Exercising

On Thursday, April 2^nd 2009, Mr. Marsigit commanded us to do some Mathematics exercises. There were 5 questions that we must do.

1. Explain how to prove that the square root of 2 is irrational number!
2. Explain how to show or to indicate that the sum angles of triangle are equals to 180^o!
3. Explain how you are able to get π (phi) !
4. Explain how you’re able to find out the are of region bounded by the graph of y = x² and y = x + 2 !
5. Explain how you’re able to determine to the intersection point between the circle x² + y² = 20 and y = x + 1 !

The answers:

1. We’ll approve that the square root of 2 is an irrational number. First, we must know that an irrational number is a number that the value cannot be expressed exactly as a fraction a/b, where a and b are integers and relatively prime.

Let’s assume that the square root of 2 is a rational number, so we can state the square root of 2 equals a per b. If x is a positive integer, x² is even if and only if x is even.

Then, we can say that the square root of 2 is equals a per b. Multiply each side with b and we get new equation, a equals to b times the square root of 2. Multiply each sides by itself and we get that a square is equals to b square.

From this equation we ensure that a square is even because a square is twice of integer. Because of a square is even, so it’s sure that a is also an even. And then, we may write a equals to 2 times c (where c is an integer) and substitute this (a equals to 2 times c) in to previous equation. We get that 4 times c square equals to 2 times b square. We can divide each side with 2 and get equation 2 times c square equals to b square.

From this last equation, we get another clue that b square is also even. Why do I say that? Because as we know that b square is equals to twice square of integer, so, b is an even, of course.

Finally, it’s clear for us that a is even and b is even. As we know that the square root of 2 is equals to a per b (where a and b is even, both of them are not relatively prime) so the square root of 2 is irrational number.

2. We can draw a triangle to prove that the sum of angles of triangle is equals 180 degree.

First, let’s assume that the point angles is A, B and C (where AC is the base). From point angle C we can make a parallel line with line AB, so there is a new angle (we call it angle D) which it’s side as big as angle A, because angle A and angle D is same side ( I mean “sehadap” –sorry, I don’t know sehadap in English-). There is an angle between angle D and angle C (we call it angle E), whish has same size with angle B because angle E and angle B is angled in the opposed. Then, it shows that angle C, angle D and angle E form a straight line which is the size is 180 degree. It’s mean that the sum of angle C plus angle D (it substituted by angle A) plus angle E (it substitude by angle B) equals to 180 degree. In the other words, the sum of angle A plus angle B plus angle C equals to 180 degree. It’s enough to prove that the sum angles of triangle is equals to 180 degree.

3. As we know that π’s (phi) value is 3.14. After I read some references, now I’ll tell the history of phi. The value of phi (π = 3.14 or 22/7) found by Egyptian people in 1650 BC. We can get the value of phi by make a circle; calculate the perimeter and the diameter. If we make any circle, we’ll get that the value of the comparison between the perimeter and the diameter is always constant: 3.14. So, if we divide the circle perimeter with its diameter it is equal to 3.14 (phi). That is why the circle perimeter formula is equal to diameter time phi.
4. First step is we must draw the graph of y equals x square. We try to substitute x with a number, for example x equals 1, so we’ll get the value of y equals one or we can write it down in a table of x and y (the substitution and the result then). We can substitute x or y with any numbers. So do the equation y equals to x plus 2. So, we are able to draw the graph and get the intercept point between those two graphs. Mathematically, the interception point can be find by assume that y (in the first equation) equals to y (in the second equation. We get a new equation, x square equals to x plus 2. By reducing each side with x plus 2, we’ll get that x square minus x minus 2 equals to zero. Then we get the values of x, x is equals negative1 or x equals 2.

The area of the region bounded by the graph of y equals to x square and y equals to x plus 2 is equals to integral of x plus 2 minus x square dx from x equals negative 1 to x equals 2 (we call the region with A). After we calculate, we get the result that A equals to four half area unit.

5. To determine the intersection point between y equals x plus 1 and x square plus y square equals twenty is, first, we change the equation form x square plus y square equals twenty into form y equals square root of twenty minus x square in bracket. Assume that y equal to y, so x plus one is equal to square root of twenty minus x square in bracket. Then, square each side and we’ll get that two x square plus two x minus nineteen equals zero. We can use abc pormula to find the values of first x and second x, and we get the first x equals minus one plus square root of thirty nine per two and second x is equal to minus one minus square root of thirty nine per two: both of them are the intercept points.