Examples of solving mathematics' problems.
1. We will prove that the square root of two is irrational number by showing the contradiction of this problem. So, we will suppose that the square root of two is rational, that is the square root of two equal a over b where a and b are relatively prime integers. It means that they have no common positive integral factor other than unity. The operation of the square root of two equal a over b becomes a equal b times the square root of two if we multiply each sides with b. Then, if we quadratic each sides we will get a square equal two times b square. A square is twice b square, while first we have assumed that a and b have to be relatively prime. So, it is impossible to find some numbers that relatively prime while the square number of one of them is twice than other square. So, it is proved that the square root of two is irrational number.
2. We will show how to indicate that the sum angles of triangles is equal to one hundred and eighty degree by drawing a triangle first. Then, we cut the triangle by scissors. Named each angles of the triangle by A, B, and C. So, we have angle A, angle B, and angle C now. What we have to do then is that cut one by one the angles of the triangle. So that we get three different angles. The next step is integrating those three different angles by making them closer each other in a line like in a jigsaw game. We get a straight line which is composed from angle A, angle B, and angle C. We know that a straight line has one hundred and eighty degree, which is the sum of those three angles of the triangle. So, it is proved that the sum angles of triangles is equal to one hundred and eighty degree.
3. We will find where phi from is. There are some steps to get there. The first is that we have to prepare wire forty-four centimeters in length and make a circle with radius of seven. Second, we have known that the length of is circumferences of the circle. And when we divide it by seven times two, it will generate twenty two seventh. The third steps is redoing the first and second steps with in one hundred centimeters long, one hundred and fifty four centimeters, and one hundred and forty four centimeters and then make each of those a circle. Then, divide each examples by each radius times two (for example, divide one hundred by the radius times two). The result will be close to twenty-two seventh or three point fourteen. This kind of number, which called as phi.
4. We will find out the area of region bounded by the graphs of y equal x square and y equal x plus two. If we want to see it clearly, we may sketch the graph. From which we should make x and y-axis, then we will sketch y equal x square by drawing a curve through zero point zero coordinate. Then, we sketch the graph of y equal x plus two, which is through the intersection point between y equal x square and y equal x plus two. Then, we will find the intersection point between y equal x square and y equal x plus two by substitution. So, we substitute y equal x square into y equal x plus two. It generates negative x square plus x plus two equal zero. So, we get the x is two or negative one. And the intersection point are two point four coordinate and negative one point one coordinate. The next step to find the area is integrated the area that bounded by y equal x square and y equal x plus two. The operation will be integral of negative x square plus x plus two about x from x equal negative one to x equal two. We solve the integral which equal negative one third times x cube plus half of x square plus two times x from x equal negative one to x equal two. The result is thirty-nine sixth. So, the area is equal thirty nine sixth. The solution is thirty-nine sixth.
5. We will find out the determine of the intersection point between the circle x square plus y square equal twenty-five and y equal x plus one. It is better for us to sketch the circle and line y equal x plus one. First, we may have to find the intersection point between the circle and the line. So, we substitute y equal x plus one into the circle x square plus y square equal twenty five. If we quadratic each sides, we will get x plus one in bracket square equal twenty-five minus x square. Then x plus one in bracket square equal x square plus two times x plus one equal twenty minus x square. Then the operation will be two times x square plus two times x minus twenty four equal zero. Then, if we divide each sides by two, we will get x square plus x minus twelve equal zero. Then we get x equal negative four or x equal three. So, the intersection point are three point four coordinate and negative four point negative three coordinate. So, the solution are three point four coordinate and negative four point negative three coordinate.
Median
The definition of median (of a triangle) is a line that made from a vertex of a triangle so that it divides the edge in front of its vertex into two same points.
How to draw a median?
First, make a triangle by three different point named as A, B, and C. The second step, from points B and C, make a bow, which have some radius so that the bow will intersect on two points. Then, give a name as D and E. The third step is connecting point D and E so that it cross a point in line BC and named as F. The fourth step is connecting point A and F, then it will be make a median.
If we make three medians in a triangle, it will be crossed in a point. Moreover, the point can be given a name as centroid or center of gravity. And the special character of a median if it is crossed in a point, the ratio will always two and one. Two is for longer line.
The length of median can be found by using Stewart’s postulate. If you want to know how to find it, it will be posted in different title.
Examples:
There is an ABC triangle.CD is the median. And the length of AB is fourteen centimeters. While BC has ten centimeters long and AC is six centimeters in length. How long is CD?
The way to get the length of CD is counting by a formula like this. CD square equal to open bracket half times BC square close bracket plus open bracket half times AC square close bracket minus open bracket a quarter times AB square close bracket.
TAUTOLOGY
Tautology is plural statements, which always true for each substitution of any truth value in its single statements.
It can be separated in two groups. They are conditional tautology and biconditional tautology.
There are so some ways to decide the truth value of plural statements. They are making the table of truth values and arithmetical procedure. Plural statements can be translated into arithmetical procedure. First, negative a can be changed into one plus a. second, a and b can be changed into a plus b plus a plus open bracket a times b close bracket. Third, a or b can be changed into a times b. fourth, if a then b can be changed into arithmetical procedure as open bracket one plus a close bracket times b. Fifth, a is if and only if b can be changed into a plus b.
Example of solving problem:
1. Show that if p then in bracket p or w close bracket or r is tautology.
2. Find that open bracket if p then q close bracket if and only if open bracket if q then p close bracket is tautology or not.
The answer of number one will be like this. If p is False, than the implication will be True because the antecedent is False. While if p is True so, the implication will be True too because the consequent is True for any truth-value q and r. so, it is showed that that if p then in bracket p or w close bracket or r is tautology.
Then, we will find the answer of number two. If p is True and q is False so the implication of p to q is False while if q then p is True, the biimplication above will be False. So, open bracket if p then q close bracket if and only if open bracket if q then p close bracket is not tautology.
The standard deviation will be posted in different title. Thanks.
Rabu, 15 April 2009
Solving Indefinite Integration Equation
Solving Indefinite Integration Equation
A. We will find y, from the known statement of dy over dx equal four times x square. The first way to get y is that we should multiply each sides with dx. So that, we get dx times dy over dx equal four times x square times dx. Moreover, it is became dy equal four times x square times dx. Then, to get y, we should integrate each sides. So that, indefinite integral of dy equal indefinite integral of four times x square times dx. In addition, the result is y equal four third times x cube plus constant. Therefore, the solving problem is y equal four third times x cube plus constant.
B. The next videos explain about adding problems. Here are some of the examples.
1. We will find x where x minus five equals three. The first step is adding each sides with a certain number that can cause x equal some number. Here, we can add five to each side. Then, the operation will be x minus five plus five equal three plus five. So that, we get x equal eight. The solution is eight.
2. We will find a where seven equals four times “a” minus one. First, we should add a number (one) to each side. The operation will be seven plus one equal four times a minus one plus one. Then, we get eight equal four times “a”. The next step is multiply each side so that we can get some number equal “a”. Here, we can multiply each other with a number which is the opposite of coefficient of “a”. Then we get eight times a quarter equal a quarter times four times a. In addition, the result is two equal a. So, the solution is two.
3. We will find x where two third times x equal eight. We should multiply each other with the opposite of two third which is half of three. (Why? The reason can be read in problem number two). Then the operation will be half of three times two third times x equal eight times half of three. It became x equal twelve. Therefore, the solution of this problem is twelve
4. We will find x where five minus two times x equal three times x plus one. The first step is eliminating one of the variables in one side. For example we may eliminate x in the right side, so that we have to subtract three times x in each sides. We get five minus five times x equal one. The second step is subtracting them with five. So that we get five minus five minus five times x equal one minus five. It gives negative five times x equal negative four. The third step is multiply them with negative one fifth. It leads negative one fifth times negative five times x equal negative one fifth times negative four. And the result is x equal four fifth. So, the solution is four fifth.
5. We will find m where three minus five times open bracket two m minus five close brackets equal negative two. The first step is open the brackets by multiply five with the number in brackets. So we get three minus ten times m plus twenty five equal negative two. Then the operation becomes negative ten times m plus twenty-eight equal negative two. Then the next step is adding each side with twenty-eight. It becomes negative ten times m equal negative thirty. Then the last step is dividing them by negative ten. Therefore, we get m equal three. So, the solution is three.
C. The next interesting video is about logarithm. We will proof one of the character of logarithm.
Logarithm base x of A plus logarithm base x of B equal logarithm base x of A times B. Then, suppose that logarithm base x of A is l and it is equivalent with x raise to the l equal A. Next, suppose that logarithm base x of B as m and we get x raise to the m equal B. Then we get logarithm base x of A over B equal n. It becomes x raise to n equal A over B which is equal x raise to l over x raise to m equal x raise to open bracket l minus m close bracket. While n is l minus m. So that is proved that logarithm base x of A over B is logarithm base x of A minus logarithm base x of B.
A. We will find y, from the known statement of dy over dx equal four times x square. The first way to get y is that we should multiply each sides with dx. So that, we get dx times dy over dx equal four times x square times dx. Moreover, it is became dy equal four times x square times dx. Then, to get y, we should integrate each sides. So that, indefinite integral of dy equal indefinite integral of four times x square times dx. In addition, the result is y equal four third times x cube plus constant. Therefore, the solving problem is y equal four third times x cube plus constant.
B. The next videos explain about adding problems. Here are some of the examples.
1. We will find x where x minus five equals three. The first step is adding each sides with a certain number that can cause x equal some number. Here, we can add five to each side. Then, the operation will be x minus five plus five equal three plus five. So that, we get x equal eight. The solution is eight.
2. We will find a where seven equals four times “a” minus one. First, we should add a number (one) to each side. The operation will be seven plus one equal four times a minus one plus one. Then, we get eight equal four times “a”. The next step is multiply each side so that we can get some number equal “a”. Here, we can multiply each other with a number which is the opposite of coefficient of “a”. Then we get eight times a quarter equal a quarter times four times a. In addition, the result is two equal a. So, the solution is two.
3. We will find x where two third times x equal eight. We should multiply each other with the opposite of two third which is half of three. (Why? The reason can be read in problem number two). Then the operation will be half of three times two third times x equal eight times half of three. It became x equal twelve. Therefore, the solution of this problem is twelve
4. We will find x where five minus two times x equal three times x plus one. The first step is eliminating one of the variables in one side. For example we may eliminate x in the right side, so that we have to subtract three times x in each sides. We get five minus five times x equal one. The second step is subtracting them with five. So that we get five minus five minus five times x equal one minus five. It gives negative five times x equal negative four. The third step is multiply them with negative one fifth. It leads negative one fifth times negative five times x equal negative one fifth times negative four. And the result is x equal four fifth. So, the solution is four fifth.
5. We will find m where three minus five times open bracket two m minus five close brackets equal negative two. The first step is open the brackets by multiply five with the number in brackets. So we get three minus ten times m plus twenty five equal negative two. Then the operation becomes negative ten times m plus twenty-eight equal negative two. Then the next step is adding each side with twenty-eight. It becomes negative ten times m equal negative thirty. Then the last step is dividing them by negative ten. Therefore, we get m equal three. So, the solution is three.
C. The next interesting video is about logarithm. We will proof one of the character of logarithm.
Logarithm base x of A plus logarithm base x of B equal logarithm base x of A times B. Then, suppose that logarithm base x of A is l and it is equivalent with x raise to the l equal A. Next, suppose that logarithm base x of B as m and we get x raise to the m equal B. Then we get logarithm base x of A over B equal n. It becomes x raise to n equal A over B which is equal x raise to l over x raise to m equal x raise to open bracket l minus m close bracket. While n is l minus m. So that is proved that logarithm base x of A over B is logarithm base x of A minus logarithm base x of B.
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