REAL NUMBERS-Notes
Maths - Notes
The 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 factor occurs.
- The prime factorisation of a natural number is unique, except for the order of its factors.
$$\begin{aligned}6&=2^{1}\times 3^{1}\\ 20&=2^{2}\times 5^{1}\end{aligned}$$ HCF (6,20) =\(2^1\quad\Rightarrow\) Product of the smallest power of each common prime factor in the numbers
HCF=2
LCM (6,20) = \(2^2 \times 3^1 \times 5^1\quad\Rightarrow\) Product of the greatest power of each prime factor involved in the number
LCM=60
\(a =p_1p_2\ldots p_n\) where \(p_1,~ p_2,~ p_3,\ldots p_n\) are primes, not necessarily distinct therefore; $$\begin{aligned}a^{2}&=\left( p_{1}p_{2}\ldots \cdot P_{n}\right) \left( p_{1}p_{2}\ldots \cdot p_{n}\right) \\ &=\left( p_{1}p_{2}\right) ^{2}\ldots \cdot \left( p_{n}\right) ^{2}\end{aligned}$$ We are given that \( p\) divides \(a^2\) therefore, from The Fundamental Theorem of Arithmetic, it follows that \(p\) is one of the prime factors of \(a^2\). However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factor of \(a^2\) is \(p_1,p_2 ,\ldots,p_n\) So, \( p\) is one of \(p_1,~p_2,\ldots,~p_n\), since $$a =p_1p_2\ldots p_n\\\Rightarrow p\text{ divides } a $$
Hence \(\sqrt 2\) is not a rational number
\(\Rightarrow\sqrt 2\) is an irrational number
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 factor occurs.
- The prime factorisation of a natural number is unique, except for the order of its factors.
Example: Consider the numbers \(4^n\) where n is a natural number, check whether there
is any value of n for which \(4^n\) ends with the digit zero.
Solution: Last digit of a number becomes zero when the number is divisible by 5, that is, a prime factorisation of the number must contain the prime 5 $$\begin{aligned}\Rightarrow 4^{n}&=\left( 2\cdot 2\right) ^{n}\\ &=\left( 2^{2}\right) ^{n}\\ &=(2\times 1)^{2n}\end{aligned}$$ $$\Rightarrow4^n \text{ have only factors of 2 and 1.}$$
Therefore, \(4^n\) never ends with the digit zero.
Solution: Last digit of a number becomes zero when the number is divisible by 5, that is, a prime factorisation of the number must contain the prime 5 $$\begin{aligned}\Rightarrow 4^{n}&=\left( 2\cdot 2\right) ^{n}\\ &=\left( 2^{2}\right) ^{n}\\ &=(2\times 1)^{2n}\end{aligned}$$ $$\Rightarrow4^n \text{ have only factors of 2 and 1.}$$
Therefore, \(4^n\) never ends with the digit zero.
Example: Find the LCM and HCF of 6 and 20 by the prime factorisation method
Solution:$$\begin{aligned}6&=2^{1}\times 3^{1}\\ 20&=2^{2}\times 5^{1}\end{aligned}$$ HCF (6,20) =\(2^1\quad\Rightarrow\) Product of the smallest power of each common prime factor in the numbers
HCF=2
LCM (6,20) = \(2^2 \times 3^1 \times 5^1\quad\Rightarrow\) Product of the greatest power of each prime factor involved in the number
LCM=60
Example: Find the HCF and LCM of 6, 72, 120 using the prime factorisation method
Solution: $$\begin{aligned}6&=2\times 3\\ 72&=2^{3}\times 3^{2}\\ 120&=2^{3}\times 3^{1}\times 5^{1}\end{aligned}$$ Smallest power of common factor 2, 3 are \(2^1 \text{ and } 3^1\) $$\begin{aligned}HCF&=2^{1}\times 3^{1}\\ &=6\end{aligned}$$ Greatest power of prime factors 2, 3 and 5 are \(2^3,~3^2 \text{ and } 5^1\) $$\begin{aligned}LCM&=2^{3}\times 3^{2}\times 5\\ &=8\times 9\times 5\\ &=360\end{aligned}$$Theorem: Let \(p\) be a prime number, if \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer
Proof: Let the prime factorization of \(a\) be as follows\(a =p_1p_2\ldots p_n\) where \(p_1,~ p_2,~ p_3,\ldots p_n\) are primes, not necessarily distinct therefore; $$\begin{aligned}a^{2}&=\left( p_{1}p_{2}\ldots \cdot P_{n}\right) \left( p_{1}p_{2}\ldots \cdot p_{n}\right) \\ &=\left( p_{1}p_{2}\right) ^{2}\ldots \cdot \left( p_{n}\right) ^{2}\end{aligned}$$ We are given that \( p\) divides \(a^2\) therefore, from The Fundamental Theorem of Arithmetic, it follows that \(p\) is one of the prime factors of \(a^2\). However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factor of \(a^2\) is \(p_1,p_2 ,\ldots,p_n\) So, \( p\) is one of \(p_1,~p_2,\ldots,~p_n\), since $$a =p_1p_2\ldots p_n\\\Rightarrow p\text{ divides } a $$
Theorem: \(\sqrt 2\) is irrational
Proof: let us assume that \(\sqrt 2\) is a rational number $$\sqrt{2}=\dfrac{a}{b},b\neq 0$$ \(a \text{ and } b\) are coprime, so Squaring both sides \[ \begin{align} \left( \sqrt{2}\right) ^{2} &= \left( \frac{a}{b}\right) ^{2} \notag \\ 2 &= \frac{a^{2}}{b^{2}} \notag \\ a^{2} &= 2b^{2} \tag{1} \end{align} \] \(\Rightarrow 2 \text{ divides } a^2 \\\Rightarrow 2 \text{ divides } a \\\) We can write any number \(c\) $$a=2c$$ Squaring both sides $$\begin{align}a^{2}&=4c^{2}\\a^{2}&=2b^{2}\\ 2b^{2}&=4c^{2}\\\tag{2} \Rightarrow b^{2}&=2c^{2}\end{align}$$ \(\Rightarrow 2 \text{ divides } b^2 \\\Rightarrow 2 \text{ divides } b \\\) \(\Rightarrow\quad\)there is at leat one number 2 which divides both \(a \text{ and } b\), which contradict our assumption that \(a, ~b\) are coprime.Hence \(\sqrt 2\) is not a rational number
\(\Rightarrow\sqrt 2\) is an irrational number
In class IX, we have learnt that
- the sum or difference of a rational number and an irrational number is irrational Number and
- The product and quotient of a non-zero rational and irrational Number is irrational
- the sum or difference of a rational number and an irrational number is irrational Number and
- The product and quotient of a non-zero rational and irrational Number is irrational
Summary
- The 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.
- If \(p\) is a prime and \(p\) divides \(a^2 \), then \(p\) divides a where a is a positive integer
