Revisiting Irrationals, Rational And Their Decimal Expansions

Irrational numbers:
All those numbers which can not be expressed as fractions i.e a number is an irrational number if it can not be written as $\frac{p}{q},q\ne 0$.

In the view of decimal expansion:
All those numbers whose decimal expansion is neither terminating nor repeating.

Eg:$\sqrt{2},\sqrt{3},\sqrt[3]{5},10100100010000...,$etc.

Proving irrationality of numbers:
In order to prove the irrationality of numbers, we will use the method of contradiction and following results:

• The sum or difference of a rational and irrational number is an irrational number.
• The product and quotient of a non-zero rational number and an irrational number is an irrational number.
• If $p$ be a prime number and $a$ be a positive integer. If $p^2$ divides $a$, then $p$ divides $a$.

For example: If $25$ divides $100$ i.e.$5^2$ divides $100$, then $5$ also divides $100$.

Remarks:

• For any prime number $p$,$\sqrt{p}$ is an irrational number.

Rational numbers:
All those numbers which can be expressed as fractions i.e. a number is a rational number if it can be written as $\frac{p}{q},q\ne 0$.

In the view of decimal expansion:
All those numbers whose decimal expansion is either terminating or non-terminating repeating.
Eg:$\frac{2}{1},\frac{1}{10001},\frac{113}{71},\frac{69}{131}$,0.02,0.3412,0.121212, . . ., 0.010203, etc.

Theorems to determine the nature of decimal expansions of rational numbers

Theorem 1.
Let $x$ be a rational number whose decimal expansion terminates. Then,$x$ can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are co-primes,and the prime factorization of $q$ is of the form $2^m{5}^n$, where $m,n$ are non-negative integers.
Eg:Let $x=0.242$
$\Rightarrow x=\frac{232}{1000}=\frac{121}{500}=\frac{121}{2^2\times 5^3}$
Thus,$q=2^2\times 5^3$, which is in the form of$2^m{5}^n$, where$m=2,n=3$.

Theorem 2.
Let $x=\frac{p}{q}$ be a rational number such that the prime factorization of $q$ is of the form $2^m{5}^n$, where $m,n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.
Eg:Let $x=\frac{39}{125}$
$\Rightarrow x=\frac{39}{5^3}=\frac{39\times 2^3}{5^3\times 2^3}=\frac{312}{1000}=0.312$.
Thus,$x$ has a terminating decimal expansion.

Theorem 3.
Let $x=\frac{p}{q}$,where $p$ and $q$ are co-primes, be a rational number such that $q$ is not of the form $2^m{5}^n$, where $m$ and $n$ are non-negative integers. Then the decimal expansion of $x$ is non-terminating and repeating.
Eg: Let $x=\frac{7}{3}$
$\Rightarrow x=2.333333...$
Thus,$x$ has decimal expansion which is non-terminating and repeating.

Euclid's Division Lemma and Fundamental Theorem of Arithmetic

Real numbers
Collection of all those numbers which are either rational or irrational.

Euclid's Division Lemma:
Let $a$ and $b$ be any two positive integers. Then there exist unique integers $q$ and $r$ such that
$a=bq+r$, where $0\le r<b$

Application of Euclid's Division Lemma to find HCF
In order calculate HCF of any two positive integers, say $a$ and $b$ with $a>b$,we may follow the following steps:
Step 1.Apply Euclid's division lemma to $a$ and $b$ to obtain $q_1$ and $r_1$ such that $a=b{q}_1+r_1,0\le r_1<b$.
Step 2.If $r_1=0$,then $b$ is the HCF of $a$ and $b$.
Step 3.If $r_1\ne 0,$ apply Euclid's division lemma to $b$ and $r_1$ and obtain two whole numbers $q_1$ and $r_2$ such that $b=q_1{r}_1+r_2$.
Step 4.If $r_2=0$,then $r_1$ is the HCF of $a$ and $b$.
Step 5.If $r_2\ne 0$,then applied Euclid's division lemma to$r_1$ and $r_2$ and continue the above process till the remainder$r_n$is zero.The divisor at this stage i.e. $r_{n-1}$, or the non-zero remainder at the previous stage, is the HCF of $a$ and $b$.

Eg:Use Euclid's division lemma to find the HCF of $240$ and $35$.
Solution:Given integers are $240$ and $35$. Clearly,$240>35$. Applying Euclid's division lemma to $240$ and $35$,we get $240=35\times 6+30$
Since, the remainder $30\ne 0$. So, we apply the division to the divisor $35$and remainder $30$ to get
$35=30\times 1+5$
Now, we apply division lemma to the new divisor $30$ and new remainder $5$ to get
$30=5\times 6+0$
The remainder at this stage is zero. So the divisor at this stage or the remainder at the previous stage. So $5$ is the HCF of $240$ and $25$.

Fundamental theorem of arithmetic:
Every composite number can be factorized as a product of primes and this factorization is unique expect for the order in which the prime factors occur.
In general, a composite number,$k$ say, can be factorized as $k=a_1{a}_2{a}_3{a}_4....a_n,$ where $a_1,a_2,a_3,a_4,..,a_n$ are prime numbers and written in ascending order i.e.$a_1<a_2<a_3<...<a_n$.
Also, if any prime number occurs more than one time in the factorization, then we will combine the same primes and get powers of primes.
Eg:The number $156$ can be expressed as $156=2\times 2\times 3\times 13=2^2\times 3\times 13$.
Also, by fundamental theorem of arithmetic, we can say if there is any other prime factorization of $156$ other than mentioned above such as $2\times 13\times 3\times 2$ or $3\times 2\times 13\times 2$,etc. all are equal.

Application of Fundamental Theorem of Arithmetic to obtain HCF and LCM

In order to find HCF and LCM of two numbers, we may follow the following criteria:
Step 1.Factorize each of the given positive integers and express them as a product of powers of primes in ascending order of magnitudes of primes.
Step 2.To calculate the HCF, identify the common prime factors and find the smallest exponent of these common factors. Now raise these common prime factors to their smallest exponents and multiply them to get HCF.
To find LCM, list all the prime factors occurring in the prime factorization of given positive integers but take all the primes only once. For each of these factors, find the greatest exponent and raise each prime to the greatest exponent and multiply them to get LCM.

Eg:Find HCF and LCM of $15,36$ and $75$ by prime factorization method.
Sol:Given positive integer are $15,36,75$ and their prime factorization is given as
$15=3\times 5=2^0\times 3^1\times 5^1$
$36=2\times 2\times 3\times 3=2^2\times 3^2\times 5^0$
And $75=3\times 5\times 5=2^0\times 3\times 5^2$
Therefore,HCF$=2^0\times 3^1\times 5^0=3$
And LCM$=2^2\times 3^2\times 5^2=900$.

Remarks:

• We may also use the following result to find any one of HCF or LCM if we are given the other:

If $a$ and $b$ are the two positive integers, then
HCF$(a,b)\times$LCM$(a,b)=a\times b$