The aim of the proof is to show that Pythagoras’ theorem is true for all right-angled triangles. The triangle shown above could be any right-angled triangle because its lengths are unspecified, and represented by the letters x, y and z.
Also above, four identical right-angled triangles are combined with one tilted square to build a large square. It is the area of this large square which is the key to the proof.
The area of the large square can be calculated in two ways.
Method 1: Measure the area of the large square as a whole. The length of each side is x + y. Therefore, the area of the large square = (x + y)2.
Method 2: Measure the area of each element of the large square. The area of each triangle is 1⁄2xy, i.e.1⁄2 × base × height. The area of the tilted square is z2. Therefore,
area of large square = 4 × (area of each triangle) + area of tilted square
Methods 1 and 2 give two different expressions. However, these two expressions must be equivalent because they represent the same area. Therefore,
The brackets can be expanded and simplified. Therefore,
The 2xy can be cancelled from both sides. So we have
which is Pythagoras’ theorem!
The argument is based on the fact that the area of the large square must be the same no matter what method is used to calculate it. We then logically derive two expressions for the same area, make them equivalent, and eventually the inevitable conclusion is that x2 + y2 = z2, i.e. the square on the hypotenuse, z2, is equal to the sum of the squares on the other two sides, x2 + y2.
This argument holds true for all right-angled triangles. The sides of the triangle in our argument are represented by x, y and z, and can therefore represent the sides of any right-angled triangle.
Euclid’s aim was to prove that √2 could not be written as a fraction. Because he was using proof by contradiction, the first step was to assume that the opposite was true, that is to say, that √2 could be written as some unknown fraction. This hypothetical fraction is represented by p⁄q, where p and q are two whole numbers.
Before embarking on the proof itself, all that is required is a basic understanding of some properties of fractions and even numbers.
(1) If you take any number and multiply it by 2, then the new number must be even. This is virtually the definition of an even number.
(2) If you know that the square of a number is even, then the number itself must also be even.
(3) Finally, fractions can be simplified: 16⁄24 is the same as 8⁄12; just divide the top and bottom of 16⁄24 by the common factor 2. Furthermore, 8⁄12 is the same as 4⁄6, and in turn 4⁄6 is the same as 2⁄3. However, 2⁄3, cannot be simplified any further because 2 and 3 have no common factors. It is impossible to keep on simplifying a fraction forever.
Now, remember that Euclid believes that √2 cannot be written as a fraction. However, because he adopts the method of proof by contradiction, he works on the assumption that the fraction p⁄q does exist and then he explores the consequences of its existence:
If we square both sides, then
This equation can easily be rearranged to give
Now from point (1) we know that p2 must be even. Furthermore, from point (2) we know p itself must also be even. But if p is even, then it can be written as 2m, where m is some other whole number. This follows from point (1). Plug this back into the equation and we get
Divide both sides by 2, and we get
But by the same arguments we used before, we know that q2 must be even, and so q itself must also be even. If this is the case, then q can be written as 2n, where n is some other whole number. If we go back to the beginning, then
The 2m⁄2n can be simplified by dividing top and bottom by 2, and we get
We now have a fraction m⁄n, which is simpler than p⁄q.
However, we now find ourselves in a position whereby we can repeat exactly the same process on m⁄n, and at the end of it we will generate an even simpler fraction, say g⁄h. This fraction can then be put through the mill again, and the new fraction, say e⁄f, will be simpler still. We can put this through the mill again, and repeat the process over and over again, with no end. But we know from point (3) that fractions cannot be simplified forever. There must always be a simplest fraction, but our original hypothetical fraction p⁄q does not seem to obey this rule. Therefore, we can justifiably say that we have reached a contradiction. If √2 could be written as a fraction the consequence would be absurd, and so it is true to say that √2 cannot be written as a fraction. Therefore √2 is an irrational number.
Let us call the length of Diophantus’ life L. From the riddle we have a complete account of Diophantus’ life which is as follows:
1⁄6 of his life, L⁄6, was spent as a boy,
L⁄12 was spent as a youth,
L⁄7 was then spent prior to marriage,
5 years later a son was born,
L⁄2 was the life span of the son,
4 years were spent in grief before he died.
The length of Diophantus’ life is the sum of the above:
We can then simplify the equation as follows:
Diophantus died at the age of 84 years.
In order to weigh any whole number of kilograms from 1 to 40 most people will suggest that six weights are required: 1, 2, 4, 8, 16, 32 kg. In this way, all the weights can easily be achieved by placing the following combinations in one pan:
However, by placing weights in both pans, such that weights are also allowed to sit alongside the object being weighed, Bachet could complete the task with only four weights: 1, 3, 9, 27 kg. A weight placed in the same pan as the object being weighed effectively assumes a negative value. Thus, the weights can be achieved as follows:
A Pythagorean triple is a set of three whole numbers, such that one number squared added to another number squared equals the third number squared. Euclid could prove that there are an infinite number of such Pythagorean triples.
Euclid’s proof begins with the observation that the difference between successive square numbers is always an odd number:
Every single one of the infinity of odd numbers can be added to a particular square number to make another square number. A fraction of these odd numbers are themselves square, but a fraction of infinity is also infinite.
Therefore there are also an infinity of odd square numbers which can be added to one square to make another square number. In other words there must be an infinite number of Pythagorean triples.
The dot conjecture states that it is impossible to draw a dot diagram such that every line has at least three dots on it.
Although this proof requires a minimal amount of mathematics, it does rely on some geometrical gymnastics, and so I would recommend careful contemplation of each step.
First consider an arbitrary pattern of dots and the lines which connect every dot to every other one. Then, for each dot, work out its distance to the closest line, excluding any lines which go through it. Thereby identify which of all the dots is closest to a line.
Below is a close-up of such a dot D which is closest to a line L. The distance between the dot and the line is shown as a dashed line and this distance is smaller than any other distance between any other line and a dot.
It is now possible to show that line L will always have only two dots on it and that therefore the conjecture is true, i.e. it is impossible to draw a diagram such that every line has three dots on it.
To show that line L must have two dots, we consider what would happen if it had a third dot. If the third dot, DA, existed outside the two dots originally shown, then the distance shown as a dotted line would be shorter than the dashed line which was supposed to be the shortest distance between a dot and a line. Therefore dot DA cannot exist.
Similarly, if the third dot, DB, exists between the two dots originally shown, then once again the distance shown as a dotted line would be shorter than the dashed line which was supposed to be the shortest distance between a dot and a line. Therefore dot DB cannot exist either.
In summary, any configuration of dots must have a minimum distance between some dot and some line, and the line in question must have only two dots. Therefore for every configuration there will always be at least this one line with only two dots – the conjecture is true.
The following is a classic demonstration of how easy it is to start off with a very simple statement and then within a few apparently straightforward and logical steps show that 2 = 1.
First, let us begin with the innocuous statement
Then multiply both sides by a, giving
Then add a2 – 2ab to both sides:
This can be simplified to
Finally, divide both sides by a2 – ab, and we get
The original statement appears to be, and is, completely harmless, but somewhere in the step-by-step manipulation of the equation there was a subtle but disastrous error which leads to the contradiction in the final statement.
In fact, the fatal mistake appears in the last step in which both sides are divided by a2 – ab. We know from the original statement that a = b, and therefore dividing by a2 – ab is equivalent to dividing by zero.
Dividing anything by zero is a risky step because zero will go into any finite quantity an infinite number of times. By creating infinity on both sides we have effectively torn apart both halves of the equation and allowed a contradiction to creep into the argument.
This subtle error is typical of the sort of blunder which caught out many of the entrants for the Wolfskehl Prize.
The following axioms are all that are required as the foundation for the elaborate structure of arithmetic:
1. For any numbers m, n
2. For any numbers m, n, k,
3. For any numbers m, n, k
4. There is a number 0 which has the property that, for any number n,
5. There is a number 1 which has the property that, for any number n,
6. For every number n, there is another number k such that
7. For any numbers m, n, k,
From these axioms other rules can be proved. For example, by rigorously applying the axioms and assuming nothing else, we can rigorously prove the apparently obvious rule that
To begin with we state that
Then by Axiom 6, let l be a number such that, k + l = 0, so
Then, by Axiom 2,
Bearing in mind that k + l = 0, we know that
By applying Axiom 4, we can at last declare what we set out to prove:
Let us examine Mr Black’s options. First, Mr Black could aim at Mr Grey. If he is successful then the next shot will be taken by Mr White. Mr White has only one opponent left, Mr Black, and as Mr White is a perfect shot then Mr Black is a dead man.
A better option is for Mr Black to aim at Mr White. If he is successful then the next shot will be taken by Mr Grey. Mr Grey hits his target only two times out of three and so there is a chance that Mr Black will survive to fire back at Mr Grey and possibly win the truel.
It appears that the second option is the strategy which Mr Black should adopt. However, there is a third and even better option. Mr Black could aim into the air. Mr Grey has the next shot and he will aim at Mr White, because he is the more dangerous opponent. If Mr White survives then he will aim at Mr Grey because he is the more dangerous opponent. By aiming into the air, Mr Black is allowing Mr Grey to eliminate Mr White or vice versa.
This is Mr Black’s best strategy. Eventually Mr Grey or Mr White will die and then Mr Black will aim at whoever survives. Mr Black has manipulated the situation so that, instead of having the first shot in a truel, he has first shot in a duel.
Mathematicians find it useful to have neat formulae which give the sum of various lists of numbers. In this case the challenge is to find a formula which gives the sum of the first n counting numbers.
For example, the sum of just the first number is 1, the sum of the first two numbers is 3 (i.e. 1 + 2), the sum of the first three numbers is 6 (i.e. 1 + 2 + 3), the sum of the first four numbers is 10 (i.e. 1 + 2 + 3 + 4), and so on.
A candidate formula which seems to describe this pattern is:
In other words if we want to find the sum of the first n numbers, then we simply enter that number into the formula above and work out the answer.
Proof by induction can prove that this formula works for every number up to infinity.
The first step is to show that the formula works for the first case, n = 1. This is fairly straightforward, because we know that the sum of just the first number is 1, and if we enter n = 1 into the candidate formula we get the correct result:
The first domino has been toppled.
The next step in proof by induction is to show that if the formula is true for any value n, then it must also be true for n + 1. If
then,
After rearranging and regrouping the terms on the right, we get
What is important to note here is that the form of this new equation is exactly the same as the original equation except that every appearance of n has been replaced by (n + 1).
In other words, if the formula is true for n, then it must also be true for n + 1. If one domino falls, it will always knock over the next one. The proof by induction is complete.