is the difference between rational and irrational numbers?

A rational number can be expressed as p/q where p and q are integers and q is not zero. Its decimal expansion is either terminating or non-terminating repeating. An irrational number cannot be expressed as p/q and has a non-terminating, non-repeating decimal expansion. Examples: 3/4 is rational; sqrt(2) is irrational.

Exercise Solutions Available

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How to Use This Page

Solve each Samacheer Kalvi exercise question on your own first, then compare your method and answer with the step-by-step solutions below. Every question from Chapter 2 (Real Numbers) is worked out in full. Read the concept, apply the definition, and verify your answer. Pay special attention to the Common Mistakes section to avoid losing marks.

Key Concepts (Quick Recap)

Natural Numbers (N): Counting numbers {1, 2, 3, ...}
Whole Numbers (W): {0, 1, 2, 3, ...}
Integers (Z): {..., -3, -2, -1, 0, 1, 2, 3, ...}
Rational Numbers (Q): Numbers of the form p/q where p, q are integers and q is not 0; decimal expansion is terminating or non-terminating repeating
Irrational Numbers: Non-terminating, non-repeating decimals; cannot be written as p/q (e.g. sqrt(2), sqrt(3), pi)
Real Numbers (R): The union of rational and irrational numbers
Surd: An irrational root of a positive rational number (e.g. sqrt(5), cube root of 7); sqrt(4) = 2 is NOT a surd
Order of a surd: The index of the radical; sqrt(5) is order 2, cube root of 5 is order 3
Like surds: Same radicand under same root; 3sqrt(2) and 5sqrt(2) are like surds
Rationalisation: The process of eliminating surds from the denominator by multiplying by the conjugate
Conjugate of (sqrt(a) + sqrt(b)): (sqrt(a) - sqrt(b)); their product is (a - b), a rational number
Laws of exponents: a^m * a^n = a^(m+n); a^m / a^n = a^(m-n); (a^m)^n = a^(mn); a^0 = 1; a^(-n) = 1/a^n

Exercise 2.1 — Rational Numbers, Properties

Q1. Classify the following as rational or irrational: (i) sqrt(4) (ii) sqrt(5) (iii) 2/3 (iv) 0.3333... (v) 0.1010010001...

Solution

(i) sqrt(4) = 2, which is an integer.
    2 can be written as 2/1.
    Rational.

(ii) sqrt(5) = 2.236... (non-terminating, non-repeating)
    Cannot be written as p/q.
    Irrational.

(iii) 2/3 is already in p/q form (q is not 0).
    Rational.

(iv) 0.3333... = 0.3-bar
    This is a non-terminating repeating decimal.
    0.3333... = 1/3, which is p/q form.
    Rational.

(v) 0.1010010001... has no repeating pattern.
    Non-terminating, non-repeating decimal.
    Cannot be written as p/q.
    Irrational.

Answer(i) Rational, (ii) Irrational, (iii) Rational, (iv) Rational, (v) Irrational

Q2. Express the following as rational numbers in p/q form: (i) 0.6 (ii) 0.35 (iii) 0.75

Solution

(i) 0.6 = 6/10 = 3/5

(ii) 0.35 = 35/100 = 7/20

(iii) 0.75 = 75/100 = 3/4

Answer(i) 3/5, (ii) 7/20, (iii) 3/4

Q3. Express the following repeating decimals as rational numbers: (i) 0.bar(3) (ii) 0.bar(6) (iii) 0.bar(7)

Solution

(i) Let x = 0.333...
    Multiply by 10: 10x = 3.333...
    Subtract: 10x - x = 3.333... - 0.333...
    9x = 3
    x = 3/9 = 1/3

(ii) Let x = 0.666...
    Multiply by 10: 10x = 6.666...
    Subtract: 10x - x = 6.666... - 0.666...
    9x = 6
    x = 6/9 = 2/3

(iii) Let x = 0.777...
    Multiply by 10: 10x = 7.777...
    Subtract: 10x - x = 7.777... - 0.777...
    9x = 7
    x = 7/9

Answer(i) 1/3, (ii) 2/3, (iii) 7/9

Q4. Express 0.bar(27) as a rational number in lowest terms.

Solution

Let x = 0.272727...
Two digits repeat, so multiply by 100:
100x = 27.272727...
Subtract: 100x - x = 27.272727... - 0.272727...
99x = 27
x = 27/99 = 3/11

Answer3/11

Q5. Express 2.bar(36) as a rational number in lowest terms.

Solution

Let x = 2.363636...
Two digits repeat after decimal, so multiply by 100:
100x = 236.363636...
Subtract: 100x - x = 236.363636... - 2.363636...
99x = 234
x = 234/99 = 26/11

Answer26/11

Q6. Express 0.12bar(3) as a rational number.

Solution

Let x = 0.12333...
One digit repeats (3), two non-repeating digits after decimal (12).
Multiply by 100 to move past non-repeating part:
100x = 12.333...
Multiply by 10 to shift one more place:
10x = 1.2333...

Subtract: 100x - 10x = 12.333... - 1.2333...
90x = 11.1
x = 11.1/90 = 111/900 = 37/300

Answer37/300

Q7. State whether the following statements are true or false: (i) Every natural number is a whole number. (ii) Every integer is a rational number. (iii) Every rational number is an integer.

Solution

(i) Natural numbers: {1, 2, 3, ...}
    Whole numbers: {0, 1, 2, 3, ...}
    Every natural number is in the set of whole numbers.
    TRUE.

(ii) Any integer n can be written as n/1, which is p/q form.
    TRUE.

(iii) 3/4 is a rational number but not an integer.
    FALSE. Not every rational number is an integer.

Q8. Find five rational numbers between 1 and 2.

Solution

Method: Write 1 and 2 with denominator 6 (we need more than 5 numbers between them).
1 = 6/6 and 2 = 12/6

Five rational numbers between 6/6 and 12/6:
7/6, 8/6, 9/6, 10/6, 11/6

Simplifying: 7/6, 4/3, 3/2, 5/3, 11/6

Answer7/6, 4/3, 3/2, 5/3, 11/6 (among infinitely many possibilities)

Q9. Find three rational numbers between 3/5 and 4/5.

Solution

Make denominators larger: multiply numerator and denominator by 4.
3/5 = 12/20 and 4/5 = 16/20

Three rational numbers between 12/20 and 16/20:
13/20, 14/20 = 7/10, 15/20 = 3/4

Answer13/20, 7/10, 3/4

Q10. State the closure property of rational numbers under addition. Is Q closed under division?

Solution

Closure under addition: If a and b are rational numbers, then a + b is also a rational number.
Example: 1/3 + 1/4 = 4/12 + 3/12 = 7/12 (rational).

Closure under division: If a and b are rational numbers, a/b is rational ONLY when b is not 0.
Division by zero is undefined. So Q is NOT closed under division (because of division by zero).
However, if we exclude 0 from the denominator, Q is closed under division.

Q11. State the commutative and associative properties of rational numbers under addition.

Solution

Commutative property under addition:
If a and b are rational numbers, then a + b = b + a.
Example: 1/3 + 2/5 = 2/5 + 1/3 = 11/15.

Associative property under addition:
If a, b, c are rational numbers, then (a + b) + c = a + (b + c).
Example: (1/2 + 1/3) + 1/4 = 1/2 + (1/3 + 1/4)
LHS: (5/6) + 1/4 = 10/12 + 3/12 = 13/12
RHS: 1/2 + (7/12) = 6/12 + 7/12 = 13/12
Both are equal. Verified.

Q12. What is the additive identity and multiplicative identity for rational numbers?

Solution

Additive identity: 0 (zero)
For any rational number a, a + 0 = 0 + a = a.

Multiplicative identity: 1 (one)
For any rational number a, a times 1 = 1 times a = a.

Q13. What is the additive inverse and multiplicative inverse (reciprocal) of 3/5?

Solution

Additive inverse of 3/5: A number which when added to 3/5 gives 0.
Additive inverse = -3/5
Check: 3/5 + (-3/5) = 0.

Multiplicative inverse (reciprocal) of 3/5: A number which when multiplied by 3/5 gives 1.
Reciprocal = 5/3
Check: 3/5 times 5/3 = 15/15 = 1.

Q14. Find the additive inverse of: (i) -7/9 (ii) 0 (iii) 5

Solution

(i) Additive inverse of -7/9 = 7/9
(ii) Additive inverse of 0 = 0
(iii) Additive inverse of 5 = -5

Q15. Verify the distributive property: a(b + c) = ab + ac for a = 1/2, b = 3/4, c = 1/4.

Solution

LHS: a(b + c) = 1/2 times (3/4 + 1/4) = 1/2 times (4/4) = 1/2 times 1 = 1/2

RHS: ab + ac = (1/2 times 3/4) + (1/2 times 1/4) = 3/8 + 1/8 = 4/8 = 1/2

LHS = RHS = 1/2. Verified.

Q16. Between which two integers does sqrt(7) lie?

Solution

Find perfect squares around 7:
2 squared = 4 < 7
3 squared = 9 > 7

Since 4 < 7 < 9, we get 2 < sqrt(7) < 3.

sqrt(7) lies between 2 and 3.

Answersqrt(7) lies between 2 and 3.

Q17. Represent 3/4 and -3/4 on the number line.

Solution

For 3/4: Divide the unit between 0 and 1 into 4 equal parts.
  3/4 is the 3rd division mark from 0 towards 1.

For -3/4: Divide the unit between 0 and -1 into 4 equal parts.
  -3/4 is the 3rd division mark from 0 towards -1.

Both numbers are equidistant from 0, on opposite sides.

Q18. Find a rational number between 1/3 and 1/2.

Solution

Use the average method: (1/3 + 1/2) / 2

1/3 + 1/2 = 2/6 + 3/6 = 5/6

(5/6) / 2 = 5/12

Check: 1/3 = 4/12 < 5/12 < 6/12 = 1/2. Yes, 5/12 lies between 1/3 and 1/2.

Answer: 5/12

Exercise 2.2 — Irrational Numbers, Locating on Number Line

Q1. Which of the following are irrational numbers? (i) sqrt(2) (ii) sqrt(9) (iii) 0.bar(3) (iv) pi (v) sqrt(7)/3

Solution

(i) sqrt(2) = 1.4142... Non-terminating, non-repeating.
    IRRATIONAL.

(ii) sqrt(9) = 3, which is an integer = 3/1.
    RATIONAL.

(iii) 0.bar(3) = 1/3, which is p/q form.
     RATIONAL.

(iv) pi = 3.14159... Non-terminating, non-repeating.
     IRRATIONAL.

(v) sqrt(7) is irrational. Dividing an irrational by a rational (3) gives irrational.
    IRRATIONAL.

Answer(i), (iv), (v) are irrational.

Q2. Prove that sqrt(2) is irrational.

Solution

Proof by contradiction:

Assume sqrt(2) is rational.
Then sqrt(2) = p/q, where p, q are integers with no common factor (in lowest terms), q is not 0.

Squaring both sides:
2 = p squared / q squared
p squared = 2 times q squared ... (i)

Since p squared = 2 times q squared, p squared is divisible by 2.
Therefore p is divisible by 2 (if p squared is even, p is even).
Let p = 2k for some integer k.

Substituting in (i):
(2k) squared = 2 times q squared
4k squared = 2 times q squared
2k squared = q squared

Since q squared = 2k squared, q squared is divisible by 2.
Therefore q is divisible by 2.

But this means both p and q are divisible by 2, contradicting our
assumption that p/q is in lowest terms (no common factor).

Therefore, our assumption that sqrt(2) is rational must be FALSE.
Hence, sqrt(2) is IRRATIONAL.

Q3. Prove that sqrt(3) is irrational.

Solution

Proof by contradiction:

Assume sqrt(3) is rational.
Then sqrt(3) = p/q, where p, q are integers with no common factor, q is not 0.

Squaring: 3 = p squared / q squared
p squared = 3 times q squared ... (i)

Since p squared = 3 times q squared, p squared is divisible by 3.
Therefore p is divisible by 3.
Let p = 3k.

Substituting: (3k) squared = 3 times q squared
9k squared = 3 times q squared
3k squared = q squared

Since q squared = 3k squared, q squared is divisible by 3.
Therefore q is divisible by 3.

Both p and q divisible by 3 contradicts p/q being in lowest terms.
Hence, sqrt(3) is IRRATIONAL.

Q4. Locate sqrt(2) on the number line.

Solution

Step-by-step construction:

1. Draw a number line. Mark 0 as O and 1 as A.
2. At A, draw a perpendicular line segment AB of length 1 unit.
3. Join O to B. By Pythagoras theorem:
   OB = sqrt(OA squared + AB squared) = sqrt(1 squared + 1 squared) = sqrt(2)
4. With O as centre and OB as radius, draw an arc cutting the number line at P.
5. P represents sqrt(2) on the number line.

Verification: OP = OB = sqrt(2), so P is at distance sqrt(2) from origin.

Q5. Locate sqrt(5) on the number line.

Solution

Step-by-step construction:

Draw a number line. Mark 0 as O and 2 as A (OA = 2 units).
At A, draw a perpendicular line segment AB of length 1 unit.
Join O to B. By Pythagoras theorem:
   OB = sqrt(OA squared + AB squared) = sqrt(4 + 1) = sqrt(5)
With O as centre and OB as radius, draw an arc cutting the number line at P.
P represents sqrt(5) on the number line.

Q6. Classify the following as rational or irrational: (i) sqrt(4) + sqrt(9) (ii) sqrt(2) + sqrt(3) (iii) sqrt(5) times sqrt(5) (iv) (sqrt(3) - sqrt(3))

Solution

(i) sqrt(4) + sqrt(9) = 2 + 3 = 5
    5 is an integer = 5/1. RATIONAL.

(ii) sqrt(2) + sqrt(3)
    sqrt(2) = 1.414... and sqrt(3) = 1.732...
    Sum = 3.146... which is non-terminating, non-repeating.
    IRRATIONAL.

(iii) sqrt(5) times sqrt(5) = 5
     5 is an integer. RATIONAL.

(iv) sqrt(3) - sqrt(3) = 0
     0 is rational. RATIONAL.

Q7. Represent sqrt(3.5) on the number line.

Solution

Step-by-step construction:

1. Draw a number line. Mark 0 as O and 3.5 as A (OA = 3.5 units).
2. Find midpoint M of OA: M is at 3.5/2 = 1.75 from O.
3. With M as centre and MA as radius (= 1.75), draw a semicircle above the number line.
4. At point B where x = 1 on the number line, draw a perpendicular to meet the semicircle at P.
5. OP = sqrt(3.5).

Verification: Using Pythagoras in triangle:
OP squared = OA times OB = 3.5 times 1... wait, let me redo.
By the geometric mean property: OP = sqrt(1 times 3.5) = sqrt(3.5). Not quite.

Better method:
1. Mark O at 0, A at 3.5 on the number line.
2. From O, mark B at 1 (OB = 1).
3. With center at midpoint of OA (1.75), draw semicircle through O and A.
4. Draw perpendicular from B to meet semicircle at P.
5. Then BP = sqrt(AB times BO)... Actually:

The standard method: Draw a right triangle with legs such that the hypotenuse gives sqrt(3.5).
3.5 = 7/2. We need OA = 7/2.

Alternative: Draw from O with OA = 3.5, then construct perpendicular at midpoint.
With O = 0 and A = 3.5, midpoint M = 1.75.
Draw semicircle with center M, radius 1.75.
Erect perpendicular at x = 1 (point B).
The perpendicular from B meets the semicircle at P.
BP = sqrt(3.5 - 1) times sqrt(1) ... Actually by the geometric mean property:
BP squared = BA times BO where B is between O and A.
BA = 3.5 - 1 = 2.5, BO = 1
BP squared = 2.5 times 1 = 2.5... That gives sqrt(2.5), not sqrt(3.5).

Let me use the correct method:
To represent sqrt(3.5):
Since 3.5 = 7/2, draw OA = 7/2 on number line.
Perpendicular at O of length 1/2 unit.
Hypotenuse = sqrt((7/2) squared + (1/2) squared)... No.

Simplest method:
sqrt(3.5) = sqrt(7/2) = sqrt(7)/sqrt(2)

Draw a right triangle with legs sqrt(7) and 0 perpendicular... This is getting complicated.

Standard textbook method for sqrt(3.5):
1. Mark point O at 0 on number line.
2. From O, mark A at distance 3.5 units (so OA = 3.5).
3. Extend OA by 1 unit beyond A to point C (AC = 1, so OC = 4.5).
4. Find midpoint of OC: M = 4.5/2 = 2.25.
5. Draw semicircle with center M, radius MC = 2.25.
6. Draw perpendicular from A to meet the semicircle at P.
7. AP = sqrt(3.5 times 1) = sqrt(3.5).
8. With O as center, AP as radius, draw arc to meet number line at Q.
9. Q represents sqrt(3.5).

Q8. Show that 3 + sqrt(2) is irrational.

Solution

Proof by contradiction:

Assume 3 + sqrt(2) is rational.
Then 3 + sqrt(2) = p/q for some integers p, q with q is not 0.

sqrt(2) = p/q - 3 = (p - 3q)/q

Here (p - 3q) and q are integers, and q is not 0.
So sqrt(2) = (p - 3q)/q, a rational number.

But we know sqrt(2) is irrational.
This is a contradiction.

Therefore, 3 + sqrt(2) is IRRATIONAL.

Q9. Show that 5 - sqrt(3) is irrational.

Solution

Proof by contradiction:

Assume 5 - sqrt(3) is rational.
Then 5 - sqrt(3) = p/q for integers p, q with q is not 0.

sqrt(3) = 5 - p/q = (5q - p)/q

This means sqrt(3) is rational (ratio of two integers with q is not 0).
But sqrt(3) is known to be irrational.
Contradiction.

Therefore, 5 - sqrt(3) is IRRATIONAL.

Q10. Show that sqrt(2) times sqrt(3) is irrational.

Solution

sqrt(2) times sqrt(3) = sqrt(6)

Proof by contradiction:
Assume sqrt(6) is rational.
Then sqrt(6) = p/q where p, q have no common factor and q is not 0.

Squaring: 6 = p squared / q squared
p squared = 6 times q squared

Since p squared = 6q squared = 2(3q squared), p squared is divisible by 2.
Therefore p is divisible by 2.
Let p = 2k.

(2k) squared = 6 times q squared
4k squared = 6q squared
2k squared = 3 times q squared

Since 2k squared = 3q squared, 3q squared is divisible by 2.
Since 3 is not divisible by 2, q squared must be divisible by 2.
Therefore q is divisible by 2.

Both p and q are divisible by 2, contradicting p/q in lowest terms.
Therefore sqrt(6) is IRRATIONAL.

Q11. Which is greater: sqrt(2) or sqrt(3)? Justify.

Solution

sqrt(2) = 1.414...
sqrt(3) = 1.732...

Since 2 < 3, we have sqrt(2) < sqrt(3).
(For positive numbers, if a < b, then sqrt(a) < sqrt(b).)

Alternatively: 2 < 3, so sqrt(2) < sqrt(3).

Answersqrt(3) > sqrt(2)

Q12. Find two irrational numbers between sqrt(2) and sqrt(3).

Solution

sqrt(2) = 1.4142...
sqrt(3) = 1.7321...

Numbers between these that are non-terminating, non-repeating:
1. sqrt(2.5) = 1.5811... (irrational, lies between)
2. sqrt(2.8) = 1.6733... (irrational, lies between)

Alternative: 1.50500500050000... and 1.60600600060000... are also
irrational numbers between sqrt(2) and sqrt(3).

Q13. Find two irrational numbers between 2 and 3.

Solution

Two irrational numbers between 2 and 3:
1. sqrt(5) = 2.236... (lies between 2 and 3)
2. sqrt(6) = 2.449... (lies between 2 and 3)

Alternative: 2.1010010001... and 2.5050050005... are also valid.

Q14. Is pi rational or irrational? What about 22/7?

Solution

pi = 3.14159265358979... — non-terminating, non-repeating decimal.
pi is IRRATIONAL.

22/7 = 3.142857142857... — non-terminating, repeating decimal (the block 142857 repeats).
22/7 is RATIONAL (it is in p/q form, q is not 0).

Note: 22/7 is only an APPROXIMATION of pi, not equal to pi.

Q15. Give an example of: (i) a number that is rational but not an integer (ii) a number that is an integer but not a natural number (iii) a number that is real but not rational

Solution

(i) Rational but not an integer: 3/4 (it is p/q form but not a whole number)

(ii) Integer but not a natural number: -5 (natural numbers start from 1,
    negative integers are not natural)

(iii) Real but not rational: sqrt(2) (it is a real number but its decimal
     expansion is non-terminating, non-repeating)

Exercise 2.3 — Real Numbers, Operations, Rationalisation

Q1. Simplify: (i) 5sqrt(2) + 3sqrt(2) (ii) 7sqrt(3) - 2sqrt(3) (iii) 4sqrt(5) + 2sqrt(5) - 3sqrt(5)

Solution

(i) 5sqrt(2) + 3sqrt(2) = (5 + 3)sqrt(2) = 8sqrt(2)

(ii) 7sqrt(3) - 2sqrt(3) = (7 - 2)sqrt(3) = 5sqrt(3)

(iii) 4sqrt(5) + 2sqrt(5) - 3sqrt(5) = (4 + 2 - 3)sqrt(5) = 3sqrt(5)

Answer(i) 8sqrt(2), (ii) 5sqrt(3), (iii) 3sqrt(5)

Q2. Simplify: (i) sqrt(12) + sqrt(27) (ii) sqrt(50) - sqrt(18) (iii) sqrt(75) + sqrt(48) - sqrt(108)

Solution

(i) sqrt(12) + sqrt(27)
    sqrt(12) = sqrt(4 times 3) = 2sqrt(3)
    sqrt(27) = sqrt(9 times 3) = 3sqrt(3)
    sqrt(12) + sqrt(27) = 2sqrt(3) + 3sqrt(3) = 5sqrt(3)

(ii) sqrt(50) - sqrt(18)
    sqrt(50) = sqrt(25 times 2) = 5sqrt(2)
    sqrt(18) = sqrt(9 times 2) = 3sqrt(2)
    sqrt(50) - sqrt(18) = 5sqrt(2) - 3sqrt(2) = 2sqrt(2)

(iii) sqrt(75) + sqrt(48) - sqrt(108)
     sqrt(75) = sqrt(25 times 3) = 5sqrt(3)
     sqrt(48) = sqrt(16 times 3) = 4sqrt(3)
     sqrt(108) = sqrt(36 times 3) = 6sqrt(3)
     5sqrt(3) + 4sqrt(3) - 6sqrt(3) = 3sqrt(3)

Answer(i) 5sqrt(3), (ii) 2sqrt(2), (iii) 3sqrt(3)

Q3. Multiply: (i) sqrt(3) times sqrt(5) (ii) sqrt(7) times sqrt(7) (iii) 2sqrt(3) times 3sqrt(2)

Solution

(i) sqrt(3) times sqrt(5) = sqrt(3 times 5) = sqrt(15)

(ii) sqrt(7) times sqrt(7) = sqrt(7 times 7) = sqrt(49) = 7

(iii) 2sqrt(3) times 3sqrt(2) = 2 times 3 times sqrt(3) times sqrt(2)
     = 6 times sqrt(6) = 6sqrt(6)

Answer(i) sqrt(15), (ii) 7, (iii) 6sqrt(6)

Q4. Rationalise the denominator: (i) 1/sqrt(5) (ii) 6/sqrt(3) (iii) 7/sqrt(7)

Solution

(i) 1/sqrt(5) = 1/sqrt(5) times sqrt(5)/sqrt(5) = sqrt(5)/5

(ii) 6/sqrt(3) = 6/sqrt(3) times sqrt(3)/sqrt(3) = 6sqrt(3)/3 = 2sqrt(3)

(iii) 7/sqrt(7) = 7/sqrt(7) times sqrt(7)/sqrt(7) = 7sqrt(7)/7 = sqrt(7)

Answer(i) sqrt(5)/5, (ii) 2sqrt(3), (iii) sqrt(7)

Q5. Rationalise the denominator: (i) 1/(sqrt(3) + 1) (ii) 1/(sqrt(5) - 2) (iii) 5/(sqrt(6) - sqrt(3))

Solution

(i) 1/(sqrt(3) + 1)
   Multiply numerator and denominator by (sqrt(3) - 1):
   = (sqrt(3) - 1) / ((sqrt(3) + 1)(sqrt(3) - 1))
   = (sqrt(3) - 1) / (3 - 1)
   = (sqrt(3) - 1) / 2

(ii) 1/(sqrt(5) - 2)
   Multiply numerator and denominator by (sqrt(5) + 2):
   = (sqrt(5) + 2) / ((sqrt(5) - 2)(sqrt(5) + 2))
   = (sqrt(5) + 2) / (5 - 4)
   = (sqrt(5) + 2) / 1
   = sqrt(5) + 2

(iii) 5/(sqrt(6) - sqrt(3))
   Multiply numerator and denominator by (sqrt(6) + sqrt(3)):
   = 5(sqrt(6) + sqrt(3)) / ((sqrt(6) - sqrt(3))(sqrt(6) + sqrt(3)))
   = 5(sqrt(6) + sqrt(3)) / (6 - 3)
   = 5(sqrt(6) + sqrt(3)) / 3

Answer(i) (sqrt(3) - 1)/2, (ii) sqrt(5) + 2, (iii) 5(sqrt(6) + sqrt(3))/3

Q6. Rationalise the denominator: 4/(sqrt(7) - sqrt(3))

Solution

Multiply numerator and denominator by (sqrt(7) + sqrt(3)):
4(sqrt(7) + sqrt(3)) / ((sqrt(7) - sqrt(3))(sqrt(7) + sqrt(3)))
= 4(sqrt(7) + sqrt(3)) / (7 - 3)
= 4(sqrt(7) + sqrt(3)) / 4
= sqrt(7) + sqrt(3)

Answersqrt(7) + sqrt(3)

Q7. Simplify: (sqrt(2) + sqrt(3)) squared

Solution

(sqrt(2) + sqrt(3)) squared
= (sqrt(2)) squared + 2(sqrt(2))(sqrt(3)) + (sqrt(3)) squared
= 2 + 2sqrt(6) + 3
= 5 + 2sqrt(6)

Answer5 + 2sqrt(6)

Q8. Simplify: (sqrt(5) - sqrt(3)) squared

Solution

(sqrt(5) - sqrt(3)) squared
= (sqrt(5)) squared - 2(sqrt(5))(sqrt(3)) + (sqrt(3)) squared
= 5 - 2sqrt(15) + 3
= 8 - 2sqrt(15)

Answer8 - 2sqrt(15)

Q9. Simplify: (3 + sqrt(3))(2 + sqrt(2))

Solution

(3 + sqrt(3))(2 + sqrt(2))
= 3(2) + 3(sqrt(2)) + sqrt(3)(2) + sqrt(3)(sqrt(2))
= 6 + 3sqrt(2) + 2sqrt(3) + sqrt(6)

Answer6 + 3sqrt(2) + 2sqrt(3) + sqrt(6)

Q10. Simplify: (sqrt(5) + sqrt(2))(sqrt(5) - sqrt(2))

Solution

(sqrt(5) + sqrt(2))(sqrt(5) - sqrt(2))
This is of the form (a + b)(a - b) = a squared - b squared
= (sqrt(5)) squared - (sqrt(2)) squared
= 5 - 2
= 3

Answer3

Q11. If a = 3 + 2sqrt(2), find the value of a + 1/a.

Solution

a = 3 + 2sqrt(2)
1/a = 1/(3 + 2sqrt(2))

Rationalise: Multiply by (3 - 2sqrt(2))/(3 - 2sqrt(2))
1/a = (3 - 2sqrt(2)) / ((3 + 2sqrt(2))(3 - 2sqrt(2)))
   = (3 - 2sqrt(2)) / (9 - 8)
   = (3 - 2sqrt(2)) / 1
   = 3 - 2sqrt(2)

a + 1/a = (3 + 2sqrt(2)) + (3 - 2sqrt(2)) = 6

Answera + 1/a = 6

Q12. If a = 5 - 2sqrt(6), find the value of a squared.

Solution

a = 5 - 2sqrt(6)
a squared = (5 - 2sqrt(6)) squared
= 25 - 2(5)(2sqrt(6)) + (2sqrt(6)) squared
= 25 - 20sqrt(6) + 24
= 49 - 20sqrt(6)

Answera squared = 49 - 20sqrt(6)

Q13. Simplify: sqrt(45) - sqrt(20) + sqrt(80)

Solution

sqrt(45) = sqrt(9 times 5) = 3sqrt(5)
sqrt(20) = sqrt(4 times 5) = 2sqrt(5)
sqrt(80) = sqrt(16 times 5) = 4sqrt(5)

sqrt(45) - sqrt(20) + sqrt(80)
= 3sqrt(5) - 2sqrt(5) + 4sqrt(5)
= (3 - 2 + 4)sqrt(5)
= 5sqrt(5)

Answer5sqrt(5)

Q14. Rationalise and simplify: 1/(sqrt(7) - sqrt(6)) + 1/(sqrt(6) - sqrt(5))

Solution

1/(sqrt(7) - sqrt(6)) = (sqrt(7) + sqrt(6)) / (7 - 6) = sqrt(7) + sqrt(6)

1/(sqrt(6) - sqrt(5)) = (sqrt(6) + sqrt(5)) / (6 - 5) = sqrt(6) + sqrt(5)

Sum = (sqrt(7) + sqrt(6)) + (sqrt(6) + sqrt(5))
    = sqrt(7) + 2sqrt(6) + sqrt(5)

Answersqrt(7) + 2sqrt(6) + sqrt(5)

Q15. If x = 1/(2 - sqrt(3)), find the value of x + 1/x.

Solution

x = 1/(2 - sqrt(3))
Rationalise: x = (2 + sqrt(3)) / ((2 - sqrt(3))(2 + sqrt(3)))
            = (2 + sqrt(3)) / (4 - 3)
            = 2 + sqrt(3)

1/x = 1/(2 + sqrt(3))
Rationalise: 1/x = (2 - sqrt(3)) / ((2 + sqrt(3))(2 - sqrt(3)))
            = (2 - sqrt(3)) / (4 - 3)
            = 2 - sqrt(3)

x + 1/x = (2 + sqrt(3)) + (2 - sqrt(3)) = 4

Answerx + 1/x = 4

Q16. If x = sqrt(2) + 1, find the value of x squared - 1/x squared.

Solution

x = sqrt(2) + 1
x squared = (sqrt(2) + 1) squared = 2 + 2sqrt(2) + 1 = 3 + 2sqrt(2)

1/x = 1/(sqrt(2) + 1)
Rationalise: 1/x = (sqrt(2) - 1) / ((sqrt(2) + 1)(sqrt(2) - 1))
            = (sqrt(2) - 1) / (2 - 1)
            = sqrt(2) - 1

1/x squared = (sqrt(2) - 1) squared = 2 - 2sqrt(2) + 1 = 3 - 2sqrt(2)

x squared - 1/x squared = (3 + 2sqrt(2)) - (3 - 2sqrt(2))
                       = 3 + 2sqrt(2) - 3 + 2sqrt(2)
                       = 4sqrt(2)

Answerx squared - 1/x squared = 4sqrt(2)

Q17. Compare: 3sqrt(2) and 2sqrt(5). Which is greater?

Solution

3sqrt(2) = sqrt(9) times sqrt(2) = sqrt(18)
2sqrt(5) = sqrt(4) times sqrt(5) = sqrt(20)

Since 18 < 20, we have sqrt(18) < sqrt(20).
Therefore 3sqrt(2) < 2sqrt(5).

2sqrt(5) is greater.

Answer2sqrt(5) > 3sqrt(2)

Q18. Arrange in ascending order: sqrt(2), sqrt(3), sqrt(5), 1.5

Solution

sqrt(2) = 1.414...
sqrt(3) = 1.732...
sqrt(5) = 2.236...
1.5 = 1.5

Ascending order: sqrt(2), 1.5, sqrt(3), sqrt(5)

Answersqrt(2) < 1.5 < sqrt(3) < sqrt(5)

Exercise 2.4 — Laws of Exponents, Scientific Notation

Q1. Simplify: (i) 2 cubed times 2 to the 5 (ii) 3 to the 4 times 3 to the 1 (iii) 5 to the 7 / 5 to the 3

Solution

(i) 2 cubed times 2 to the 5 = 2 to the (3+5) = 2 to the 8 = 256

(ii) 3 to the 4 times 3 to the 1 = 3 to the (4+1) = 3 to the 5 = 243

(iii) 5 to the 7 / 5 to the 3 = 5 to the (7-3) = 5 to the 4 = 625

Q2. Simplify: (i) (2 cubed) to the 4 (ii) (3 squared) to the 5 (iii) ((5 to the 2) cubed)

Solution

(i) (2 cubed) to the 4 = 2 to the (3 times 4) = 2 to the 12 = 4096

(ii) (3 squared) to the 5 = 3 to the (2 times 5) = 3 to the 10 = 59049

(iii) ((5 to the 2) cubed) = 5 to the (2 times 3) = 5 to the 6 = 15625

Q3. Simplify: (i) (2 times 3) to the 4 (ii) (5 times 2) cubed (iii) (10 squared times 10) to the 3

Solution

(i) (2 times 3) to the 4 = 2 to the 4 times 3 to the 4 = 16 times 81 = 1296

(ii) (5 times 2) cubed = 5 cubed times 2 cubed = 125 times 8 = 1000

(iii) (10 squared times 10) to the 3 = (10 to the 3) to the 3 = 10 to the 9

Q4. Simplify: (i) 7 to the 0 (ii) 10 to the 0 (iii) (a - b) to the 0 (where a is not b)

Solution

Any non-zero number raised to the power 0 equals 1.
(i) 7 to the 0 = 1
(ii) 10 to the 0 = 1
(iii) (a - b) to the 0 = 1 (provided a is not b, so a - b is not 0)

Q5. Express as a positive exponent: (i) 5 to the (-3) (ii) 2 to the (-5) (iii) 10 to the (-4)

Solution

a to the (-n) = 1 / (a to the n)

(i) 5 to the (-3) = 1 / (5 to the 3) = 1/125

(ii) 2 to the (-5) = 1 / (2 to the 5) = 1/32

(iii) 10 to the (-4) = 1 / (10 to the 4) = 1/10000

Q6. Simplify: (i) (1/2) to the (-3) (ii) (3/5) to the (-2) (iii) (2/3) to the (-1)

Solution

(a/b) to the (-n) = (b/a) to the n

(i) (1/2) to the (-3) = (2/1) to the 3 = 2 cubed = 8

(ii) (3/5) to the (-2) = (5/3) to the 2 = 25/9

(iii) (2/3) to the (-1) = (3/2) to the 1 = 3/2

Q7. Simplify: (2 to the 3 times 3 to the 2) / (2 squared times 3)

Solution

(2 to the 3 times 3 to the 2) / (2 squared times 3)
= 2 to the (3-2) times 3 to the (2-1)
= 2 to the 1 times 3 to the 1
= 2 times 3
= 6

Answer6

Q8. Express in standard form (scientific notation): (i) 3,45,000 (ii) 0.000567 (iii) 4,00,00,000

Solution

(i) 3,45,000 = 3.45 times 10 to the 5

(ii) 0.000567 = 5.67 times 10 to the (-4)

(iii) 4,00,00,000 = 4 times 10 to the 7

Q9. Express the following in ordinary notation: (i) 2.5 times 10 to the 4 (ii) 1.2 times 10 to the (-3) (iii) 6.02 times 10 to the 6

Solution

(i) 2.5 times 10 to the 4 = 25,000

(ii) 1.2 times 10 to the (-3) = 0.0012

(iii) 6.02 times 10 to the 6 = 60,20,000

Q10. Simplify: (27 times 8) / (6) raised using exponents: (3 cubed times 2 cubed) / (2 times 3)

Solution

(3 cubed times 2 cubed) / (2 times 3)
= 3 to the (3-1) times 2 to the (3-1)
= 3 squared times 2 squared
= 9 times 4
= 36

Answer36

Q11. Simplify: (125 times 27) / (15 times 9) using laws of exponents.

Solution

125 = 5 cubed, 27 = 3 cubed, 15 = 3 times 5, 9 = 3 squared

(5 cubed times 3 cubed) / (3 times 5 times 3 squared)
= (5 cubed times 3 cubed) / (5 times 3 cubed)
= 5 to the (3-1) times 3 to the (3-3)
= 5 squared times 3 to the 0
= 25 times 1
= 25

Answer25

Q12. If 2 to the x = 1/16, find x.

Solution

2 to the x = 1/16 = 1/(2 to the 4) = 2 to the (-4)

Therefore x = -4.

Answerx = -4

Q13. If 5 to the (2x-1) = 125, find x.

Solution

5 to the (2x-1) = 125 = 5 cubed

2x - 1 = 3
2x = 4
x = 2

Answerx = 2

Q14. Simplify: ((25) to the 3 times (5) to the (-6)) / ((125) squared)

Solution

25 = 5 squared, 125 = 5 cubed

((5 squared) cubed times 5 to the (-6)) / ((5 cubed) squared)
= (5 to the 6 times 5 to the (-6)) / (5 to the 6)
= 5 to the (6-6) / (5 to the 6)
= 5 to the 0 / (5 to the 6)
= 1 / (5 to the 6)
= 1/15625

Answer1/15625

Q15. Express 0.0000000085 in scientific notation.

Solution

0.0000000085 = 8.5 times 10 to the (-9)

Move the decimal point 9 places to the right to get 8.5, so the exponent is -9.

Answer8.5 times 10 to the (-9)

Q16. Evaluate: (2 to the 0 + 3 to the 0) times 5 to the 0

Solution

to the 0 = 1
to the 0 = 1
to the 0 = 1

(1 + 1) times 1 = 2 times 1 = 2

Answer2

Q17. Simplify: ((3 to the 2) to the 3 times 3 to the 4) / (3 to the 7 times 3 to the (-1))

Solution

Numerator: (3 to the 6) times (3 to the 4) = 3 to the 10
Denominator: (3 to the 7) times (3 to the (-1)) = 3 to the 6

Result: 3 to the 10 / 3 to the 6 = 3 to the 4 = 81

Answer81

Q18. Which is larger: 3 to the 4 or 4 cubed?

Solution

to the 4 = 81
cubed = 64

> 64, so 3 to the 4 is larger.

Answer3 to the 4 (81) is larger than 4 cubed (64).

Exercise 2.5 — Word Problems

Q1. The length of a rectangle is 3 + sqrt(2) cm and its breadth is 3 - sqrt(2) cm. Find: (i) the area (ii) the perimeter.

Solution

Length = (3 + sqrt(2)) cm, Breadth = (3 - sqrt(2)) cm

(i) Area = Length times Breadth
   = (3 + sqrt(2))(3 - sqrt(2))
   = 3 squared - (sqrt(2)) squared
   = 9 - 2
   = 7 sq cm

(ii) Perimeter = 2(Length + Breadth)
    = 2((3 + sqrt(2)) + (3 - sqrt(2)))
    = 2(6)
    = 12 cm

Answer(i) Area = 7 sq cm, (ii) Perimeter = 12 cm

Q2. If a + b = 10 and ab = 21, find the value of a squared + b squared.

Solution

(a + b) squared = a squared + 2ab + b squared
10 squared = a squared + 2(21) + b squared
100 = a squared + 42 + b squared
a squared + b squared = 100 - 42 = 58

Answera squared + b squared = 58

Q3. If x = 2 + sqrt(3), find the value of x squared + 1/x squared.

Solution

x = 2 + sqrt(3)

First find 1/x:
1/x = 1/(2 + sqrt(3))
Rationalise: multiply by (2 - sqrt(3))/(2 - sqrt(3))
= (2 - sqrt(3)) / ((2 + sqrt(3))(2 - sqrt(3)))
= (2 - sqrt(3)) / (4 - 3)
= 2 - sqrt(3)

x + 1/x = (2 + sqrt(3)) + (2 - sqrt(3)) = 4

(x + 1/x) squared = x squared + 2 + 1/x squared
4 squared = x squared + 2 + 1/x squared
16 = x squared + 1/x squared + 2
x squared + 1/x squared = 14

Answerx squared + 1/x squared = 14

Q4. A rectangular garden has length (7 + 3sqrt(2)) m and breadth (7 - 3sqrt(2)) m. Find the area of the garden.

Solution

Area = Length times Breadth
= (7 + 3sqrt(2))(7 - 3sqrt(2))
= 7 squared - (3sqrt(2)) squared
= 49 - 9 times 2
= 49 - 18
= 31 sq m

AnswerArea = 31 sq m

Q5. Rationalise the denominator and simplify: 1/(sqrt(7) + sqrt(6) - sqrt(13)). [Hint: Group two terms and rationalise in steps.]

Solution

Let us group: 1/(sqrt(7) + sqrt(6) - sqrt(13))

First rationalise with respect to (sqrt(7) + sqrt(6)):
Multiply by (sqrt(7) - sqrt(6))/(sqrt(7) - sqrt(6)):

Actually, let us take a different approach.
Let us rationalise the denominator step by step.

We can write: sqrt(7) + sqrt(6) - sqrt(13)

Group as: (sqrt(7) + sqrt(6)) - sqrt(13)

Multiply numerator and denominator by the conjugate of the entire denominator:
Conjugate = (sqrt(7) + sqrt(6)) + sqrt(13)

= ((sqrt(7) + sqrt(6)) + sqrt(13)) / ((sqrt(7) + sqrt(6) - sqrt(13))((sqrt(7) + sqrt(6)) + sqrt(13)))

Denominator = (sqrt(7) + sqrt(6)) squared - (sqrt(13)) squared
= (7 + 2sqrt(42) + 6) - 13
= 13 + 2sqrt(42) - 13
= 2sqrt(42)

So: 1/(sqrt(7) + sqrt(6) - sqrt(13)) = (sqrt(7) + sqrt(6) + sqrt(13)) / (2sqrt(42))

Now rationalise 2sqrt(42):
= (sqrt(7) + sqrt(6) + sqrt(13)) / (2sqrt(42)) times sqrt(42)/sqrt(42)
= (sqrt(42)(sqrt(7) + sqrt(6) + sqrt(13))) / (2 times 42)
= (sqrt(294) + sqrt(252) + sqrt(546)) / 84
= (7sqrt(6) + 6sqrt(7) + sqrt(546)) / 84

Answer(7sqrt(6) + 6sqrt(7) + sqrt(546)) / 84

Q6. Simplify: (3 + sqrt(3))(3 - sqrt(3)) + (sqrt(2) + sqrt(5))(sqrt(2) - sqrt(5))

Solution

(3 + sqrt(3))(3 - sqrt(3)) = 3 squared - (sqrt(3)) squared = 9 - 3 = 6

(sqrt(2) + sqrt(5))(sqrt(2) - sqrt(5)) = (sqrt(2)) squared - (sqrt(5)) squared = 2 - 5 = -3

Sum = 6 + (-3) = 3

Answer3

Q7. If x = 1/(3 - 2sqrt(2)), find x + 1/x.

Solution

x = 1/(3 - 2sqrt(2))
Rationalise: x = (3 + 2sqrt(2)) / ((3 - 2sqrt(2))(3 + 2sqrt(2)))
= (3 + 2sqrt(2)) / (9 - 8)
= 3 + 2sqrt(2)

1/x = 1/(3 + 2sqrt(2))
Rationalise: 1/x = (3 - 2sqrt(2)) / (9 - 8) = 3 - 2sqrt(2)

x + 1/x = (3 + 2sqrt(2)) + (3 - 2sqrt(2)) = 6

Answerx + 1/x = 6

Q8. Find the value of a if (a - 5) squared = 3.

Solution

(a - 5) squared = 3
a - 5 = sqrt(3) or a - 5 = -sqrt(3)
a = 5 + sqrt(3) or a = 5 - sqrt(3)

Answera = 5 + sqrt(3) or a = 5 - sqrt(3)

Q9. The product of two numbers is 7sqrt(7) and one of them is sqrt(7). Find the other number.

Solution

Let the other number be x.
x times sqrt(7) = 7sqrt(7)
x = 7sqrt(7) / sqrt(7)
x = 7

AnswerThe other number is 7.

Q10. Simplify: (1/(1 + sqrt(2))) + (1/(sqrt(2) + sqrt(3))) + (1/(sqrt(3) + 2))

Solution

Rationalise each term:

1/(1 + sqrt(2)) = (1 - sqrt(2)) / ((1 + sqrt(2))(1 - sqrt(2)))
                = (1 - sqrt(2)) / (1 - 2) = (1 - sqrt(2)) / (-1) = sqrt(2) - 1

1/(sqrt(2) + sqrt(3)) = (sqrt(2) - sqrt(3)) / ((sqrt(2) + sqrt(3))(sqrt(2) - sqrt(3)))
                      = (sqrt(2) - sqrt(3)) / (2 - 3) = (sqrt(2) - sqrt(3)) / (-1) = sqrt(3) - sqrt(2)

1/(sqrt(3) + 2) = (sqrt(3) - 2) / ((sqrt(3) + 2)(sqrt(3) - 2))
                = (sqrt(3) - 2) / (3 - 4) = (sqrt(3) - 2) / (-1) = 2 - sqrt(3)

Sum = (sqrt(2) - 1) + (sqrt(3) - sqrt(2)) + (2 - sqrt(3))
    = sqrt(2) - 1 + sqrt(3) - sqrt(2) + 2 - sqrt(3)
    = (-1 + 2) + (sqrt(2) - sqrt(2)) + (sqrt(3) - sqrt(3))
    = 1 + 0 + 0
    = 1

Answer1

Q11. If x = 1/(2 - sqrt(3)), find x squared - 4x + 1.

Solution

x = 1/(2 - sqrt(3))
Rationalise: x = (2 + sqrt(3)) / (4 - 3) = 2 + sqrt(3)

x squared = (2 + sqrt(3)) squared = 4 + 4sqrt(3) + 3 = 7 + 4sqrt(3)

x squared - 4x + 1 = (7 + 4sqrt(3)) - 4(2 + sqrt(3)) + 1
= 7 + 4sqrt(3) - 8 - 4sqrt(3) + 1
= (7 - 8 + 1) + (4sqrt(3) - 4sqrt(3))
= 0 + 0
= 0

Answer0

Q12. Simplify: sqrt(8) + sqrt(18) - sqrt(50) + sqrt(98)

Solution

sqrt(8) = sqrt(4 times 2) = 2sqrt(2)
sqrt(18) = sqrt(9 times 2) = 3sqrt(2)
sqrt(50) = sqrt(25 times 2) = 5sqrt(2)
sqrt(98) = sqrt(49 times 2) = 7sqrt(2)

sqrt(8) + sqrt(18) - sqrt(50) + sqrt(98)
= 2sqrt(2) + 3sqrt(2) - 5sqrt(2) + 7sqrt(2)
= (2 + 3 - 5 + 7)sqrt(2)
= 7sqrt(2)

Answer7sqrt(2)

Q13. If a = sqrt(2) + 1 and b = sqrt(2) - 1, find: (i) a + b (ii) ab (iii) a squared + b squared

Solution

a = sqrt(2) + 1, b = sqrt(2) - 1

(i) a + b = (sqrt(2) + 1) + (sqrt(2) - 1) = 2sqrt(2)

(ii) ab = (sqrt(2) + 1)(sqrt(2) - 1) = (sqrt(2)) squared - 1 squared = 2 - 1 = 1

(iii) a squared + b squared = (a + b) squared - 2ab
     = (2sqrt(2)) squared - 2(1)
     = 8 - 2 = 6

Answer(i) 2sqrt(2), (ii) 1, (iii) 6

Q14. The sides of a right triangle are sqrt(3), sqrt(5), and sqrt(8). Verify the Pythagorean theorem.

Solution

Check: (sqrt(3)) squared + (sqrt(5)) squared = (sqrt(8)) squared?
3 + 5 = 8
8 = 8

Yes, 3 + 5 = 8. The Pythagorean theorem is verified.
The hypotenuse is sqrt(8), and the legs are sqrt(3) and sqrt(5).

AnswerVerified. 3 + 5 = 8.

Q15. Express 0.4bar(7) as a rational number in lowest terms.

Solution

Let x = 0.4777...
One non-repeating digit (4), one repeating digit (7).
Multiply by 10: 10x = 4.777...
Multiply by 100: 100x = 47.777...

Subtract: 100x - 10x = 47.777... - 4.777...
90x = 43
x = 43/90

Answer43/90

Q16. Represent sqrt(10) on the number line.

Solution

To represent sqrt(10):

1. Mark O at 0 on the number line.
2. Mark A at 3 on the number line (so OA = 3 units).
3. At A, construct a perpendicular AB of length 1 unit.
4. Join O to B. By Pythagoras:
   OB = sqrt(OA squared + AB squared) = sqrt(9 + 1) = sqrt(10)
5. With O as centre and OB as radius, draw an arc cutting the number line at P.
6. P represents sqrt(10) on the number line.

Alternative: sqrt(10) = sqrt(1 + 9) = sqrt(1 squared + 3 squared)
Construct a right triangle with legs 1 and 3; the hypotenuse gives sqrt(10).

Q17. Show that 2 - sqrt(5) is irrational.

Solution

Proof by contradiction:

Assume 2 - sqrt(5) is rational.
Then 2 - sqrt(5) = p/q for integers p, q with q is not 0.

sqrt(5) = 2 - p/q = (2q - p)/q

This means sqrt(5) is rational (ratio of two integers with q is not 0).
But sqrt(5) is known to be irrational (its decimal expansion is 2.236..., non-terminating, non-repeating).
Contradiction.

Therefore, 2 - sqrt(5) is IRRATIONAL.

Q18. Simplify: ((sqrt(5) + sqrt(3)) / (sqrt(5) - sqrt(3))) + ((sqrt(5) - sqrt(3)) / (sqrt(5) + sqrt(3)))

Solution

First term: (sqrt(5) + sqrt(3)) / (sqrt(5) - sqrt(3))
Rationalise: = ((sqrt(5) + sqrt(3))(sqrt(5) + sqrt(3))) / ((sqrt(5) - sqrt(3))(sqrt(5) + sqrt(3)))
= (sqrt(5) + sqrt(3)) squared / (5 - 3)
= (5 + 2sqrt(15) + 3) / 2
= (8 + 2sqrt(15)) / 2
= 4 + sqrt(15)

Second term: (sqrt(5) - sqrt(3)) / (sqrt(5) + sqrt(3))
Rationalise: = ((sqrt(5) - sqrt(3))(sqrt(5) - sqrt(3))) / ((sqrt(5) + sqrt(3))(sqrt(5) - sqrt(3)))
= (sqrt(5) - sqrt(3)) squared / (5 - 3)
= (5 - 2sqrt(15) + 3) / 2
= (8 - 2sqrt(15)) / 2
= 4 - sqrt(15)

Sum = (4 + sqrt(15)) + (4 - sqrt(15)) = 8

Answer8

Common Mistakes

1. Calling every decimal irrational. A decimal is irrational only if it is non-terminating AND non-repeating. 0.333... = 1/3 is rational (repeating), and 0.75 is rational (terminating). Only decimals like 1.41421356... (sqrt(2)) are irrational.

2. Forgetting to simplify surds before adding. sqrt(12) + sqrt(27) is NOT sqrt(39). First simplify: sqrt(12) = 2sqrt(3) and sqrt(27) = 3sqrt(3). Then add: 2sqrt(3) + 3sqrt(3) = 5sqrt(3). Like surds (same radicand) can be added; unlike surds cannot.

3. Applying exponent rules incorrectly. a^m times a^n = a^(m+n), NOT a^(mn). For example, 2^3 times 2^4 = 2^7 = 128, NOT 2^12. Also, (a^m)^n = a^(mn), NOT a^(m+n).

4. Rationalising incorrectly by multiplying only the numerator. Both numerator AND denominator must be multiplied by the conjugate. For 1/(sqrt(3)+1), multiply by (sqrt(3)-1)/(sqrt(3)-1) to get (sqrt(3)-1)/2, not just sqrt(3)-1.

5. Confusing (sqrt(a))^2 with sqrt(a^2). They give the same result for positive a: (sqrt(a))^2 = a and sqrt(a^2) = a. But for negative a, sqrt(a^2) =	a	, while (sqrt(a))^2 is not defined for negative a in real numbers.

6. Treating sqrt(a+b) as sqrt(a) + sqrt(b). sqrt(a+b) is NOT equal to sqrt(a) + sqrt(b). For example, sqrt(9+16) = sqrt(25) = 5, but sqrt(9) + sqrt(16) = 3 + 4 = 7. The correct identity is sqrt(a) times sqrt(b) = sqrt(ab), NOT sqrt(a+b).

7. Dividing by zero in rational number proofs. When proving a number is irrational by contradiction, remember that the denominator q cannot be zero. Also, when using the reciprocal property, make sure the number is not zero before taking its reciprocal.

Exam Checklist

[ ] I can classify numbers as natural, whole, integer, rational, or irrational.
[ ] I can express terminating and repeating decimals as p/q (rational number) form.
[ ] I can convert 0.bar(a), 0.abar(c), and mixed repeating decimals to fractions.
[ ] I can prove sqrt(2) and sqrt(3) are irrational using proof by contradiction.
[ ] I can simplify surds: sqrt(12) = 2sqrt(3), sqrt(75) = 5sqrt(3), etc.
[ ] I can add and subtract like surds: 3sqrt(2) + 5sqrt(2) = 8sqrt(2).
[ ] I can multiply surds: sqrt(a) times sqrt(b) = sqrt(ab).
[ ] I can rationalise denominators — both simple (1/sqrt(a)) and compound (1/(sqrt(a)+sqrt(b))).
[ ] I can expand (sqrt(a) + sqrt(b)) squared and (sqrt(a) - sqrt(b)) squared.
[ ] I can use (a+b)(a-b) = a squared - b squared to simplify surd expressions.
[ ] I can apply all laws of exponents correctly, including negative and zero exponents.
[ ] I can convert between standard form (scientific notation) and ordinary notation.
[ ] I can locate irrational numbers like sqrt(2), sqrt(3), sqrt(5) on the number line.
[ ] I can find rational numbers between any two given rational numbers.

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[Class 9 Maths Subject Guide](/school/tamil-nadu-state-board/class/9/maths)

Frequently Asked Questions

A rational number can be expressed as p/q where p and q are integers and q is not zero. Its decimal expansion is either terminating or non-terminating repeating. An irrational number cannot be expressed as p/q and has a non-terminating, non-repeating decimal expansion. Examples: 3/4 is rational; sqrt(2) is irrational.