REAL NUMBERS-Notes

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.

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

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

Summary

1. 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.
2. If \(p\) is a prime and \(p\) divides \(a^2 \), then \(p\) divides a where a is a positive integer