Numbers #12

Subitizing: The ability to recognize(or guess) the number of a small group of objects without counting.

The name subitizing comes from the Latin word “subitus” which means “sudden”.

Subitizing can be seen in many every-day activities. One of them is a six-pack soda. No matter how they are lined up, we know that the number of soda bottles is 6. We inherit this knowledge without counting the bottles. And if we decide to drink one of them, we automatically know(without having to count them) that the number of soda bottles left is 5.

You don’t have to count the dots on the surfaces. You just know that it is 5.

Another example of subitizing can be given from the game backgammon. Assume that two dices are rolled and you identify them as 2 and 5. The process of identifying the dices can be measured in milliseconds. This can be even shortened as you spend more time playing the game. In short; subitizing is a skill that can be developed if one spends time and work on it.

Research studies showed that 6-month olds can differentiate, visually (a top bounces 3 times) and from sounds (clapping hands 3 times), between 1, 2, and even 3. In other words; humans start developing the number concept when they are just infants.

Kebab Truck & Subitizing
Subitizing is hidden behind the number of customer groups in the game of Kebab Truck. As the game is played, scores become higher and higher. The reason behind this is that players’ subitizing skills are improving.

Let’s check this scene from Kebab Truck:

In the beginning, you will be making certain moves during the game. Nevertheless, in time, your moves will differ substantially. The biggest reason behind this is that your subitizing skills were improved while you were playing the game.

Kebab Truck also helps the players to develop their basic arithmetic skills. These improvements are not limited to adding and subtracting the number of customers. Once you understand how the scoring system formulated, you will realize that (to maximize your scoring) multiplication is an important part of this game as well.

Real Mathematics – Numbers #11

A very long time ago in Mesopotamia, a few hundreds of people lived together in the village of Badaks. Badaks were very hard-working people, and they were among the first farmer communities. Their lives depended on two things more than anything: Their farms and sheep.

Monday syndrome in Badaks village…

There was a lot of sheep in the village of Badaks. Thanks to them, people of Badaks were able to protect themselves from cold weather. Their milk and meat were also important to Badaks as food sources. Because of their importance, the person in charge of the sheep had to be wise and trustworthy.

Zaylin a.k.a. the protector of sheep!

Zaylin, head of Badaks, was in charge of this crucial duty.

Every day, with the first sunlight, Zaylin took the sheep out of their pens for them to explore the hills and graze the green grass of the village of Badaks. Before the sun is gone, Zaylin had to gather the sheep and be sure that every sheep returned to the pens.

Omg! Where are the rest of you?!

Even though Badaks were one of the most progressive communities of their time, they didn’t know the use of numbers like the rest of humanity.

At this point, Zaylin had a bit of a problem: How did he know that he returned with the same number of sheep as left in the morning? Don’t get me wrong; Zaylin was an intelligent person for his time. But like everybody else, he didn’t know how to count.

One wonders…

Put yourself in Zaylin’s shoes: Is it possible to detect if you lost any sheep or not when you finish a day without counting or any use of numbers?

M. Serkan Kalaycıoğlu

Real Mathematics – Numbers #10

A game for kids who would like to get better at arithmetic operations, decimal system and numbers in overall:

European Championship

  • Only materials needed for this game are a twelve-faced dice, pen/pencil and a piece of paper.
    IMG_6524
  • Game consists of encounters between two players.
  • In each encounter players roll the dice four times in order.
  • Outcome of a rolled twelve-faced dice is like the following:
    20190129_135510
  • For every player only ambition is to write the biggest possible four-digit number.
  • Difference of players’ four-digit numbers decides the winner.

Scoring of the game

Player with the bigger number would get:

  • 4 points if the difference is a four-digit number.
  • 3 points if the difference is a three-digit number.
  • 2 points if the difference is a two-digit number.
  • 1 point if the difference is a single-digit number.

If the difference is zero; meaning that the numbers are equal to one another, then both players get no points.

Every encounter finishes when one of the players gets to 7 points.

League

In case there is enough number of students, it is possible to construct a league version of the game that finishes after playoffs. For instance if there were 20 students we could divide them into 4 groups with 5 teams. In each group every player would play 4 games and after the group stage group leaders would go onto the playoffs where the champion can be decided after semi-final and final games.

World Cup

In this version of the game players would roll the dice three times and write the biggest possible three-digit number. Although this time winner gets to be decided like following:

  • If the difference is odd, biggest number wins.
  • If the difference is even, smallest number wins.
  • Winner gets 3 points as loser gets nothing. Differences are kept as averages.
  • If numbers are the same, players get 1 point each.

M. Serkan Kalaycıoğlu

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.

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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 #1

Hippasus: First Victim of the Science Mob

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Pythagoras is a very well known historic figure. Even though most of the people know him through the geometry theorem attributed to him, he had accomplished more than just a theorem. He was also the head of the first known science mob in the history.

Pythagorean Theorem: In a right-angled triangle square of the perpendicular sides add up to the square of the third side of the triangle that is also known as the hypotenuse.

Pythagoras was born in the island of Samos. He had an enormous reputation as a mathematician throughout the ancient Greece. His followers (Pythagoreans) chose to live as their leader. They were a tight and closed group that ate neither meat nor beans and isolated themselves from having any kind of possession.

According to Pythagoras universe was built on the numbers. Every number had a character and everything that is happening around us could be explained with numbers. He believed that numbers have categories such as beautiful, ugly, masculine, feminine, perfect and such. For instance 10 was the best number because it contained the summation of the first four numbers: 1+2+3+4=10.

Pythagoreans also believe that every number is rational: Meaning that each number can be represented as a division of two other numbers. (E.g. 10/2 = 5)

Oath Breaker

One day one of Pythagoras’ followers broke his oath and asked the forbidden question: What is the length of the hypotenuse of an equilateral right-angled triangle?

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Geogebra shows that the hypotenuse is around 1,41 units. This is not the exact value of the length as this length can never be measured.

Hippasus was a devoted Pythagorean. One day he sailed away with his brothers. When he was in the open sea, he started thinking about the problem of the right-angled equilateral triangle. In the end he claimed that he found irrational numbers. This was an oath breaker as it was forbidden to question Pythagoras’ words. Hippasus never came back from that trip, and Pythagoreans continued to keep the existence of the irrational numbers as secret.

Incommensurables: Do they exist?

According to the Pythagorean Theorem: Length of hypotenuse on a right-angled equilateral triangle.

kc3b6k2

If Hippasus was wrong, √2 was a rational number which means √2 can be written as the division of two other numbers. Let’s say that this is true and a/b is equal to √2.

Ps: a and b are relatively prime. This means that a/b can’t be simplified; they are the smallest numbers for that ratio.

kc3b6k21

Let’s square both sides so that we are free from the square root.

kc3b6k22

Now send the denominator to the left side of the equality.

kc3b6k23

This actually means that two squares that have side b add up to another square that has side a.

kc3b6k24

Hence, we just need to show that when we add two identical squares, we can get another square.

karekc3b6k1

Since the little squares add up to the large square, let’s try to put them inside the large one.

karekc3b6k2

As seen above, little squares intersect in the middle and leave gaps on the corners. If we stick to our initial assertion, this intersection must have same area as the gaps. But there is something absurd here, because this intersection is a square. Also the gaps are identical squares that add up to the intersection.

If I call sides of the little squares d, and the big square c:

kc3b6k25

This result is the same as our starting point. We just found ourselves in a loop which means that our initial assertion was wrong. √2 can’t be shows as a ratio of a/b. Hence, √2 is not a rational number.

One Wonders…

  1. Try to prove that √2 is an irrational number, using Euclid’s tools which are compass and an unmarked ruler.
  2. How can we understand if √3 is rational or not? (Hint: Try to prove geometrically like I did in the article.)

M. Serkan Kalaycıoğlu

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.

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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 – Numbers #5

The word fraction has a Latin root “fractio” which means “to break”. Let’s take a slightly different approach for fractions. Instead of breaking, we will use “folding”.

Fractions with Papers

In order to understand the four arithmetical operations in fractions, we can use a standard A4 paper. First we will fold the paper into two halves. Every half represents the fraction ½. We can continue folding one of the pieces in two halves. In the end we will have the following papers which we can use for four mathematical operations:

img_4511
a. 1

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b. 1/2+1/2

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c. 1/2+1/4+1/4

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d. 1/2+1/4+1/8+1/8

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e. 1/2+1/4+1/8+1/16+1/16

Addition

We are aware of the fact that all five of the versions are equal to one another and they each add up to 1. If we take a look at versions b and c we can conclude that ½ is equal to ¼ + ¼. If we substitute 1/2s in the version b we can find the following result:

1 = ½ + ½ = ¼ + ¼ + ¼ + ¼.

Subtraction

Let’s continue from the previous. If we subtract ¼ from both sides of ½ = ¼ + ¼, then we get:

½ – ¼ = ¼ + ¼ – ¼

½ – ¼ = ¼

Multiplication

Let’s start with another A4 paper and take its whole as 1. This time we will fold the paper and use the marks on it.

img_4511

Assume that we are trying to find ½ * ¾.

Check the first fraction and fold the paper into two halves since it is ½.

img_4514-e1535716599583

Then check the second fraction and fold the paper into four equal parts.

img_4515

Since second fraction is ¾, mark 3 parts of the paper.

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Now unfold the A4 paper completely. It shows that there are 8 equal parts and 3 of them are marked. Hence solution is 3/8.

img_4520

One wonders…

  1. Using the paper method show the connection(s) between 1/8 and 1/32.
  2. Find 1/8 – 1/32 with the paper method.

Game

Imagine a gambling game in which you don’t have to gamble your money. In the worst case scenario, you will win nothing. A dream for gamblers, isn’t it?

Let’s say you get 128.000 dollars and 6 red and black cards (3 for each). Here is how the game goes: In every hand you must bet half of your total money. When you pick a red card, you will get twice what you played. When you pick a black card, you will lose all the money you bet in that hand.

When the game is finished (after all 6 cards are played) if you end up with more than 128.000 dollars you will get that surplus.

What is your strategy to win? Explain your answer.

M. Serkan Kalaycıoğlu

Real Mathematics – Numbers #4

In schools we start learning mathematics with learning what numbers are. Unfortunately numbers are taken for granted and being overlooked just because it starts in the elementary school. The truth is this part of mathematics is a joint work of countless civilizations that lasted thousands of years. Although, categorizing and defining all those information were done only in the near past. This means that things we learn in the first few years of school have so much more depth than we think they have.

Especially fractions (or rational numbers) weren’t used in Europe in the sense we understand them today until the 17th century. In fact for a long time people thought of fractions not as numbers but as two numbers being divided to one another.

Rhind

Ancient Egyptians were one of the first known civilizations that used fractions. They created one of the most important and oldest documents in the history of civilizations using papyrus trees. Around 4000 years ago they started writing valuable information on papyrus leafs. Rhind papyrus is one of those documents. It is believed to be written around 1800 BC. Thanks to Rhind, we can understand how ancient Egyptians used fractions.

440px-rhind_mathematical_papyr

It is uncanny how commonly they used fractions in Rhind. Although they were obsessed with unit fractions as they found ways to describe every fraction with them.

Unit Fraction: Fractions that have 1 on their numerators.

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In the ancient Egypt they used a shaped that looks like an open mouth (or an eye). This shape was the notation for the unit fraction. Denominator of the fraction would be placed under the mouth.

Table of 2/n

Inside Rhind there is a method for describing fractions in the form of 2/n (when n is odd) with two unit fractions. Table starts with 2/3 and ends with 2/101. In the papyrus it says that 2/3 is equal to ½ + 1/6. For the rest of the papyrus a formula was given in order to describe fractions in the form of 2/3k: It is 1/2k + 1/6k.

Let’s try it for 2/9. 9 is 3k, hence k=3. This gives 2/9 = 1/6 + 1/18. Ingenious, isn’t it?

Next number on the table is 2/5 = 1/3 + 1/15. This is also a general formula just like the previous one. Any fraction in the form of 2/5k can be shown as 1/3k + 1/15k.

Fraction Line

Besides ancient Egyptians, Babylonians used fractions too. But their choice of symbols was so confusing, it is impossible to understand which number is written. You could only check the rest of the calculation (if there is any) and guess which number is being used.

babilke121

In the Babylon civilization number system has base 60. The one on the left is 12, other one is 15. But they can also mean 12+(15/60) too. Lack of symbol for fractions caused a lot of problems in this civilization.

Around 1500 years ago Indian mathematicians were shining. They found the number system we use today and even the number zero was invented (or discovered). Their brilliance was key for fractions too as they showed fractions one under the other. Muslims were the ones who thought of putting the fraction line between numbers.

In the end we owe our modern notation to Indian and Muslim mathematicians.

hintk

The way 7/15 was shown in the old Indian symbols.

One wonders…

Try to find answers to following questions about the Rhind:

  1. Why did they only consider odd numbers in the denominator of the fractions?
  2. Find 2/7 and 2/11 using their methods.
  3. What happens after 11?
  4. Try to come up with a general formula for 3, 5, 7 and 11.

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.

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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