THE RESEARCH OF MATHEMATICS

The research of mathematics is important to student of mathematics. To get the research of mathematics we must have a good ways. One of the ways is to search the references about the mathematics such as, a browsing internet, search the related books, etc.
We should have knowledge about it to improve our knowledge about mathematics and to know the procedure or the steps to found the new formula, axioms, and theorems in mathematics. Theorem in which there is no contradiction in side.
It is very easy to establish mathematics system. Let X₁, X₂, X₃, ..........are real members.
The formal mathematics:
1. Numbers theory
2. Group theory
3. Ring theory
4. Field theory
5. Euclidean geometry
6. Non. Euclidean geometry
7. Number system







ABSTRACT
Mathematics is an exact science, where in all the calculations can
accountability. The nature of mathematics are deductive axiomatic,
means to learn mathematics depart from the base of understanding,
statement of the base, then axiom. Approach from the base of understanding,
statements and the axiom base can result in a theorem that
becomes the basis for finding a solution of a problem, then comes
theorem-Theorem others. Many branches of mathematics, one
only is the field of geometry. The usefulness of the science of geometry in
design, measurement of height, and other applications.
There are several fundamental theorem in the science of geometry, one of them
is the Pythagorean theorem. This theorem was discovered by a mathematician
named Pythagoras. In this case the author tried to analyze through
Pythagorean theorem and properties derived from the Pythagorean theorem
can answer the problem of measurement segment that was not possible
manually.











INTRODUCTION
A. Background
The nature of mathematics are deductive axiomatic, that is to
study of mathematics starting from understanding the base, the base statement
the definition agreed upon, then the axiom is a statement
accepted without proof. The approach of understanding the base, the base statement and axioms can come up with a theorem that is the basis of find solution.
Many branches of mathematics, one of which is the field
geometry. The geometry of the usefulness of science in engineering,
measurement of height, and other applications. There are several fundamental theorems is the science of geometry, one is the Pythagorean theorem only. This theorem was discovered by mathematician named Phytagoras. Pythagoras was born around the year 582 BC on the island of Samos, Greece. Pythagoras discovered a geometrical formula
simply about the relationship side of the triangle. This formula later known as the Pythagorean Theorem. In many building design cannot be separated from the Pythagorean theorem, in the physical sciences Pythagorean theorem is very helpful in measuring the height of a building, in mathematics itself Pythagorean theorem is
assist in making a line that can not be measured with a ruler.

B. Purpose
Based on the above problem formulation, the goal in This research is proving that the Pythagorean theorem and properties derived from the Pythagorean theorem can be answered a problem of measurement segment that was not possible manually





DISCUSSION
1. Pythagorean
In mathematics, the Pythagorean Theorem is a relationship in Euclidean geometry among the three sides of a right triangle. This theorem is called by a Greek philosopher and mathematician, 6th century BC, Pythagoras. Pythagoras is often regarded as the inventor of this theorem, although the actual facts of this theorem was known by mathematicians of India (in Sulbasutra Baudhayana and Katyayana), Greek, and Babylonian Tionghoa long before Pythagoras was born. Pythagoras get credit because it was he who first proved the universal truth of this theorem by mathematical proofs.
2. Pythagorean Theorem
Since the fourth century AD, Pythagoras has commonly been given credit for discovering the Pythagorean theorem, a theorem in geometry that states that in a right-angled triangle the square of the hypotenuse (the side opposite the right angle), c, is equal to the sum of the squares of the other two sides, b and a—that is, a2 + b2 = c2.
While the theorem that now bears his name was known and previously utilized by the Babylonians and Indians, he, or his students, are often said to have constructed the first proof. It must, however, be stressed that the way in which the Babylonians handled Pythagorean numbers, implies that they knew that the principle was generally applicable, and knew some kind of proof, which has not yet been found in the (still largely unpublished) cuneiform sources. Because of the secretive nature of his school and the custom of its students to attribute everything to their teacher, there is no evidence that Pythagoras himself worked on or proved this theorem. For that matter, there is no evidence that he worked on any mathematical or meta-mathematical problems. Some attribute it as a carefully constructed myth by followers of Plato over two centuries after the death of Pythagoras, mainly to bolster the case for Platonic meta-physics, which resonate well with the ideas they attributed to Pythagoras. This attribution has stuck, down the centuries up to modern times. The earliest known mention of Pythagoras's name in connection with the theorem occurred five centuries after his death, in the writings of Cicero and Plutarch.
3. Proof

Now we start with four copies of the same triangle. Three of these have been rotated 90°, 180°, and 270°, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4•ab/2), we get
c² = (a - b)² + 2ab
= a² - 2ab + b² + 2ab
= a² + b²



CONCLUTION

• Right-angled side is the side that forms a right angle.
While is the side before the right angle.
• Square of the hypotenuse equals the sum of squares square is called Pythagoreans theorem.

1.Determine the integration factors of the following differential equation and the find the solution
(3x2y + 2xy + y3) dx + (x2+ Y2) dy =0

2.Known : sin -1 x =
∫_0^x▒1/√((1-t2)) dt

Determine the first four tribes taknol in Mac Laurin series for sin -1

3.Known :
g =((█(x@y@z))= (█(1@-3@5))+ a (█(-6@4@7)) ) a parameter
Determine the line g and h has an intersection

4.Chanted a coin in a row as much as 6 times, how many chances at least one comes face?

5.Consider the following data for 120 mathematics at the college concerning the languages French, German, Russian
65 study French
45 study German
42 study Russian
20 study French and German
15 study German and Russian
8 study all three languages
Let F, G, and R denote the sets of students studying French, German, and Russian, respectively. We wish to find the number of students who study at least one of the three languages, and to fill the correct number of students in each of the eight regions


Answers:
1.(3x2y + 2xy + y3) dx + (x2+ Y2) dy =0
M(x,y) = 3x2y + 2xy + y3
N(x,y) = x2+ Y2
∂M(x,y) / ∂y = 3x2+ 2x + 3y2
∂N(x,y)/ ∂x = 2x
Formula to get the integrations factors is :
(1/N(x,y)) [(∂M(x,y)/∂y)] - [(∂N(x,y)/∂x)]
Substitute the equation 1 and 2 to formulas;
1/(x2 + y2) ((3x2+ 2x + 3y2) – (2x))
1/(x2 + y2) (3x2 + 3y2) =3
You can get μ(x) = e∫3 dx = e3x
Multiply e3x to M (x,y) and M (x,y) be M* (x,y)
M* (x,y) = 3x2ye3x + 2xy e3x + y3e3x
Differential M* (x,y) to y, so:
∂M*(x,y) / ∂y = 3x2e3x + 2xe3x + 3y2e3x
And so N(x,y) multiply with e3x
N*(x,y) = x2e3x + y2e3x
Differential N*(x,y) to x so:
∂N*(x,y)/∂x = 2xe3x + 3e3xx2 + 3y2e3x
Cause the different equation is exact differential equation, so:
∂F/∂x=M* AND ( ∂F)/∂y=N*
To get the solution we must be :
F (x,y) = ∫(3x2ye3x + 2xye3x + y3e3x) dx + Q (y)
= yx2e + 1/3 y3 e3x + Q(y)
∂F(x,y) /∂y = x2e3x + y2 e3x +Q’(y)
∂F/∂y = N*
e2 e3x + y2 e3x = x2e3x + y2e3x + Q’(y)
Q’(y) = 0
Q(y) = c
So the solution from this problem is
F(x,y) = yx2e + 1/3 y3 e3x + c

2. Known : sin -1 x =
∫_0^x▒1/√((1-t2)) dt
We have a formula :
(1 + X)P = 1 + (█(p@1))x + (█(p@2))x2 +…
Where : + (█(p@k)) =( p (p-1) (p-2))/k!
So we can apply that
(1-t2)-1/2 = 1 + (-1/2) (-t2) + (-1/2) (-3/2) (-t2)2 + (-1/2) (-t2) (-5/2) (-t2)3+…
= 1 + t2/2 + (3/8) t4 + (15/48) t6 +…
= t2/2 + (3/8) t4 + (5/16) t6 +…
Sin -1 x = ∫_0^x▒〖1 〗+ t2/2 + (3/8)t4 + (5/16)t6 +…
= x + x3/6 + (3/40)x5 + (5/112)x7


3.Known :
g = (█(x@y@z)) = (█(1@-3@5))+ a(█(-6@4@7)) a parameter
x = -2 -9b
h = y = 8-3b b parameter (2)
z = -6 +4b
to prof gnh = N
N € g => coordinate N must on the g
N € h => coordinate N must on the h
x = 1 – 6a
y = -3 + 4a
z = 5 + 7a xN= 1 – 6a
substitute coordinate N to (1) => yN = -3 +4a (3)
zN = 5 + 7a



substitute coordinate N to (2) => xN = -2 -9b
h = yN = 8-3b b parameter (4)
zN = -6 +4b
left segment of (3) equals left segment of (4)
right segment of (3) equals right segment of (4)
1 – 6a = -2 – 9b  3 = 6a – 9b
1 = 2a – 3b …………. (5)
-3 + 4a = 8-3b  4a + 3b =11……………(6)
5 + 7a = -6 + 4b  7a – 4b = -11……….(7)
So:
2a – 3b = 1
(8) 4a + 3b=11
7a – 4b = -11
Eliminate (5) and (6) => b = 3
Substitute b to (5) => a = 5
Substitute value a and b to (7)
7a – 4b = -11
7.5 – 4.3 = -11
35 - 12 ≠ -11
If that no similar gnh = empty seT
N if (8) have a solution

4.Suppose that E happened at least once come home. Sample space S contains the sample is 26 = 64 points. Because each reflection can produce two kinds of results (fronts or near), we know that P (E) = 1 – P (E’). if E’ represents the event that no face appeared. This will only happen in a way, which is all behind chanting produces. Be P(E’) = 1/64 so P(E)=1 - 1/64 = 63/64

5.(F U G U R) = n(F) + n(G) + n(R) –n(F slices G) – n(F slices R) - n( G slices R)
+ n(F sliceS G slices R)
= 65+45+42-20-25-15+8 = 100
That is, n(F U G U R)=100 students at least one of the three languages
Now, we use this result to fill in the venn diagram. We have 8 study all the three languages
20 – 8 =12 study French and german but not Russian
25 – 8 = 17 study French and Russian but not german
15 – 8 = 7 study german and Russian but not French
65 – 12 – 8 -17 =28 study only French
45 – 12 – 8 – 7 =18 study only german
42 – 17 – 8 -7 = 10 study only Russian
120 – 100 =20 do not study at the languages
Observe that 28 + 18 + 10 = 56 students only one of the languages

Problems solving created by salindri murbarani 08305141034
Solutions the problems created by Narita yuri