give examples for euclids axioms and postulates
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EUCLID'S AXIOMS :-
Euclid organized all the known works of mathematics at his time into Elements, a collection of thirteen books. Each book contains a series of propositions or theorems, varying in number from about ten to hundred. These propositions or theorems are preceded by definitions. In Book I, twenty-three definitions are followed by five postulates. Five common notions or axioms are listed after the postulates. He assumed certain properties that were not to be proved. Those properties were universal truths. Euclid divided them into Axioms and Postulates. Some of those properties were not specific to geometry, and he called them common notions (or axioms).
Some of Euclid’s axioms are stated as follows.
1. Things which are equal to the same thing are also equal to one another.
2. If equals be added to equals, then the wholes are equal.
3. If equals be subtracted from equals, then the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.
6. Things which are double of the same things are equal to one another.
7. Things which are halves of the same things are equal to one another.
These axioms have a lot of practical applications, which we shall discuss one by one. Let us start with the first axiom.
Suppose the area of a rectangle is equal to the area of a triangle and the area of that triangle is equal to the area of a square. By applying Euclid’s first axiom, we can say that the areas of the triangle and the square are equal. Similarly, if a = b and b = c, then we can also say that a = c.
Let us now discuss the second axiom. Let us consider a line segment AD in which, AB = CD.
If we add BC to both sides of this relation (equals are added), then according to Euclid’s second axiom, we can say that, AB + BC = CD + BC i.e., AC = BD.
Consider the rectangles ABCD and PQRS drawn in the given figure.
Suppose that the areas of the rectangles ABCD and PQRS are equal. If we remove triangle XYZ from each of the two rectangles as shown in the figure, then we can say that the areas of remaining portions of the two triangles are equal. We derived this from Euclid’s third axiom.
Euclid’s fourth axiom is sometimes used in geometrical proofs. Let us consider a point Q that lies between points P and R of a line segment PR as shown in the figure.
From this figure, we can notice that (PQ + QR) coincides with the line segment PR. Thus, by using Euclid’s fourth axiom, which states that “things which coincide with one another are equal to one another”, we can write, PQ + QR = PR.
Using the same figure that we used in the fourth axiom, we can see that PQ is a part of line segment PR. By using Euclid’s fifth axiom, we can say that the whole i.e., line segment PR is greater than the part i.e., PQ. Mathematically, we can write it as PR > PQ.
Euclid’s sixth and seventh axioms are interrelated. Let us consider two identical circles with radii r1 and r2 respectively.
Suppose their diameters are d1 and d2 respectively.
As these are identical circles, their radii are equal.
∴ r1 = r2
∴ r1 = r2
Using Euclid’s sixth axiom, we can say 2r1 = 2r2
Using Euclid’s sixth axiom, we can say 2r1 = 2r2
∴ d1 = d2
∴ d1 = d2
Therefore, we can say that when the radii of two circles are equal, then their diameters are also equal.
If instead of taking the radii as equal, we say that the diameters of the two circles are equal, then using the seventh axiom, we can say that the radii of the two circles are also equal.
Let us solve some examples to understand the concept better.
Show that a line segment has only one mid-point.
Let us suppose that a line segment PQ has two mid-points R and S.
Case I: When R is the mid-point, we have, PR = RQ.
We will use one of Euclid’s axioms, according to which, when equals are added to equals, then the wholes are equal.
On adding PR to both sides, we obtain
2PR = PR + RQ
⇒ PQ = 2PR … (1)
Case II: When S is the mid-point, we have PS = SQ.
Again, we will use Euclid’s axiom that when equals are added to equals, the wholes are equal.
On adding PS to both sides, we obtain
2PS = PS + SQ
⇒ PQ = 2PS … (2)
On comparing (1) and (2), we obtain
2PS = 2PR
PS = PR (Things which are double of the same things are equal to one another.)
This implies that the points S and R lie at the same place on line PQ. Thus, our assumption is wrong. Hence, a line segment can have only one mid-point.
In the given line segment AD, AC = BD. Prove that AB = CD.
It is given that AD is a line segment with AC = BD.
We will use one of Euclid’s axioms, according to which, when equals are subtracted from equals, then the remainders are equal.
On subtracting BC from both sides, we obtain
AC – BC = BD – BC
∴ AB = CD
EUCLID'S FIRST FOUR POSTULATES :-
Along with the definitions that Euclid gave in Elements, he assumed certain properties that were not to be proved. Five of those properties were specifically geometric, and he called them postulates.
Let us discuss the first four postulates one by one.
Postulate 1: It is possible to draw a straight line from any point to any other point.
Postulate one suggests that if we have two points P and Q on a plane, then we can draw at least one line that can simultaneously pass through these two points. Euclid does not mention that only one line can pass through two points, but he assumes the same. The fact that there can be only one line passing through two given points is illustrated in the following figure.
Postulate 2: A terminated line can be produced indefinitely.
This postulate can be considered as an extension of postulate 1. According to this postulate, we can make a different straight line from a given line by extending its points on either sides of the plane.
In the following figure, MN is the original line, while M′N′ is the new line formed by extending the original line in either direction.
Postulate 3: It is possible to describe a circle with any centre and radius.
According to Euclid, a circle is a plane figure consisting of a set of points that are equidistant from a reference point. It can be drawn with the knowledge of its centre and radius.
The shapes of circles do not change when different radii are considered. Only their sizes change.
Postulate 4: All right angles are equal to one another.
A right angle is unique in the sense that it measures exactly 90°. Hence, all right angles are of the measure 90° irrespective of the lengths of their arms. Hence, all right angles are equal to each other.
Remark: Unlike right angles, acute and obtuse angles are not unique in the sense that their measures lie between 0° to 89° and 91° to 179° respectively. Hence, the measure of one acute angle is not the same as the measure of another acute angle. Similarly, each obtuse angle has a different measure.