Maharashtra Board Class 9 Maths Chapter 2 Set 2.2 Algebra Standard Part 1 Real Numbers Solutions

Question 1. Show that \( 4\sqrt{2} \) is an irrational number.
Answer: Let us assume that \( 4\sqrt{2} \) is a rational number. So, we can find co-prime integers \( a \) and \( b \) (\( b \neq 0 \)) such that:
\( 4\sqrt{2} = \frac{a}{b} \)
\( \implies \sqrt{2} = \frac{a}{4b} \)
Since \( a \) and \( b \) are integers, \( 4b \) is also an integer (as \( b \neq 0 \)), which means \( \frac{a}{4b} \) is a rational number. Therefore, \( \sqrt{2} \) must also be a rational number. However, this contradicts the well-known fact that \( \sqrt{2} \) is an irrational number. This method of proof is known as proof by contradiction, which is a powerful tool in mathematics. Thus, our assumption that \( 4\sqrt{2} \) is rational is incorrect. Hence, \( 4\sqrt{2} \) is an irrational number.
In simple words: To prove \( 4\sqrt{2} \) is irrational, we first pretend it is a normal fraction (rational). But when we simplify it, we get \( \sqrt{2} \) equals a fraction, which we know is impossible because \( \sqrt{2} \) is definitely irrational. Since this leads to a contradiction, our starting guess must be wrong, meaning \( 4\sqrt{2} \) is indeed irrational.

🎯 Exam Tip: When proving numbers irrational, always start with the assumption that the number is rational and show that it leads to a contradiction. Clearly state that the RHS is rational while the LHS is irrational.

Question 1. Prove that \( 4\sqrt{2} \) is an irrational number.
Answer: Let us assume that \( 4\sqrt{2} \) is a rational number. This method of proof is known as proof by contradiction, which is a powerful tool in mathematics.
So, we can find co-prime integers ‘a’ and ‘b’ (\( b \neq 0 \)) such that:
\( 4\sqrt{2} = \frac{a}{b} \)
\( \implies b(4\sqrt{2}) = a \)
\( \implies 32b^2 = a^2 \) ... (i) [Squaring both the sides]
\( \implies b^2 = \frac{a^2}{32} \)
Since 32 divides \( a^2 \), so 32 divides ‘a’ as well.
So, we write \( a = 32c \), where c is an integer.
\( \implies a^2 = (32c)^2 \)
\( \implies 32b^2 = 32 \times 32c^2 \) ... [From (i)]
\( \implies b^2 = 32c^2 \)
\( \implies c^2 = \frac{b^2}{32} \)
Since 32 divides \( b^2 \), so 32 divides ‘b’.
\( \implies \) 32 divides both a and b.
Thus, a and b have at least 32 as a common factor.
But this contradicts the fact that a and b have no common factor other than 1.
\( \implies \) Our assumption that \( 4\sqrt{2} \) is a rational number is wrong.
\( \implies 4\sqrt{2} \) is an irrational number.
In simple words: To prove \( 4\sqrt{2} \) is irrational, we first pretend it is a normal fraction. By doing some algebra, we show this assumption leads to an impossible situation where both the top and bottom numbers share a common factor of 32, which contradicts our starting rule.

🎯 Exam Tip: Always clearly state that 'a' and 'b' are co-prime integers at the beginning of the proof, as this is the key property that leads to the contradiction.

Question 2. Prove that \( 3 + \sqrt{5} \) is an irrational number.
Answer: Let us assume that \( 3 + \sqrt{5} \) is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (\( b \neq 0 \)) such that:
\( 3 + \sqrt{5} = \frac{a}{b} \)
\( \implies \sqrt{5} = \frac{a}{b} - 3 \)
Since, a and b are integers, \( \frac{a}{b} - 3 \) is a rational number and so \( \sqrt{5} \) is a rational number.
But this contradicts the fact that \( \sqrt{5} \) is an irrational number. This contradiction shows that our initial assumption cannot hold true under any circumstances.
\( \implies \) Our assumption that \( 3 + \sqrt{5} \) is a rational number is wrong.
\( 3 + \sqrt{5} \) is an irrational number.
In simple words: We start by assuming \( 3 + \sqrt{5} \) is a rational fraction. If we subtract 3 from both sides, it means \( \sqrt{5} \) must also be rational, which we already know is false, proving our starting assumption was wrong.

🎯 Exam Tip: When dealing with the sum of a rational and an irrational number, isolate the square root term on one side of the equation to easily show the contradiction.

Question 3. Represent the numbers \(\sqrt{5}\) and \(\sqrt{10}\) on a number line.
Answer:
i. Draw a number line and take point A at 2.
Draw AB perpendicular to the number line such that AB = 1 unit.
In \(\Delta OAB\), \(m\angle OAB = 90^\circ\)
\(\therefore (OB)^2 = (OA)^2 + (AB)^2\) ... [Pythagoras theorem]
\(= (2)^2 + (1)^2\)
\(\dots (OB)^2 = 5\)
\(\therefore OB = \sqrt{5}\) units. ... [Taking square root of both sides]
With O as centre and radius equal to OB, draw an arc to intersect the number line at C. This geometric construction helps us locate irrational numbers precisely on a continuous scale.
The coordinate of the point C is \(\sqrt{5}\).

 

Diagram details for \(\sqrt{5}\):

• Point O represents 0 on the number line.
• Point A represents 2 on the number line.
• Point B is 1 unit vertically above A.
• Point C represents \(\sqrt{5}\) on the number line where the arc from B meets the line.

ii. Draw a number line and take point P at 3.
Draw PR perpendicular to the number line such that PR = 1 unit.
In \(\Delta OPR\), \(m\angle OPR = 90^\circ\)
\(\therefore (OR)^2 = (OP)^2 + (PR)^2\) ... [Pythagoras theorem]
\(= (3)^2 + (1)^2\)
\(\therefore (OR)^2 = 10\)
\(\dots OR = \sqrt{10}\) units. ... [Taking square root of both sides]
With O as centre and radius equal to OR, draw an arc to intersect the number line at Q.
The coordinate of the point Q is \(\sqrt{10}\).

Diagram details for \(\sqrt{10}\):

• Point O represents 0 on the number line.
• Point P represents 3 on the number line.
• Point R is 1 unit vertically above P.
• Point Q represents \(\sqrt{10}\) on the number line where the arc from R meets the line.

In simple words: To find \(\sqrt{5}\) and \(\sqrt{10}\) on a number line, we build right-angled triangles with a height of 1 unit. By using Pythagoras' theorem, the hypotenuse of these triangles becomes exactly \(\sqrt{5}\) and \(\sqrt{10}\) units long, which we then swing down onto the number line using a compass.

🎯 Exam Tip: Always label the points clearly and mention the Pythagoras theorem in your steps to secure full marks.

Question 4. Write any three rational numbers between the two numbers given below.
i. 0.3 and – 0.5
Answer:
i. The two given numbers are 0.3 and –0.5. We can write these as fractions or decimals to easily find numbers between them. Any three rational numbers between 0.3 and –0.5 are 0.2, 0.1, and 0. (Other possible correct answers include –0.1, –0.2, –0.3, and –0.4).
In simple words: Rational numbers are just numbers that can be written as fractions or simple decimals. Since 0.3 is positive and -0.5 is negative, we can easily pick simple numbers like 0.2, 0.1, and 0 that lie between them on the number line.

🎯 Exam Tip: When finding rational numbers between decimals, convert them to simpler decimals or fractions to easily identify the intermediate values.

Question 1. Find three rational numbers between the given numbers:
(i) 0.3 and -0.5
(ii) -2.3 and -2.33
(iii) 5.2 and 5.3
(iv) -4.5 and -4.6
Answer:
(i) 0.3 and -0.5
0.3 = 0.30 and -0.5 = -0.50
We know that,
\( 0.30 > 0.29 > \dots > 0.10 > \dots > -0.10 > \dots > -0.30 > \dots > -0.50 \)
\( \implies \) the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

Alternate Method:
A rational number between two rational numbers \( a \) and \( b \) is \( \frac{a + b}{2} \).
First rational number = \( \frac{0.3 + (-0.5)}{2} = \frac{0.3 - 0.5}{2} = \frac{-0.2}{2} = -0.1 \)
Second rational number = \( \frac{0.3 + (-0.1)}{2} = \frac{0.3 - 0.1}{2} = \frac{0.2}{2} = 0.1 \)
Third rational number = \( \frac{-0.1 + (-0.5)}{2} = \frac{-0.1 - 0.5}{2} = \frac{-0.6}{2} = -0.3 \)
\( \implies \) the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

(ii) -2.3 and -2.33
-2.3 = -2.300 and -2.33 = -2.330
We know that,
\( -2.300 > -2.301 > \dots > -2.310 > \dots > -2.320 > \dots > -2.330 \)
\( \implies \) the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325.

(iii) 5.2 and 5.3
5.2 = 5.20 and 5.3 = 5.30
We know that,
\( 5.20 < 5.21 < 5.22 < 5.23 < \dots < 5.30 \)
\( \implies \) the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23.

(iv) -4.5 and -4.6
-4.5 = -4.50 and -4.6 = -4.60
We know that,
\( -4.50 > -4.51 > -4.52 > -4.53 > \dots > -4.60 \)
\( \implies \) the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.53.
In simple words: To find numbers between two decimals, we can add extra zeros at the end to make them look like longer numbers (for example, turning 0.3 into 0.30). This makes it easy to see and pick any three numbers that lie in between them.

🎯 Exam Tip: Always convert the given decimals to have the same number of decimal places by adding zeros. This makes finding the intermediate rational numbers much easier and prevents calculation errors.

Question (iv). Find three rational numbers between -4.5 and -4.6.
Answer: We have \( -4.5 = -4.50 \) and \( -4.6 = -4.60 \). We know that, \( -4.50 > -4.51 > -4.52 > \dots > -4.55 > \dots > -4.60 \). We can find infinitely many such numbers by extending the decimal places further. \( \therefore \) the three rational numbers between \( -4.5 \) and \( -4.6 \) are \( -4.51 \), \( -4.52 \) and \( -4.55 \).
In simple words: To find numbers between -4.5 and -4.6, we can write them as -4.50 and -4.60. Then, we can easily pick any numbers in between, like -4.51, -4.52, and -4.55.

🎯 Exam Tip: Adding a zero at the end of decimal numbers makes it much easier to visualize and find the rational numbers lying between them.