Algebra : Foundation with Numerical Arithmetics : Algebra : First Principles (Summary)

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Algebra : First Principles (Summary)

what you'll learn...

overview

Algebraic expressions are representation of quantities with variables, numbers, and arithmetic operations between them.

The expressions are modified as per the PEMA precedence and CADI Laws and properties of Numerical Arithmetics.

recap

Summary of what we have learned so far:

• Numerical Expressions can be modified as per Laws and Properties of Arithmetics (LPA) without changing the value of the expressions.

• Numerical Equations consists of numerical expressions on left hand side and right hand side, and states them to be equal.

• Multiple Numerical equations can be added, subtracted, multiplied, divided into new equations.

• A numerical expression can be modified as per Laws and Properties of Arithmetics to arrive at an equivalent numerical expression, which together is a numerical identity.

• A Numerical Inequation and an equation can be combined, by addition, subtraction, multiplication, and division -- into a derived inequation.

Each of the above form the basis for multiple parts of algebra.

algebraic expression

We learned that "Numerical expressions are statements of quantities with numbers and arithmetic operations".

Algebraic expressions are statements of quantities with variables, numbers, and arithmetic operations.

eg: $x^{2} - x + 4$ $x^2-x+4$ is an algebraic expression with variable $x$ $x$ .

polynomials

We learned that "Numerical Expressions can be modified as per Laws and Properties of Arithmetics (LPA) without changing the value of the numerical expressions".

Algebraic expressions can be modified as per the LPA without changing the value of the algebraic expressions.

This leads to

• definition of polynomials

• simplification of polynomials

• addition and subtraction of polynomials

• multiplication and division of polynomials

• factorization of polynomials

example: polynomials

An example illustrating use of LPA in algebraic expressions.

eg: addition of two polynomials $x^{2} - x$ $x^2-x$ and $x - 3$ $x-3$ $(x^{2} - x) + (x - 3)$ $(x^2-x)+(x-3)$

as per closure property of addition $(x - 3)$ $(x-3)$ is a number, and as per associative property of addition
$= x^{2} + (- x + (x - 3))$ $=x^2+(-x+(x-3))$

as per associative property of addition
$= x^{2} + (- x + x) - 3$ $=x^2+(-x+x)-3$

as per additive inverse property
$= x^{2} + 0 - 3$ $=x^2+0-3$

as per additive identity property
$= x^{2} - 3$ $=x^2-3$

This example is quite simple for the elaborate solution. The elaborate solution is presented to illustrate the way the LPA is used in modifying algebraic expressions. For complex problems, this procedure helps to clarify if the steps are correct.

algebraic equations

We learned that "Multiple Numerical equations can be added, subtracted, multiplied, divided into new equations".

Algebraic equations are modified to arrive at solutions with modifications of algebraic expressions and equations as per the LPA.

An example illustrating use of LPA in algebraic equations.

$3 x + 5 = 11$ $3x+5 = 11$

add the known equation $- 5 = - 5$ $-5=-5$ to the algebraic equation $\Rightarrow (3 x + 5) - 5 = 11 - 5$ $=> (3x+5) -5 = 11 -5$

as per associative law
$\Rightarrow 3 x + (5 - 5) = 11 - 5$ $=> 3x+(5 -5) = 11-5$

as per PEMA precedence
$\Rightarrow 3 x + 0 = 6$ $=> 3x+0 = 6$

as per additive identity property $3 x + 0 = 3 x$ $3x+0 = 3x$
$\Rightarrow 3 x = 6$ $=> 3x = 6$

divide by the equation $3 = 3$ $3=3$
$\Rightarrow (3 x) \div 3 = 6 \div 3$ $=> (3x)-:3 = 6 -:3$

as per the commutative law of multiplication $3 x = x \times 3$ $3x=x xx 3$
$\Rightarrow (x \times 3) \div 3 = 6 \div 3$ $=> (x xx 3)-:3 = 6-:3$

as per multiplicative inverse property and associative law of multiplication
$\Rightarrow x \times (3 \times \frac{1}{3}) = 6 \div 3$ $=> x xx (3 xx 1/3) = 6-:3$

as per PEMA precedence
$\Rightarrow x \times 1 = 2$ $=> x xx 1 = 2$

as per multiplicative identity
$\Rightarrow x = 2$ $=> x=2$

This example is quite simple for the elaborate solution. This elaborate solution is presented to illustrate the way the LPA is used in handling equations. When the problem is complex, this procedure helps to clarify if the steps are correct.

algebraic identities

We learned that "A numerical expression can be modified as per Laws and Properties of Arithmetics to arrive at an equivalent numerical expression, which together is a numerical identity".

Algebraic expressions are modified to arrive at an equivalent algebraic expression, which together is an algebraic identity.

An example illustrating use of LPA in deriving algebraic identities.

$(a + b) (a - b)$ $(a+b)(a-b)$

as per closure law of addition $a + b$ $a+b$ is a number and as per distributive law $= (a + b) \times a - (a + b) \times b$ $=(a+b) xx a - (a+b)xxb$

as per distributive law
$= a \times a + b \times a - a \times b - b \times b$ $=a xx a +b xx a - axx b - bxxb$

as per commutative law of multiplication
$= a \times a + a \times b - a \times b - b \times b$ $=a xx a +a xx b - axx b - bxxb$

as per closure law of multiplication $a \times b$ $a xx b$ is a number and as per additive inverse property $a \times b - a \times b = 0$ $axxb - axxb = 0$
$= a \times a + 0 - b \times b$ $=a xx a + 0 - bxxb$

as per additive identity property
$= a^{2} - b^{2}$ $=a^2-b^2$

This results in the identity $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b) = a^2-b^2$ .

algebraic inequations

We learned that "A Numerical Inequation and an equation can be combined, by addition, subtraction, multiplication, and division -- into a derived inequation."

Algebraic inequations can be modified with known numerical equations or other given algebraic equations.

An example illustrating use of LPA in algebraic inequations.

Consider $3 x + 2 < 8$ $3x+2<8$

Considering an equation $- 2 = - 2$ $-2=-2$ and adding that to the inequation
$\Rightarrow (3 x + 2) - 2 < 8 - 2$ $=> (3x+2)-2 <8-2$

by associative law of addition
$\Rightarrow 3 x + (2 - 2) < 8 - 2$ $=> 3x+(2-2)<8-2$

by additive inverse property
$\Rightarrow 3 x + 0 < 6$ $=> 3x+0<6$

by additive identity property
$\Rightarrow 3 x < 6$ $=> 3x<6$

considering the equation $3 = 3$ $3=3$ and dividing the inequation by this
$\Rightarrow 3 x \div 3 < 6 \div 3$ $=> 3x-:3<6-:3$

by commutative law of multiplication $3 \times x = x \times 3$ $3 xx x = x xx 3$
$\Rightarrow (x \times 3) \div 3 < 6 \div 3$ $=> (x xx 3) -:3 < 6-:3$

by associative law of multiplication
$\Rightarrow x \times (3 \times \frac{1}{3}) < 6 \div 3$ $=> x xx (3 xx 1/3) < 6-:3$

by multiplicative identity and inverse properties
$\Rightarrow x < 2$ $=> x< 2$

This example is quite simple to give such an elaborate solution. The elaborate solution is presented to illustrate the way the LPA is used in handling algebraic inequations. When the problem is complex, this procedure helps to clarify if the steps are correct.

summary

Algebra - First Principles : Algebra deals with algebraic expressions.

Algebraic expressions are representation of quantities with variables, numbers, and arithmetic operations between them.

The expressions are modified as per the PEMA precedence and CADI Laws and properties of Arithmetics.

The algebraic experessions gives raise to various topics

• polynomials (one expression)

• algebraic equations (two expressions with statement of equality )

• algebraic identities (one expression modified into another form)

• algebraic in-equations (two expressions with statement of one being less / more / etc.)

Outline

The outline of material to learn "Algebra Foundation" is as follows.

Note: click here for detailed outline of Foundation of Algebra

→ Numerical Arithmetics

→ Arithmetic Operations and Precedence

→ Properties of Comparison

→ Properties of Addition

→ Properties of Multiplication

→ Properties of Exponents

→ Algebraic Expressions

→ Algebraic Equations

→ Algebraic Identities

→ Algebraic Inequations

→ Brief about Algebra