Real Numbers – Class 10 Mathematics (Lecture 2)
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State the Fundamental Theorem of Arithmetic.
Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Example: \(60 = 2^2 \times 3 \times 5\) is the unique prime factorisation of 60.
How to find HCF and LCM using prime factorisation method?
HCF (a,b) = Product of the smallest power of each common prime factor in the numbers.
LCM (a,b) = Product of the greatest power of each prime factor involved in the numbers.
Relationship: \(HCF(a,b) \times LCM(a,b) = a \times b\)
A school has 3 sections with 62, 84, and 108 students respectively. What is the minimum number of rooms required if each room has the same maximum capacity?
Solution:
We need to find HCF of 62, 84, and 108 to determine maximum students per room.
HCF(62, 84, 108) = 12
Number of rooms required = \(\frac{62}{12} + \frac{84}{12} + \frac{108}{12} = 5 + 7 + 9 = 21\) rooms
Find the smallest number which when increased by 20 is exactly divisible by 90 and 144.
Solution:
First, find LCM(90, 144):
\(90 = 2 \times 3^2 \times 5\)
\(144 = 2^4 \times 3^2\)
\(LCM = 2^4 \times 3^2 \times 5 = 720\)
Let the number be \(x\)
\(x + 20 = 720\)
\(x = 720 – 20 = 700\)
Answer: The smallest number is 700.
The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, find the other number.
Solution:
Let HCF = \(h\), LCM = \(14h\)
Given: \(h + 14h = 600\)
\(15h = 600\) ⇒ \(h = 40\)
LCM = \(14 \times 40 = 560\)
Let the other number be \(b\)
Using: \(280 \times b = HCF \times LCM\)
\(280 \times b = 40 \times 560\)
\(b = \frac{40 \times 560}{280} = 80\)
Answer: The other number is 80.
Two milk containers contain 398L and 436L of milk. After transferring to another drum, 7L and 11L remain in the containers respectively. What is the maximum capacity of the drum?
Solution:
Milk transferred from first container: \(398 – 7 = 391L\)
Milk transferred from second container: \(436 – 11 = 425L\)
Maximum capacity of drum = HCF(391, 425)
\(391 = 17 \times 23\)
\(425 = 17 \times 25\)
HCF(391, 425) = 17
Answer: Maximum capacity of drum is 17L.
Prove that \(7 \times 11 \times 13 + 13\) is a composite number.
Proof:
\(7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1)\)
\(= 13(77 + 1) = 13 \times 78\)
\(= 13 \times 2 \times 3 \times 13 = 2 \times 3 \times 13^2\)
Since it has more than two prime factors (2, 3, and 13), it is a composite number.
Find the greatest 6-digit number exactly divisible by 18, 24, and 36.
Solution:
First, find LCM(18, 24, 36):
\(18 = 2 \times 3^2\)
\(24 = 2^3 \times 3\)
\(36 = 2^2 \times 3^2\)
LCM = \(2^3 \times 3^2 = 8 \times 9 = 72\)
Greatest 6-digit number = 999999
Divide 999999 by 72: \(999999 ÷ 72 = 13888\) remainder 63
Subtract remainder: \(999999 – 63 = 999936\)
Answer: 999936
State Theorem 1.2: If a prime number \(p\) divides \(a^2\), then \(p\) divides \(a\).
Theorem 1.2: Let \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.
Proof Idea: If \(p\) divides \(a^2\), then \(p\) must be a prime factor of \(a^2\). Since \(p\) is prime, it must also be a prime factor of \(a\).
This theorem is used in proving irrationality of numbers like \(\sqrt{2}\).
Prove that \(\sqrt{2}\) is irrational.
Proof by Contradiction:
Assume \(\sqrt{2}\) is rational, so \(\sqrt{2} = \frac{p}{q}\) where \(p\) and \(q\) are coprime integers (HCF = 1).
Squaring both sides: \(2 = \frac{p^2}{q^2}\) ⇒ \(2q^2 = p^2\)
This means 2 divides \(p^2\), so by Theorem 1.2, 2 divides \(p\).
Let \(p = 2m\) for some integer \(m\).
Substitute: \(2q^2 = (2m)^2 = 4m^2\) ⇒ \(q^2 = 2m^2\)
This means 2 divides \(q^2\), so 2 divides \(q\).
But if 2 divides both \(p\) and \(q\), then HCF(\(p, q\)) ≥ 2, contradicting our assumption that they are coprime.
Therefore, \(\sqrt{2}\) is irrational.
Prove that \(\sqrt{3} + \sqrt{5}\) is irrational.
Proof:
Assume \(\sqrt{3} + \sqrt{5}\) is rational = \(r\)
Then \(\sqrt{5} = r – \sqrt{3}\)
Squaring: \(5 = r^2 – 2r\sqrt{3} + 3\)
\(2r\sqrt{3} = r^2 – 2\)
If \(r\) is rational and non-zero, then \(\sqrt{3} = \frac{r^2 – 2}{2r}\) is rational.
But \(\sqrt{3}\) is known to be irrational. Contradiction.
Therefore, \(\sqrt{3} + \sqrt{5}\) is irrational.
Prove that \(5 – \frac{3}{7}\sqrt{3}\) is irrational.
Proof:
Assume \(5 – \frac{3}{7}\sqrt{3}\) is rational = \(r\)
Then \(\frac{3}{7}\sqrt{3} = 5 – r\)
\(\sqrt{3} = \frac{7(5 – r)}{3}\)
If \(r\) is rational, then the right side is rational.
But \(\sqrt{3}\) is irrational. Contradiction.
Therefore, \(5 – \frac{3}{7}\sqrt{3}\) is irrational.