On Thursday, April 2nd 2009, Mr. Marsigit commanded us to do some Mathematics exercises. There were 5 questions that we must do.
- Explain how to prove that the square root of 2 is irrational number!
- Explain how to show or to indicate that the sum angles of triangle are equals to 180o!
- Explain how you are able to get π (phi) !
- Explain how you’re able to find out the are of region bounded by the graph of y = x2 and y = x + 2 !
- Explain how you’re able to determine to the intersection point between the circle x2 + y2 = 20 and y = x + 1 !
The answers:
- 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.
- 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.
- 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.
- 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.
- 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.