Rational Numbers

Rational Numbers:

We know that the numbers of the form of $\frac{m}{n}$, where m and n are integers and n ? 0,are called rational numbers.

Rational Numbers has the following results:

• If m and n are integers and n ? 0, then $\frac{m}{n}$ is not necessarily an integer.
• Every integer is a rational number, for if m $\in$ I, then we can write m as $\frac{m}{1}$, which is a rational number.
• $\frac{p}{q}$= $\frac{m}{n}$ implies that mq = np.
• $\frac{m}{n}$ is said to be in its lowest terms, if GCD( m, n ) = 1. if $\frac{m}{n}$ is not in its lowest terms, we reduce it to lowest terms by dividing each numerators and denominators by GCD( m , n) .
• We define $\frac{m}{n}$ + $\frac{p}{n}$ = $\frac{m + p}{n}$ and $\frac{m}{n}$ + $\frac{p}{q}$ = $\frac{pn + qm}{qn}$

Properties of Rational Numbers.

Let Q be the set of all rational numbers. Then,

For Addition:

1. The sum of two rational number is always a rational number.

2. Addition of rational numbers is commutative as well as associative.

3. 0 $\in$ Q such that $\frac{m}{n}$ + 0 = $\frac{m}{n}$ + $\frac{0}{1}$ = $\frac{0}{1}$ + $\frac{m}{n}$ = $\frac{m}{n}$.

4. If $\frac{m}{n}$ $\in$ Q, then – $\frac{m}{n}$ $\in$ Q such that $\frac{m}{n}$ + (-$\frac{m}{n}$) = (-$\frac{m}{n}$) + $\frac{m}{n}$ = 0

Thus –$\frac{m}{n}$ is additive Identity of $\frac{m}{n}$.

For Multiplication:

1. The product of two rational number is always a rational number

2. Multiplication of two rational numbers is commutative as well as associative.

3. Multiplication distributes addition in rational number.

4. 1 is multiplicative identity, as $\frac{m}{n}$ × $\frac{1}{1}$ = $\frac{1}{1}$ × $\frac{m}{n}$ = $\frac{m}{n}$ for all $\frac{m}{n}$ $\in$ Thus 1 is the multiplicative Identity in rational numbers.

5. For a non zero rational number $\frac{m}{n}$, there exist a rational number $\frac{n}{m}$ such that $\frac{m}{n}$ × $\frac{n}{m}$ = $\frac{n}{m}$ × $\frac{m}{n}$ = 1.

Here $\frac{n}{m}$ is called the reciprocal or multiplicative inverse of $\frac{m}{n}$.

Rational Expressions

An expression of from $\frac{p(x)}{q(x)}$, where and q(x) are polynomials and q(x) is non-zero, is called a Rational Expression.

We call p(x) the numerator q(x) is the denominator of rational expression.

Clearly, $\frac{p(x)}{q(x)}$ need not to be a polynomial. Every polynomial is a rational number since if p(x) is a polynomial, then we can write it p(x) as $\frac{p(x)}{1}$,

Thus every polynomial is a rational number but every rational number needs not to be a polynomial.

$\frac{3x-4}{5x^2+2x+3}$ is a rational expression whose numerator is a linear polynomial and the denominator is a quadratic polynomial.

$\frac{x^4 - 3x^2 -6x - 4}{5x^3 + 2x + 3}$ is a rational expression whose numerator is a polynomial of a degree 4 and the denominator is a polynomial of a degree 3.

Greatest Common Divisor (GCD) or Highest Common Factor (HCF)

The GCD of two polynomial p(x) and g(x) is that common divisor which has the highest degree among all common divisors and in which the coefficient of highest- degree term is positive.

We will understand it by a few examples from online algebra 1 examples.

Example 1. Find the GCD of the polynomial 2x² -18 and x² -2x -3.

Solution: Let p(x) = 2x² -18 = 2(x² -9) = 2(x + 3)(x - 3)

and g(x) = x² -2x -3= (x + 1)(x - 3)

Clearly the Greatest Common Divisor (GCD) or Highest Common Factor(HCF) of p(x) and g(x) = (x - 3).

Example 2. Find the GCD of the polynomial 20x² -9x + 1 and 5x² - 6x + 1.

Solution: Let p(x) = 20x² -9x + 1 = (5x - 1)(4x - 1)

and g(x) = 5x² - 6x + 1 = (5x - 1)(x - 1)

Clearly the Greatest Common Divisor (GCD) or Highest Common Factor(HCF) of p(x) and g(x) = (5x - 1).

Example 3. Find the GCD of the polynomial 2x² -x - 1 and 4x² + 8x + 3.

Solution: Let p(x) = 2x² -x - 1 = (x - 1)(2x + 1)

and g(x) = 4x² + 8x + 3 = (2x + 3)(2x + 1)

Clearly the Greatest Common Divisor (GCD) or Highest Common Factor(HCF) of p(x) and g(x) = (2x + 1).

Example 4. Find the GCD of the polynomial 24(6x⁴ -x³ - 2x²) and 20(2x⁶ + 3x⁵ + x⁴)

Solution: Let p(x) = 24(6x⁴ - x³ - 2x²) = 24x² (6x² -x - 2) = 24x²(3x - 2)(2x + 1) = 2³ × 3 × x²(3x - 2)(2x + 1)

and g(x) = 20(2x⁶ + 3x⁵ + x⁴)= 20x⁴ (2x² + 3x + 1) = 20x⁴(x + 1)(2x + 1) = 2² × 5 × x⁴(x + 1)(2x + 1)

The Greatest Common Divisor (GCD) or Highest Common Factor(HCF) of p(x) and g(x) = 2²x²(2x + 1) = 4x²(2x + 1)

Least Common Multiple (LCM)

The LCM of two polynomials p(x) and g(x) is the polynomial of lowest degree which has both p(x) and g(x) as divisors and coefficients of highest-degree term has the same sign as that of the highest-degree term of p(x).g(x).

The LCM of two polynomials p(x) and g(x) = $\frac{p(x).g(x)}{(GCD) or (HCF) of p(x) and g(x)}$

we will understand it by a few examples.

We know that the integers and polynomials have identical properties for addition and multiplication

Properties of Integers:

Let I be the set of all Integers.Then,

For Addition:

1. The sum of two integers is always an Integer.

2. (a + b) + c = a + (b + c) for all a, b and c $\in$ I. [ Associative law ]

3. a + b + c = b + a for all a, b $\in$ I. [ Commutative law ]

4. 0 is an integer such that 0 + a = a + 0 = a for all a $\in$ I

Thus 0 is the additive Identity in Integers.

5. For each a $\in$ I, there exist –a $\in$ I such a + (-a) = -a + a = 0

Here -a is called additive inverse of a.

For Multiplication:

1. The product of two integers is always an Integer.

2. (a b) c = a (b c) for all a, b and c $\in$ I. [ Associative law ]

3. a b = b a for all a, b $\in$ I. [ Commutative law ]

4. 1 is an integer such that 1 * a = a * 1 = 1 for all a $\in$ I

Thus 1 is the multiplicative Identity in Integers.

5. a .[b + c] = a.b + a.c for all a, b and c $\in$ I.[ Distributive law ]

Properties of Polynomials:

Let P be the set of all polynomials. Then,

For Addition:

1. The sum of two polynomials is always a polynomial.

2. [p(x) + q(x)] + r(x) = p(x) + [q(x) + r(x)] for all p(x), q(x)] and r(x) $\in$ P. [Associative law]

3. p(x) + q(x) = q(x) + p(x) for all p(x), q(x) $\in$ P.[ Commutative law ]

4. 0 is an integer such that 0 + p(x) = p(x) + 0 = p(x) for all p(x) $\in$ P

Thus Zero polynomial is the additive Identity in polynomials.

5. For each p(x) $\in$ P, there exist – p(x) $\in$ P such p(x) + {-p(x)} = - p(x) + p(x) = 0

Here - p(x) is called additive inverse of p(x).

For Multiplication:

1. The product of two polynomials is always a polynomial.

2. [p(x) q(x)] r(x) = p(x)[(q(x) r(x)] for all p(x), q(x) and r(x), $\in$ P.[ Associative law ]

3. p(x) . q(x) = q(x) . p(x)] for all p(x), q(x) $\in$ P.[ Commutative law ]

4. 1 is a constant polynomial such that 1 × p(x) = p(x) × 1 = 1 for all p(x) $\in$ P

5. p(x) .[ q(x) + r(x)] = p(x) . q(x) + p(x) . r(x) for all p(x) , q(x) and r(x) $\in$ P.[ Distributive law ]

Example1. Write a rational expression whose numerator is a quadratic polynomial with zeros 2 and -3 and whose denominator is a cubic polynomial with zeros 1, 9 and -1.

Solution: Required rational expression: $ \frac{(x-2)(x+3)}{(x-1)(x-9)(x+1)}$ = $\frac{x^2+x-6}{x^3-9x^2-x+9}$

Rational Expression in its lowest term:

If p(x) and q(x) are polynomials with integral coefficient such that GCD [p(x) q(x)] = 1 and q(x) 0, then we say that p(x)/ p(x) is its lowest form. If p(x)/ p(x) is not in its lowest terms, we reduce it to lowest terms by dividing each numerators and denominators by GCD (p(x), p(x)).

Example2. For each of the following rational expression, examine whether it is in its lowest terms. If not, express each of them in their lowest terms.

(i) $ \frac{(x - 3)(x + 5)}{(x - 5)}$ (ii) $ \frac{(3x - 2)(x + 1)}{(x + 1)(2x -1)}$ (iii) $\frac{(x+2)}{x^2+5x+6}$

Solution: (i) Let p(x) = (x - 3) (x + 5)

and g(x) = (x - 5)

Clearly, (GCD) of p(x) and g(x) = 1

So, $ \frac{p(x)}{q(x)}$ = $ \frac{(x - 3)(x + 5)}{(x - 5)}$ is in its lowest term.

(ii) Let p(x) = (3x - 2) (x + 1)

and g(x) = (x + 1)(2x – 1)

Clearly, (GCD) of p(x) and g(x) = (x +1)

So the given rational expression is not in its lowest term.

So, $ \frac{p(x)}{q(x)}$ = $ \frac{(3x - 2)(x + 1)}{(x + 1)(2x – 1)}$ = $ \frac{(3x - 2)}{(x + 1)}$ [canceling (x + 1) from both numerator and denominator]

So, $ \frac{p(x)}{q(x)}$ = $ \frac{(x - 3)(x + 5)}{(x - 5)}$ is in its lowest term.

(iii) Let p(x) = (x + 2)

and g(x) = x^2 + 5x + 6 = (x + 2)(x + 3)

Clearly, (GCD) of p(x) and g(x) = (x +2)

So the given rational expression is not in its lowest term.

So, $ \frac{p(x)}{q(x)}$ = $ \frac{(x + 2)}{ x^2 + 5x + 6}$ = $ \frac{1}{(x + 3)}$ [canceling (x + 2) from both numerator and denominator]