Real Mathematics – Killer Numbers #2

In the previous article I was talking about the numbers which put an end to Hippasus’ life. These numbers are not only fatal; they are also incommensurable as well. On top of these, it is impossible to write these killer numbers as ratios of two other numbers.

I believe that there are more than enough reasons to choose a name such as “irrational” for these numbers. For me, it is astonishing to accept that there are some lengths which we can’t measure although they are just in front of us.

√2: One of the most famous irrational numbers.

Whether we realize it or not we can easily spot these lengths in everything that has square shape. Just divide a square diagonally into two equal parts and you will get two right-angled equilateral triangles.

img_4552

Assume that the square had side lengths 12. This gave a right-angled equilateral triangle with perpendicular sides with length 12. If we apply the Pythagorean Theorem:

This is an irrational number.

In case you’d like to measure this length, you will see a number that has infinite decimals: 16,97056…

I wonder what would happen if I call this number 17.

√2 is Finally Rational

If 12√2=17, we would get:

pisag5

We did it! √2 can be written as a ratio of two other numbers. It means √2 is rational. From now on we can write 17/12 wherever we see √2.

Although let’s stick to geometry a little bit more and see if we really got something or not.

Proof by Contradiction

First we divide the triangle as follows:

We can see that there are two identical right-angled triangles (A and B) that have perpendicular sides with length 5 and 12, and another right-angled triangle (C) that is equilateral.

Let’s analyze the triangle C from close. It has perpendicular sides with length 5 and a hypotenuse that has length 7. Using Pythagorean Theorem we can conclude:

img_45541

25 + 25 = 49.

50 = 49.

This is a contradiction.

√2 is not rational.

One Wonders…

Check and see what would happen if we used a square that has side lengths 10.

Real Mathematics – Killer Numbers#6

Socrates’ Lesson

In the previous articles I have talked about Plato and his effect on science; particularly geometry. Thanks to his book named Meno, we know about one of the most influential philosophers of all times: Socrates.

philosophydiscourse-cropped-425x259

Meno was another book of Plato that was written as dialogues. In this book there were two main characters: Meno and Socrates.

In the beginning of the book Meno asks Socrates if virtue is teachable or not. Even though Meno is crucial for understanding Socrates’ philosophy, there is one part of the book that interests me the most.

Problem

The book gets interesting when Socrates starts asking “the boy” who was raised near Meno. At first, Socrates is asking the boy to describe shape of a square and its properties. After a series of questions Socrates asks his main problem: How can one double the area of a given square?

This is an ancient problem that is also known as “doubling the square”. The boy answers Socrates’ questions and eventually finds the area of a square with side length of 2 units. The boy also concludes that since this area has 4 units, double of such square should have 8 units. But when asked to find one side of such square, the boy gives the answer of 4 units. However after his answer the boy realizes that a square with sides of 4 units has 16 units of area, not 8.

Classical Greek Mathematics

After this point the boy follows Socrates’ descriptions in order to draw a square that has 8 units of area. At first Socrates commands the boy to draw a square that has sides 2:

1kare

This square’s area is 4 units. Then Socrates tells him to draw three identical squares:

4kare

Now Socrates tells the boy to unite these squares as follows:

tekkare

Socrates asks the boy to draw the diagonals in each square. They both know the fact that a diagonal divides a square into two equal areas:

kosekare

It is easy to see that the inner square has a total area of 8 units:

kosekare2

One side of the inner square is the diagonal from small squares. In order to find that diagonal the boy uses Pythagorean Theorem:

karepis

Conclusion

Even though he only uses a compass and an unmarked ruler, the boy found a length that is irrational thanks to Socrates’ instructions. Back in ancient Greece numbers were imagined as lengths/magnitudes. This is why as long as they constructed it neither Socrates nor the boy cared about irrationality of a length.

Pythagoras and his cult claimed that all numbers are rational and they tried to hide the facts that irrational numbers exist. But in the end philosophers like Socrates won the debate and helped mathematics to flourish into many branches.

M. Serkan Kalaycıoğlu

Real Mathematics – Killer Numbers #5

Equilateral Triangle and Irrational Number

In the previous article I asked you to prove whether it is possible or not to draw and equilateral triangle on the system of lattice points. Here is one of the possible proofs for that.

Proof by Contradiction

Let’s assume that we have an equilateral triangle that has sides of length 2 units:

akuie2

Here, we will make a critical assumption: Corners of this triangle sits on the system of lattice points. Because of that the triangle must have an area that is rational. Why?

Because Pick’s theorem says that whenever we are inside the system of lattice points the numbers of points and area of the polygon are directly related with each other. And since we can’t count irrational number of points (eg. we can’t have √3 points, can we?!), area of the polygon must be rational too.

euake3

We already know that we can find a triangle’s area: ½(height x base). Then let me draw the height of the base and find its length using Pythagoras’ theorem:

h2 + 12 = 22

h2 = 4 – 1

h2 = 3

h = √3.

We just found the height of our equilateral triangle an irrational number. From here we will find the area

1/2(2*√3) = √3 units.

CONTRADICTION

This result is a contradiction. Despite what Pick’s theorem says (polygons inside the system of lattice points must have rational areas) this result shows an irrational number. Then we can conclude that this or any equilateral triangle can never be drawn on the system of lattice points.

One wonders…

Can you find another polygon which you can’t draw on the system of lattice points?

M. Serkan Kalaycıoğlu

Real Mathematics – Killer Numbers #4

Lattice Points

Imagine a page from a squared notebook. Take all the edges of little squares and leave the vertices. If you assume that horizontally and vertically distance between two consecutive points is exactly 1 unit. I name this plane as the system of lattice points.

In the system of lattice points every point is represented by a number duo:

latis3

Pick’s Theorem: An alternative method in order to find the area of a polygon.

In the system of lattice points one can construct as many polygons as wanted with joining points together. Pick’s theorem is handy for finding the areas of these polygons.

According to the theorem only two things are needed to find the area of a given polygon: Number of points polygon has on its edges (let’s call it e) and the number of points staying inside the polygon (let’s call it i). Pick’s theorem gives the following formula for the area of a polygon sitting on the system of lattice points:

Area = i + (e/2) – 1

Example 1: Triangle.

üçge

Assume that there is a triangle as shown above. Number of points on its edges e is equal to 4 as the number of points inside the triangle i is equal to 0. Hence Pick’s theorem says the area of this triangle is:

Area = 0 + (4/2) – 1

          = 0 + 2 – 1

          = 1 unit.

We know from basic geometry that the area of a triangle is the half of height times base: (2*1)/2 = 1 unit.

Example 2: Square.

karre

In the picture we can see that e=12 and i=4. Thus the area is:

4 + (12/2) – 1 = 4 + 6 – 1 = 9 units.

We know that area of a square is the square of the lenght of its edge which makes 3*3=9 units.

Square and triangle are easy examples and perhaps you are thinking that Pick’s theorem is redundant. Then let’s continue drawing a more complex polygon and find its area.

Example 3: Polygon.

3788_1

Now Pick’s theorem shows its strength. Without the theorem we’d have to divide this polygon into various polygons and then calculate areas one by one. But with Pick’s theorem it is just counting points:

e=12 and i=72. Hence the area of the polygon can be found with:

Area = 72 + (12/2) – 1 = 72 + 6 – 1 = 77 units.

Equilateral Triangle

Up to this point it looks like we are dealing with geometry but the headlines said that it is an article about numbers. Let me change the course of the article with a question.

Q: Is it possible to draw an equilateral triangle on our points system while the corners of the triangle sits on the lattice points?

For instance let me draw the base of an equilateral triangle that has 2 units of length:

ekui
Third corner (Q) sits between lattice points.

As seen above third corner of the triangle won’t be on a lattice point.

One wonders…

Can you prove that this is the case for each and every equilateral triangle on the system of lattice points?

To be continued…

M. Serkan Kalaycıoğlu