Set Theory and Various Types of Sets

Set theory is one of the important theories in mathematics. It can help in solving of various mathematical problems. The problems can be represented in the graphical form with the help of set theory. A set is nothing but a collection of objects. There can be intersection of sets and union of sets. On performing the intersection operation the common elements of both the sets are got. The union of two sets gives all the elements present in both the sets. Basically a set can consist of various objects in it. But usually in mathematics sets usually deal with the objects which are related to mathematics.

The concept of set theory is very ancient. In the 1870’s itself good research was done on this topic and considerable progress was made. A Venn diagram is best used to represent the operations on sets. The process of intersection and union can be easily represented on the Venn diagram. There can be bigger set and a smaller which is part of it is called its subset. This can be explained with the help of an example. If a set contains the elements {1, 2, 3, 4} and there is another set which contains the elements {1, 2} then the latter set is subset of the former set. It simply means the all the elements present in the second set are also present in the first set. First set covers the whole of the second set in it.

The compliment of a set is nothing but a set containing elements which are present in the universal set but are not present in the given set. The set complement can be explained with the help of an example. If there are elements like {a, b, c, d, e} in the universal set and elements in the given set whose compliment is to be found out, are {a, c, e} then the complement set is given by { b, d}. So, the elements b, d is contained in the universal set but is not present in the set whose compliment has to be found. So, these elements are part of the required set. The compliment of a set X can be represented by the notation X’. This is a simple notation and there is another method of representing the same. Instead of the apostrophe symbol the letter ‘c’ can also be used to represent compliment.

Introduction to polygon basics

The polygon is a basically called plane figure and that is surrounded with closed path, collected of a fixed series in straight line segments with a nearer polygonal chain. And these segments are called its edges or sides. Let us discuss  the topic called polygon basics. (Source – Wikipedia)

Types of Polygon Basics

Names of polygons basics with different number of sides:

• Triangle : Triangle consists of three sides
• Quadrilateral :Quadrilateral consists of four sides
• Pentagon: Pentagon consists of five sides
• Hexagon:Hexagon consists of six sides
• Octagon:Octagon consist of eight sides.
• Nanogon : Nonagon consists of nine sides
• Decagon : Decagon consists of ten sides
• Heptagon:Heptagon consists of sevensides.
• Triskaidecagon or Tridecagon : Trikaidecagon or tridecagon consists of  thirteen sides. 

• Tetrakaidecagon or Tetradecagon : Tetrakaidecagon or tetradecagon consists of fourteen sides.
• Pendedecagon : Pendedecagon consists of fifteen sides.
• Hexdecagon: Texdecagon consists of  sixteen sides.
• Heptdecagon: Heptdecagon seventeen sides.
• Octdecagon: Octdecagon eighteen sides.
• Enneadecagon: Enneadecagon consists of nineteen sides
• Icosagon: Icosagon consists of twenty sides.
• Triacontagon : Triacontagon consists of thirty sides.
• Teracontagon: Teracontagon consists of forty sides.
• Tetracontagon: Tetracontagon consists of fifty sides.
• Pentacontagon: Tentacontagon consists of sixty sides.

 The triangle is the basic method in polygon and it has three angles. This involves three sides and three vertices.

 Triangle

Square

The octagon

The drawn diagramatic representation are the  basics form in polygon.

Problems Based on Polygon Basics

Example 1 :

Determine the side distance end to end of hexagon is 12 cm. Mention the area of the hexagon.

Solution:

            Given:

                        Side distances  (t) = 12 cm

            Formula:

                        Area of the hexagon (A) = t2 2.6

                                                = 122 x 2.6

                                                = 144 x 2.6

                                                = 374.4

Example 2:

The side distances of hexagon is 13cm. Find the area of the hexagon.

Solution:

            Given:

                        Side length (t) = 13 cm

            Formula:

                        Area of the hexagon (A) = t2 2.6

                                                = 132 x 2.6

                                                = 169 x 2.6

                                                = 439.4

              Area of the hexagon = 439.4 cm2    

Real numbers

Real Number Definition– As the name says “Real”, so these numbers are actually numbers that really exists. Any number that we think of is considered as a real num, be it positive or negative, fraction or decimal. Real num. are numbers those includes both rational and irrational number. A real-number has to have a value. If there is no value to any number then we can call that number as an imaginary number. All integers like -75, 89, and 84 etc. are considered as real num. All fractions like 3/5, 7/2, -9/7 are considered as real no. too.

Decimals along with repeating decimals are also considered as real num.
These can be any positive or negative number. We can plot All Real Numbers on the number line too.  Therefore we can order these numbers and that we cannot do in case of imaginary numbers. The name of imaginary numbers itself says that they are imaginary so we can just imagine them; they do not have a specified value. These numbers can be plotted the same way we plot the integers that is smaller numbers on left and larger numbers on right. So greater the number, more it will be towards right side of number line.
So we can call real nos. as all those numbers which are present on number line are termed as such.
Some Real Numbers Examples are pi, 34/7, 5.676767, -1034, 45.87 etc.
Some examples of imaginary numbers are square root of -34 or square root of -2, as there is a negative under the root, so the value of this number cannot be found.

Similarly value of infinity cannot be determined too .
Hence these numbers are not considered as real and are considered as an imaginary numbers. So these numbers are all integers, fractions decimals and repeating decimals numbers. We can add, subtract, multiply and divide real nos. just like another numbers. We can perform the operations on real-numbers same way as we do them on other numbers.

Now the question arises that Is 0 a Real Number– Zero is considered as an integer and all integers are real-numbers. Therefore zero is considered to be as a real-number.
 All Real Numbers Symbol is R which is used by many mathematicians. The symbol R is used to represent the set of real nos. All such numbers can be seen on the number line but we cannot find imaginary numbers on that.

Construct angles tutoring

In geometry, we have many constructions of different geometrical shapes, angles,angle bisectors etc, these constructions are really challenging and mind blowing,now let see and the steps of  few constructions of angles.

Constructions of angles can be done by two methods,one by using ruler and compass and another by using protractor. Now we see and understand the steps of angles constructions by using both the methods.

Construct Angle 60' by Two Methods

Construct angle 60' by a ruler and compass.

Construction method I: Draw a convenient line segment AB,keep the compass at A ,draw a semi circle.the semi circle should intersect the line segment AB,then mark the intersect the point as C, then again keep the compass at C,without changing the measurement of the compass draw an arc , which cut the semicircle at the point D.Now join A and D and extend the line.Hence we get angle 60'.

Construction method II:  

In this method we draw a line segment AB and keep the protractor at A ,mark 60' from right to left .Mark the new point as ,C Join AC.Now we get angle 60' using protractor.

Construct Angle 90' by Two Methods

Construct angle 60' by a ruler and compass.

Construction method I:

Draw a line segment AB,draw a semi circle at A,which intersect C,then keep the compass at C draw an arc which intersect the semi circle at D,again with the same measurement keep the compass  at D ,draw one more arc ,which intersect the semi circle at E,do not change the measurement of compass, keep the compass at D and E, cut two more arc's which intersect each other exactly at F, join A and F.The line AF exactly perpendicular to the line segment AB.Hence angle 90' constructed.

Construction method II :

Now in this method, draw a line segment AB keep the protractor at A and mark 90', name as C , join A and C , we get line AC is perpendicular to the line AB. Hence  angle 90' constructed using protractor.

Exercises on Construct Angles

1) Construct angle 120' using ruler and compass.

2) Construct angle 120' using protractor.

3) Construct angles 15' , 30', 45' , 75', 105', 135' using ruler and compass.

4) Construct angles 15' , 30', 45' , 75', 105', 135' using protractor.

Learning to Simplify Mixed Numerals

We have studied the number line. We know about the natural numbers, whole numbers, integers, real numbers and also imaginary numbers. When these numbers are combined we get interesting combinations. We need to do a study on fractions. Fractions can be numbers in the form of P/Q. P and Q are integers. These fractions now are of three types namely proper, improper and mixed fractions. Mixed fractions are bit complicated compared to the other two. So we need to study them in detail. First we need to learn the process of simplifying mixed fractions as a first step in this direction. It is a relatively simple process.  To simplify mixed fractions we convert them to improper fractions. Improper fractions are easy to handle and can be used for the purpose of calculations. After converting the mixed fraction into an improper one, we can further simplified if need be. For this we need to convert mixed fraction to decimal and solve this problem.

After getting the improper fraction, we take the numerator of the fraction as the dividend and the denominator as the divisor. We divide the dividend with the divisor till we get the remainder as zero. When the digits in the dividend get over we continue the process of division by adding zeros to it and a decimal point to the quotient. Sometimes we don’t get the remainder as zero. In that case the decimal part might contain digits that are repeating. It can be ended there. So the question arises how to solve mixed fractions after we have converted them into decimals. Solving can involve addition of two mixed fractions or it can also be subtraction of two mixed fractions. Both the processes can be relatively simple if the denominators in both the cases are the same. But if the denominators are different the process is bit different than when they are same. Now we will learn subtracting mixed fractions with different denominators to get the final answer. When the denominators are not same first convert into similar ones and then do the process of addition or subtraction. This can be done by taking the LCM of the denominators. After taking the LCM the denominators become common and the numerators can be added or subtracted to get the answer. This helped us to know what is a mixed fraction better. It contains a whole number part and a proper fraction.