Here are more detailed examples and problem-solving techniques from Chapter 1 of CBSE Class 9 Maths (Number Systems):
1. Proving Irrationality
Example 1: Prove that 2\sqrt{2} is irrational.
Steps:
- Assume 2\sqrt{2} is rational.
- Then, 2=pq\sqrt{2} = \frac{p}{q}, where pp and qq are integers, and q≠0q \neq 0.
- Squaring both sides: 2=p2q22 = \frac{p^2}{q^2}, so p2=2q2p^2 = 2q^2.
- This implies p2p^2 is even (since it is divisible by 2), and so pp is even.
- Let p=2kp = 2k. Substituting in p2=2q2p^2 = 2q^2, we get: (2k)2=2q2 ⟹ 4k2=2q2 ⟹ q2=2k2.(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2.
- This shows q2q^2 is even, so qq is also even.
- Since both pp and qq are even, they have a common factor of 2, contradicting the assumption that pp and qq are co-prime.
- Hence, 2\sqrt{2} is irrational.
2. Simplification of Expressions
Example 2: Simplify (5+2)(5−3)(\sqrt{5} + 2)(\sqrt{5} – 3).
Solution: Using the distributive property:
(5+2)(5−3)=(5)2+(−35)+(25)−(2⋅3).(\sqrt{5} + 2)(\sqrt{5} – 3) = (\sqrt{5})^2 + (-3\sqrt{5}) + (2\sqrt{5}) – (2 \cdot 3).
Simplify terms:
=5−35+25−6.= 5 – 3\sqrt{5} + 2\sqrt{5} – 6.
Combine like terms:
=−1−5.= -1 – \sqrt{5}.
3. Rationalization
Example 3: Rationalize 17+6\frac{1}{\sqrt{7} + \sqrt{6}}.
Solution: Multiply numerator and denominator by the conjugate of 7+6\sqrt{7} + \sqrt{6}, which is 7−6\sqrt{7} – \sqrt{6}:
17+6×7−67−6=7−6(7)2−(6)2.\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}.
Simplify:
=7−67−6=7−6.= \frac{\sqrt{7} – \sqrt{6}}{7 – 6} = \sqrt{7} – \sqrt{6}.
4. Decimal Expansion
Example 4: Show that 18\frac{1}{8} has a terminating decimal expansion.
Solution:
- Write 18\frac{1}{8} in decimal form: 18=0.125.\frac{1}{8} = 0.125.
- Since the decimal terminates, 18\frac{1}{8} has a terminating decimal expansion.
Example 5: Show that 13\frac{1}{3} has a non-terminating, repeating decimal expansion.
Solution:
- Divide 11 by 33: 1÷3=0.333…=0.3‾.1 \div 3 = 0.333… = 0.\overline{3}.
- The decimal repeats, so 13\frac{1}{3} has a non-terminating, repeating decimal expansion.
5. Operations with Real Numbers
Example 6: Add 2\sqrt{2} and 323\sqrt{2}.
Solution:
2+32=(1+3)2=42.\sqrt{2} + 3\sqrt{2} = (1 + 3)\sqrt{2} = 4\sqrt{2}.
Example 7: Multiply 232\sqrt{3} and 464\sqrt{6}.
Solution:
23⋅46=(2⋅4)(3⋅6)=818.2\sqrt{3} \cdot 4\sqrt{6} = (2 \cdot 4)(\sqrt{3} \cdot \sqrt{6}) = 8\sqrt{18}.
Simplify 18\sqrt{18}:
=8⋅9⋅2=8⋅32=242.= 8 \cdot \sqrt{9 \cdot 2} = 8 \cdot 3\sqrt{2} = 24\sqrt{2}.
6. Problem-Solving Practice
Problem 8: Show that 3+2\sqrt{3} + \sqrt{2} is irrational.
Solution:
- Assume 3+2\sqrt{3} + \sqrt{2} is rational.
- Let 3+2=r\sqrt{3} + \sqrt{2} = r, where rr is rational.
- Rearrange: 3=r−2\sqrt{3} = r – \sqrt{2}.
- Since rr and 2\sqrt{2} are rational, r−2r – \sqrt{2} is rational, implying 3\sqrt{3} is rational (contradiction).
- Hence, 3+2\sqrt{3} + \sqrt{2} is irrational.
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