GSEB Class 9 Maths Solutions Chapter 1 Number Systems Exercise 1.5

 

Question 1. Classify the following numbers as rational or irrational.
(i) \( 2 - \sqrt{5} \)
(ii) \( (3 + \sqrt{23}) – \sqrt{23} \)
(iii) \( \frac{2 \sqrt{7}}{7 \sqrt{7}} \)
(iv) \( \frac{1}{\sqrt{2}} \)
(v) \( 2\pi \)
Answer:
(i) 2 is a rational number and \( \sqrt{5} \) is an irrational number. When you subtract an irrational number from a rational number, the result is an irrational number.
In simple words: A rational number combined with an irrational number through subtraction always yields an irrational number.
(ii) \( (3 + \sqrt{23}) – \sqrt{23} = 3 + \sqrt{23} - \sqrt{23} = 3 \). Since 3 is a whole number, it is a rational number.
In simple words: The expression simplifies to 3, which is a rational number because it can be written as \( \frac{3}{1} \).
(iii) \( \frac{2 \sqrt{7}}{7 \sqrt{7}} \) simplifies to \( \frac{2}{7} \). As this can be written as a fraction of two integers, it represents a rational number.
In simple words: After cancelling \( \sqrt{7} \), the fraction becomes \( \frac{2}{7} \), which is a rational number.
(iv) Here, 1 is a rational number and \( \sqrt{2} \) is an irrational number. The outcome of dividing a non-zero rational number by an irrational number will always be an irrational number. Thus, \( \frac{1}{\sqrt{2}} \) is an irrational number.
In simple words: Dividing a rational number by an irrational number always gives an irrational result.
(v) Here, 2 is a rational number (which is not zero) and \( \pi \) is an irrational number. The product of a non-zero rational number and an irrational number is always an irrational number. Hence, \( 2\pi \) is an irrational number.
In simple words: When you multiply a non-zero rational number by an irrational number, the answer is always irrational.

Exam Tip: Remember the basic rules for combining rational and irrational numbers: (rational + irrational = irrational), (rational - irrational = irrational), (rational × irrational = irrational if rational is non-zero), (rational / irrational = irrational if rational is non-zero), (irrational × irrational can be rational or irrational).

 

Question 2. Simplify each of the following expressions:
(i) \( (3 + \sqrt{3}) (2 + \sqrt{2}) \)
(ii) \( (3 + \sqrt{3}) (3 – \sqrt{3}) \)
(iii) \( (\sqrt{5} + \sqrt{2})^2 \)
(iv) \( (\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2}) \)
Answer:
(i) \( (3 + \sqrt{3}) (2 + \sqrt{2}) \)
\( = 3(2 + \sqrt{2}) + \sqrt{3}(2 + \sqrt{2}) \)
\( = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{3} \times \sqrt{2} \)
\( = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6} \)
In simple words: We multiply each term in the first bracket by each term in the second bracket, then simplify the square root terms to get the final expression.
(ii) \( (3 + \sqrt{3}) (3 – \sqrt{3}) \)
\( = 3^2 - (\sqrt{3})^2 \)
\( = 9 - 3 \)
\( = 6 \)
In simple words: This uses the difference of squares formula, where \( (a+b)(a-b) = a^2-b^2 \). So, \( 3^2 - (\sqrt{3})^2 \) becomes \( 9-3 \), which is 6.
(iii) \( (\sqrt{5} + \sqrt{2})^2 \)
\( = (\sqrt{5})^2 + 2 (\sqrt{5}) (\sqrt{2}) + (\sqrt{2})^2 \)
\( = 5 + 2\sqrt{10} + 2 \)
\( = 7 + 2\sqrt{10} \)
In simple words: We expand the squared term using the formula \( (a+b)^2 = a^2+2ab+b^2 \). This leads to \( 5 + 2\sqrt{10} + 2 \), which simplifies to \( 7 + 2\sqrt{10} \).
(iv) \( (\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2}) \)
\( = (\sqrt{5})^2 – (\sqrt{2})^2 \)
\( = 5 - 2 \)
\( = 3 \)
In simple words: Similar to part (ii), this uses the difference of squares formula, \( (a-b)(a+b) = a^2-b^2 \). So, \( (\sqrt{5})^2 - (\sqrt{2})^2 \) becomes \( 5-2 \), which is 3.

Exam Tip: Memorize the algebraic identities for square roots: \( (a \pm \sqrt{b})^2 = a^2 \pm 2a\sqrt{b} + b \) and \( (\sqrt{a} \pm \sqrt{b})^2 = a \pm 2\sqrt{ab} + b \), as well as the difference of squares: \( (a-b)(a+b) = a^2-b^2 \).

 

Question 3. Recall, \( \pi \) is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, \( \pi = \frac { c }{ d } \). This seems to contradict the fact that \( \pi \) is irrational. How will you resolve this contradiction?
Answer: The definition of \( \pi \) as the ratio of a circle's circumference (\( c \)) to its diameter (\( d \)) is \( \pi = \frac { c }{ d } \). There is no actual contradiction here. This is because either \( c \) (circumference) or \( d \) (diameter) or both must be irrational numbers. If both were rational, \( \pi \) would also be rational, which we know is false. Thus, \( \pi \) remains an irrational number.
In simple words: There's no problem because for \( \pi \) to be irrational, at least one of the measurements—the circumference or the diameter—must itself be an irrational number. We cannot measure them both perfectly in rational units.

Exam Tip: Understand that \( \pi \) being irrational means you can never precisely measure both the circumference and diameter of a circle using only rational numbers; at least one measurement will be irrational.

 

Question 4. Represent \( \sqrt{9.3} \) on the number line.
Answer: We aim to find \( \sqrt{9.3} \) using a geometric method.

1. Draw a number line and mark point A. Then, from A, measure 9.3 units to mark point B.
2. Next, from B, measure 1 unit further to mark point C.
3. Find the midpoint of AC by drawing a perpendicular bisector, and label this midpoint O.
4. Use O as the center and OC as the radius to draw a semicircle.
5. At point B, draw a line perpendicular to AC that intersects the semicircle at point D. The length of BD will be \( \sqrt{9.3} \).
6. Finally, with B as the center and BD as the radius, draw an arc that meets the number line at P. This point P represents \( \sqrt{9.3} \) units from B.

AOBD is a right angled triangle. Let \( AC = x+1 \) units where \( x=9.3 \). Radius of circle \( r = \frac { AC }{ 2 } = \frac {x+1}{ 2 } \) units. So, \( OC = OD = \frac {x+1}{ 2 } \) units. We can express OB as \( OB = OC - BC = \frac {x+1}{ 2 } - 1 = \frac {x+1-2}{ 2 } = \frac {x-1}{ 2 } \) units. By Pythagoras theorem in \( \triangle OBD \), we have: \( BD^2 = OD^2 – OB^2 \) \( BD^2 = \left( \frac {x+1}{ 2 } \right)^2 – \left( \frac {x-1}{ 2 } \right)^2 \) \( = \frac{(x+1)^2 - (x-1)^2}{4} \) \( = \frac{(x^2+2x+1) - (x^2-2x+1)}{4} \) \( = \frac{x^2+2x+1 - x^2+2x-1}{4} \) \( = \frac{4x}{4} = x \) Thus, \( BD^2 = x \) \( \implies BD = \sqrt{x} \) Hence, for \( x = 9.3 \), \( BD = \sqrt{9.3} \) \( \implies BD \approx 3.049 \)In simple words: We construct a semicircle and a perpendicular line. The length of this perpendicular line from the number line to the semicircle at a specific point gives us the square root value we want to find. We then transfer this length to the number line using a compass.

Exam Tip: Remember the steps for geometric construction clearly. This method is crucial for representing irrational numbers of the form \( \sqrt{x} \) on the number line. Practice drawing the diagrams accurately.

 

Question 5. Rationalise the denominators of the following:
(i) \( \frac{1}{\sqrt{7}} \)
(ii) \( \frac{1}{\sqrt{7}-\sqrt{6}} \)
(iii) \( \frac{1}{\sqrt{5}+\sqrt{2}} \)
(iv) \( \frac{1}{\sqrt{7}-2} \)
Answer:
(i) \( \frac{1}{\sqrt{7}} \)
To rationalize \( \frac{1}{\sqrt{7}} \), we multiply both the numerator and denominator by \( \sqrt{7} \).
\( = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} \)
\( = \frac{\sqrt{7}}{7} \)
In simple words: Multiply the top and bottom by \( \sqrt{7} \) to remove the square root from the denominator.
(ii) \( \frac{1}{\sqrt{7}-\sqrt{6}} \)
To rationalize \( \frac{1}{\sqrt{7}-\sqrt{6}} \), we multiply the numerator and denominator by its conjugate, \( \sqrt{7}+\sqrt{6} \).
\( = \frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} \)
\( = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2} \)
\( = \frac{\sqrt{7}+\sqrt{6}}{7-6} \)
\( = \frac{\sqrt{7}+\sqrt{6}}{1} \)
\( = \sqrt{7}+\sqrt{6} \)
In simple words: Multiply the top and bottom by the opposite sign version of the denominator (the conjugate) to use the difference of squares rule and get rid of the square roots.
(iii) \( \frac{1}{\sqrt{5}+\sqrt{2}} \)
For \( \frac{1}{\sqrt{5}+\sqrt{2}} \), we multiply the top and bottom by its conjugate, \( \sqrt{5}-\sqrt{2} \).
\( = \frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \)
\( = \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2-(\sqrt{2})^2} \)
\( = \frac{\sqrt{5}-\sqrt{2}}{5-2} \)
\( = \frac{\sqrt{5}-\sqrt{2}}{3} \)
In simple words: To remove the square roots from the bottom, multiply both parts of the fraction by \( \sqrt{5}-\sqrt{2} \).
(iv) \( \frac{1}{\sqrt{7}-2} \)
To rationalize \( \frac{1}{\sqrt{7}-2} \), we multiply the numerator and denominator by its conjugate, \( \sqrt{7}+2 \).
\( = \frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} \)
\( = \frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2} \)
\( = \frac{\sqrt{7}+2}{7-4} \)
\( = \frac{\sqrt{7}+2}{3} \)
In simple words: Multiply the numerator and denominator by \( \sqrt{7}+2 \) to clear the square root from the bottom.

Exam Tip: When rationalizing a denominator with a binomial involving square roots (like \( a \pm \sqrt{b} \) or \( \sqrt{a} \pm \sqrt{b} \)), always multiply by its conjugate. The conjugate has the same terms but with the opposite sign between them, allowing you to use the difference of squares identity \( (x-y)(x+y) = x^2-y^2 \).