THE PROOF

I.
How to proof that the square root of two is irrational number?
To proof that the square root of two is irrational number is we can use the reductio ad absurdum. It can show that the square root of two is irrational number by make an assumption that the square root of two is not the irrational number, or it’s called rational number.
First, the definition of rational number is the number that can explain by comparing two number x and y that’s x over y, and x and y are two numbers which relatively prime. And from this statement we can make an assumption that square root of two equal x over y. And the next step is we can get x equal square root of two times y. So, if we squared the two side of this equation, we can get x square equal two times y square. And from this equation, we can make a conclusion that x square always even number because x is a number that made from two times y square and it for y equal all number. So from it we also can get that x is even number too.
The next step is we make the second assumption that x equal two times z. So we from this statement we can get two times z equal square root of two times y. if we squared these two side of this equation, we can see that four times z square equal two times y square. And if we dividing these side of this equation by two, we can get two times z square equal y square. From this equation, we can identify that y square is even number because it’s made from two times z square for z is all number, and we also know that y is even number too.
Finally from these steps, we found that x and y are always even number. So we can’t called square root of two is rational number because both of x and y is always not relatively prime number. So, square root of two is irrational number.

How to show or indicate that the sum angle of triangle is equal to one hundred and eighty degree?
The steps to show that the sum angle of triangle is equal to one hundred and eighty degree is we must draw a triangle which having x, y and z angle. And firstly we know that the sum angle of the straight line is one hundred and eighty degree too. So we can get that the sum angle of straight line is equal x angle, y angle, and z angle in triangle. To make it clearer, we can see this picture of the triangle. We can make some parallel lines to help us imagine the angles in triangle.

z
z
x y x

How to get the value of phi?

To get the value of phi we can do these steps that are we must take some measure from some circle about the diameter. And after this part of the job, we also must find the measure of each perimeter of circles. We can make some assumptions about it for example for the first circle we call it a circle, and having a diameter and also having a perimeter. And the second circle called b circle and also having b diameter and b perimeter too. And if we take some measure from a lot of circle we can called its c circle, d circle, and etc.
We can find the value of phi by comparing the perimeter a with diameter a, perimeter b with diameter b, perimeter c with diameter c, and etc.
Phi = Pa/Da + Pb/Db + Pc/Dc + … dividing by the sum of comparisons.
From this comparison, we get the mean from each comparison, and finally we find phi is 3.14 or 22/7 (twenty seventh).


Explain how you are able to find out the area of region bounded by the graph of y equal x square and y equal x plus two?
To know the region of the wanted area is we must find out the axis of x by make a substitution process to the equalities. We know that y equal x square and y equal x plus two, so we can operate the equation by y equal y, so we can see here we get x square equal x plus two too. From this equation we get x square minus x minus two equal null. So the next is we get x minus two in bracket times x plus one in bracket equal null. Because the value of x minus two in bracket times x plus one in bracket is same with get x square minus x minus two. And the last is we know that e equal two and x equal negative one and it being the axis.
The second step to know the area is we can use the integral, because the area is equal integral with lower boundary x equal negative one and the upper boundary is x equal two of x plus two in bracket minus x square dx. So, the next is we must finish this define integral, that is integral with lower boundary x equal negative one and the upper boundary is x equal two of x plus two minus x square in bracket dx, equal negative one third times x cube plus a half times x square plus two times x from x equal negative one until x equal two.
To finish this equation we must substitution the value of x equal two into this formula minus the value of x equal negative one that also substitution in this formula. So the value is negative one third times two cube plus a half times two square plus two times two in bracket minus negative one third times negative one cube plus a half times negative one square plus two times negative one in bracket.
We continue to finish it by algebraic roles as usually we did. So we get negative eight third plus two plus four in bracket minus one third plus a half minus two in bracket equal negative nine third minus a half plus eight equal five minus a half equal four and a half. And finally we get the area of the region that bounded by the graph of y equal x square and y equal x plus two is four and a half.

Explain how you’re able to determine the intersection points between the circle x square plus y square equal twenty and y equal x plus one!
To find the intersection points of the equation, we can use the substitution way. The way is: we substituting y equal x plus one to equation x square plus y square equal twenty. So we get x square plus x plus one in bracket square equal twenty. And then we get x square plus x square plus two times x plus one equal twenty. Two times x square plus two times x minus nineteen equal zero. And to find the value of x, we use the abc formula.
So, x one and two is negative b plus minus square root of b square minus four times a times c all over two times a. So we get negative two plus minus four minus four times two times negative nineteen in bracket to the power of a half all over two times two. And later we get negative two plus minus square root of one hundred and fifty six all over four. And from this equation, we know that x one is negative two plus square root of one hundred and fifty six all over four, and then x two equal negative two minus square root of one hundred and fifty six all over four.
And by substituting x one and x two to y equal x plus one, we can get the value of y one and y two. So, y one equal two plus square root of one hundred and fifty six all over four, and y two equal two minus square root of one hundred and fifty six all over four. And finally we know that the intersection points are x one point y one that is negative two plus square root of one hundred and fifty six all over four point two plus square root of one hundred and fifty six all over four, and x two point y two equal negative two minus square root of one hundred and fifty six all over four point equal two minus square root of one hundred and fifty six all over four.


II.
SIMILARITY OF TRIANGLES
Definition of similar triangles is triangles whose corresponding angles are congruent and whose corresponding sides are in proportion. We can say triangle is also planar polygonal figures. Two or more triangles are similar when and only when:
(i)Corresponding angles are equal (have the same measure).
(ii)Corresponding sides are proportional.
There are some theorems about the similarity of triangles. The theorems are always used if we found some problems about similarity of triangle, so the theorems are very important to know. These theorems are:
The following theorems are valid in Euclidean geometry:
Theorem AA (angle-angle)
If one triangle has a pair of angles that are congruent to a pair of angles in another triangle, then the two triangles are similar.
Theorem SAS (side-angle-side)
If the pairs of sides of a triangle are proportional to the pairs of sides in another triangle and if the angles included by the side-pairs are congruent, then the triangles are similar.
Theorem SSS (side-side-side)
If the sides of a triangle are proportional to the sides of another triangle, so the triangles are similar.
The applications of the similarity of triangles are it’s used to measure how long and how wide some area, it also used to make we easier to determine which triangle that similar each other, and etc.
The example problems about the similarity of triangles are:
In the triangle ABC shown below, A accent C accent is parallel to AC. Find the length y of BC accent and the length x of A accent A!
The picture is a triangle that having A, B and C angle. And there is a line that parallel with line AC and it lies and made two intersection points with line AB in A accent, and with the line BC in point C accent. And we can make assumption that A A accent is x and C C accent is y.
The picture also shown that the length of AC is twenty two centimeters, the length of C C accent is fifteen centimeters, and the length of A accent B is thirty centimeters.
Solution of problem one:
BA is a transversal that intersects the two parallel lines A accent C accent and AC, hence the corresponding angles BA accent C accent and BAC are congruent. BC is also a transversal to the two parallel lines A accent C accent and AC and therefore angles BC accent A accent and BCA are congruent. These two triangles have two congruent angles are therefore similar and the lengths of their sides are proportional. Let us separate the two triangles as shown below.
Picture:
Shown two triangles that are ABC triangle and A accent B C accent triangle. So the next we call ABC triangle is first triangle and the other triangle is second triangle. So, AB equals thirty plus x, AC equal twenty two, and BC equal y plus fifteen. And the second triangle is A accent C equal fourteen, A accent B equal thirty, and B C accent equal y.

Because we know that the angle of A in first triangle is having the same measure with the A accent angle in the second triangle, we now use the proportionality of the lengths of the side to write equations that help in solving for x and y, that is:
Thirty plus x in bracket over thirty equal twenty two over fourteen equal y plus fifteen in bracket over y
An equation in x may be written as follows. Thirty plus x in bracket over thirty equal twenty two over fourteen.
Solve the above for x. That’s four hundred and twenty plus fourteen x equal six hundred and sixty. So, fourteen times x equal six hundred and sixty minus four hundred and twenty. And we get fourteen times x equal two hundred and fourty to get the value of x is by dividing two hundred and fourty by fourteen.
So, x equal seventeen point one (rounded to one decimal place).

An equation in y may be written as follows. Twenty two over fourteen equal y plus fifteen in bracket over y. And solve the above for y to obtain, twenty two times y equal fourteen times y plus two hundred and ten. So, twenty times y minus fourteen times y equal two hundred and ten. So, eight y equals two hundred and ten, and we get y equal two hundred ten over eight, y equal twenty six point two five.


Problem two:
ABC is a right triangle. AM is perpendicular from vertex A to the hypotenuse BC of the triangle. How many similar triangles are there?
Answer:
Picture 1: triangle ABC, angle A is right angle and angle AMC equal angle AMB are right angle too. AB equal a, AC equal b, BM equal y, BM equal x, and AM equal h.
Picture 2: triangle AMB which has right angle in angle AMB, BM equal x, MA equal h, and AB equal a.
Picture 3: triangle AMC which has angle M as right angle AM equal h, MC equal y, and AC equal b.
Solution to Problem 5:
Consider triangles ABC and MBA. They have two corresponding congruent angles: the right angle and angle B. They are similar. Also triangles ABC and MAC have two congruent angles: the right angle and angle C. Therefore there are three similar triangles: ABC, MBA and MAC.

ARITHMETIC SERIES
The definition of arithmetic series is the sum of an arithmetic sequence. To make it clearer we should see this example:
A series such as three plus seven plus eleven plus fifteen plus etc until plus ninety nine or ten plus twenty plus thirty plus etc until one thousand. Which is has a constant difference between terms. The first term is a1 (a one), the common difference is d, and the number of terms is n. The sum of an arithmetic series is found by multiplying the number of terms times the average of the first and last terms.
The difference between the consecutive terms of the sequence is constant. The constant difference is called common difference. Hence the sequence is arithmetic.
If we have an arithmetic series for example:
1+2+3+…+n, (one plus two plus three plus etc until plus n)
We can write it into this formula or the sum identity is:

(sigma k from k equal one until k equal n equal a half times n times open bracket n plus one close bracket)

Then, if we have an arithmetic series that can be shown like this:
a one plus a two plus a three plus etc until a n, we can write it into this formula:

S n equal a half times a one plus a n in bracket, and we can see here that S n is same with sigma k from k equal one until k equal n equal a half times n times open bracket n plus one close bracket.
Sum = n/2(2a + (n-1)b)
Summary equal n over two times two times a plus open bracket n minus one close bracket times b in bracket.
The next formula is a n equal a one plus n minus one in bracket times d, which are a n is the n-th series, a one is first series, n is the sum of whole series, and d is difference value from that series, or d is a two minus a one equal a three minus a two, etc.
an = a1 + (n – 1)d

The application of arithmetic series is very large; we can easier to know the sum of some series for example is to predict the sum or value of data.
Example problem 1: three plus seven plus eleven plus fifteen plus etc until plus ninety nine which has a one equal three, and d equal four.


3 + 7 + 11 + 15 + ••• + 99 has a1 = 3 and d = 4. To find n, use the explicit formula for an arithmetic sequence.
We solve three plus n minus one in bracket times four equal ninety nine. So we start to calculate it, three plus four n minus four equal ninety nine. So, four times n minus one equal ninety nine. Four times n equal one hundred, so n equal one hundred over four equal twenty five.
3 + (n – 1)• 4 = 99
n = 25
So, the sum is twenty five times three plus ninety nine in bracket over two equal one thousand two hundred seventy five.

or: sum is twenty five over two times open bracket two times three plus twenty five minus one in bracket times four close bracket equal one thousand two hundred seventy five.


Example problem 2: Find the 35th partial sum of a n equal a half times n plus one.
an = (1/2)n + 1
The thirty fifth partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are:
a one equal a half times one plus one equal three second
a two equal a half times two plus one equal two
a three equal a half three plus one equal five second
a1 = (1/2)(1) + 1 = 3/2
a2 = (1/2)(2) + 1 = 2
a3 = (1/2)(3) + 1 = 5/2
The terms have a common difference d equal a half, so this is indeed an arithmetic sequence. The last term in the partial sum will be a thirty five equal a one plus thirty five minus one in bracket times d equal three second plus thirty four times a half equal thirty seven second.
a35 = a1 + (35 – 1)(d) = 3/2 + (34)(1/2) = 37/2.
Then, plugging into the formula, the 35th partial sum is:
n over two in bracket times a one plus a n in bracket equal thirty five second times three second plus thirty seven second in bracket equal thirty five second times fourty second equal three hundred and fifty.

COMPOSITE NUMBER
The most important thing that we should know before we learn about the composite number is we must know the definition of composite number. So, a composite number is a natural greater than 1 that is divisible by a number other than itself and 1 (in other words, it has more than the two factors of 1 and itself).
The integer fourteen is a composite number because it can be factored as two times seven.
And to make we learn about it deeper, we must also know about the prime number, the definition of prime number is a natural number greater than 1 that has only itself and 1 as factors.
One thing as important as the definition above is about the fundamental theorem that usually used in this material, as shown below:
Every composite number can be expressed as a product of prime numbers in one and only one way.
The prime factorization of a whole number is the number written as the productof its prime factors.
And also one method used to find the prime factorization of a composite number is called a factor tree.
For example, find the factor from twenty!
The answer is: we must make a factor tree from twenty. That is twenty equal two times ten, and in the picture of factor tree is two and ten being the branch of twenty, and it continued until twenty can be expressed just as a product of prime number. So, the next is ten equal two times five, here we find that two, two, and five are all prime number being the factors of twenty.
We also must know about the relatively prime numbers, and the definition of relatively prime number is pairs of numbers that have 1 as their greatest common divisor. For example, the greatest common divisor of seven and twenty five is 1. Thus, seven and twenty five are relatively prime numbers.
To find the greatest common divisor of two or more numbers,
Write the prime factorization of each number.
Select each prime factor with the smallest exponent that is common to each of the prime factorizations.
Form the product of the numbers from step two. The greatest common divisor is the product of these factors.
Example: Find the greatest common divisor of two hundred and sixteen and two hundred and thirty four.
Solution: Step 1. Write the prime factorization of each number.
The picture of factor tree is explain about two hundred and sixteen equal two times one hundred and eighteen, and one hundred and eighteen equal two times fifty and four, and fifty and four equal two times twenty seven, and twenty seven equal three times nine, and nine equal three times three.
And the second factor tree is explain that two hundred and thirty four equal two times one hundred and seventeen, one hundred and seventeen equal three times thirty nine, and thirty nine equal three times thirteen.
The factor tree at the left indicates that two hundred and sixteen equal two to the power of three times three to the power of three.
The factor tree at the right indicates that two hundred and thirty four equal two times three to the power of two times thirteen.
Step 2: Select each prime factor with the smallest exponent that is common to each of the prime factorizations.
Which exponent is appropriate for two and three? We choose the smallest exponent; for two we take two to the power of one and for three we take three square.
Step 3: Form the product of the numbers from step three. The greatest common divisor is the product of these factors. Greatest common divisor equal two times three square equal two times nine equal eighteen. Thus, the greatest common factor for two hundred and sixteen and two hundred and thirty four is eighteen.
The least common multiple (LCM) of two or more natural numbers is the smallest natural number that is divisible by all of the numbers.
To find the least common multiple using prime factorization of two or more numbers:
Write the prime factorization of each number.
Select every prime factor that occurs raised to the greatest power to which it occurs, in these factorizations.
Form the product of the numbers from step 2. The least common multiple is the product of these factors.
Example: Find the least common multiple of one hundred forty four and three hundred.
Solution: Step one. Write the prime factorization of each number.
One hundred and forty four equal two to the power four times three square.
Three hundred equal two square times three times five square.
Step 2: Select every prime factor that occurs, raised to the greatest power to which it occurs, in these factorizations.
One hundred and forty four equal two to the power of four times three square.
Three hundred equal two square times three times five square.
Step 3: Form the product of the numbers from step 2. The least common multiple is the product of these factors.
LCM equal two to the power of four times three square times five square equal sixteen times nine times twenty five equal three thousand and six hundred.
Hence, the LCM of one hundred and forty four and three hundred is three thousand and six hundred. Thus, the smallest natural number divisible by one hundred and fourty four and three hundred is three thousand and six hundred.
Properties:
All even numbers greater than two are composite numbers.
The smallest composite number is four.
Every composite number can be written as the product of (not necessarily distinct) primes. (Fundamental theorem of arithmetic)
Also, n minus one in bracket factorial equivalent with zero modulo n for all composite numbers, n more than five. (Wilson's theorem)
Kinds of composite numbers:
One way to classify composite numbers is by counting the number of prime factors. A composite number with two prime factors is a semi prime or 2-almost prime (the factors need not be distinct; hence squares of primes are included). A composite number with three distinct prime factors is a sphenic number. In some applications, it is necessary to differentiate between composite numbers with an odd number of distinct prime factors and those with an even number of distinct prime factors.
Some theorems about composite number are:
Theorem1)
If C is an odd number composed of two prime factors a and b, then the number of consecutive odd integers required to add up to C, is equal to the smaller of the two prime factors of C.

Theorem2)
The larger of the two prime factors of C is the average to the first and last number in the consecutive sum of odd numbers that add up to C.

Theorem3)
If C is an odd number that is composed of two prime factors a and C, then there exists at least one non-arbitrary perfect square that can be used to factor C through a finite number of algebraic operations.

Theorem4)
If C is an odd number that is composed of two prime factors a and b, then the non-arbitrary perfect square that can be used to factor C via finite number of algebraic operations is the first perfect square acquired by adding a consecutive number of odd integers starting with the number 1 to C and is equal to the average of the prime factors of C quantity squared.
Example:
Solution: Find the sum of the first one hundred even positive numbers.
The sum of the first one even positive numbers is two or one times one plus one equal one times two.
The sum of the first two even positive numbers is two plus four equal six or two times two plus one in bracket equals two times three.
The sum of the first three even positive numbers is two plus four plus six equal twelve or three times open bracket three plus one close bracket equal three times four.
The sum of the first four even positive numbers is two plus four plus six plus eight equal twenty or four times open bracket four plus one close bracket equal four times five.
Look for a pattern :
The sum of the first one hundred even positive numbers is two plus four plus six plus eight plus etc equal x or one hundred times one hundred plus one in bracket equal one hundred times one hundred and one or ten thousand and one hundred.


And finally this is the end of my fifth task and I wish it can be used well.
Best Regards….