Using geometry to solve algebraic problems

In my post The hunt for the roots, I mentioned that in the ancient world geometry was often used to solve algebraic problems. This method of solving problems was especially popular in Greece mathematics. In this post I will show some extremely simple examples of how problems were solved, and from this I will also show the reasons behind such geometrical approach.

Before we go to the math, a little clarification is necessary. It is not very easy to write posts about history of mathematics because the symbols we usually use now were totally unfamiliar in the ancient world. In this post I decided to use modern notation, with its translation to the ancient formulation of the problem when needed. Therefore, don't focus your attention on the equations - they are of little importance. In this post the main part is the methods that were used, and not the problems themselves. As the first example, lets solve the following problem:

5x+4=14

For us the solution is obvious - we just need to move 4 to the second side and divide. Geometrically this way of solution cannot be done, because we don't even know what "4" is. We need firstly to translate the problem into a geometrical problem (and since we are talking about ancient geometry, we cannot use symbols because they were invented only recently). The problem becomes:

How many line segments of length five do you need to add to a line segment of length four to get a segment of length fourteen?

In this form, the solution is trivial. All you need to do is to draw a line with length four, and then to start adding lines with length five until you get fourteen.

The answer in both cases is two, but I am sure the geometrical solution seems much less straightforward. It is also not clear what to do if the answer is not a natural number. If the answer is rational, for example in the equation 5x+4=8, we can still use the same method to get the solution. All we need is to find a segment small enough - instead of using segment with length five, we will use a segment with length 1. We will get than that we need 4 such segments, which corresponds to 4/5 of the original segment, so the answer will be 4/5.

At this point you probably noticed a very important problem in this method - it is no good for irrational numbers. If for example I want to solve the equation:

5x=

It would turn out to be impossible. The reason for this is a bit difficult to explain. To understand it, lets firstly remember that there is no solution for this equation in rational numbers. The only number that solves that equation is irrational. We don't care about this, but for the Greeks it was a major problem. To see this lets try to solve it using the method I used to solve the previous example. For this we need to divide a segment of length five into smaller segments, in such a way that by adding them we will get . So we need to find a natural number m such that there exists another natural number n with the property that . However, this is not possible. If it were we would get that is rational by definition - which we know is not true. It is also worth noting that since the concept of a negative number was unknown to the ancient Greeks, equations of the form 5x=-7 were not solved nor appeared in any way.

Before going on, there is one thing which I intentionally left hanging in the air - what is the length of the line segment? I freely used phrases such as "a line segment with length five", but does this phrase has any meaning? For any one who studied analytical geometry this should be obvious, but the Greeks didn't have analytical geometry. However, they had a philosophical explanation - They even claimed at some periods that "all is a number"... Because of this, the connection between numbers and lines was very clear to them. But what was a number from their point of view?
The number system of the Greeks was not very complicated, and consisted of two groups of numbers: The natural numbers and the positive rational numbers. This may sound not enough for us, after all we are used to using negative numbers, irrational numbers and complex numbers. However, for almost any purpose this system was more than enough. After all you don't need an irrational number of horses for example. It is also impossible to divide any material object into irrational number of parts.

As I already said, Greeks saw the connection between lines and numbers. They thought that for any positive number there was a line that represented it. But is the opposite true? Is it true that every line has a number that represents it? For a long time it was believed to be true. However, the discovery of the irrational numbers put an end to this. Since all the positive rational numbers are of the form n/m where n and m are natural numbers, it follows that if every single line is represented by a single number, there exists a number that is not rational - . Neither they not are are sating that the existence of such a line is obvious. It must be proved to exist. But this is very simple. All we need to do is to construct a right triangle with the sides having length one. The length of the third side is then found according to the Pythagoras theorem to be . The Greeks knew how to show that it is irrational, but they used a different word for this. They proved that there is no common measure to the sides of such a triangle. It doesn't matter how small the segments you use are - even if you will use a segment whose length is a nanometer, still you will find out that one side you can fully cover with such segment, but not the other. In modern notation this is written as: There are no n,m natural numbers such that n/m=.

This cause a major problem - they had a line with no number. A magnitude without representation. It took some time, but the Greeks improved their system of numbers to include irrational numbers. However, this discovery also caused them to focus more on geometry than on algebra, because they now new that there were numbers which are impossible to write. And this caused them to prove all algebraic identities using geometry, because in this form the proof was believed to be correct for all numbers.

I intended to write in this post about second degree equations, but it seems to me that it would be better to stop now and to write about the rest in another post. This one is already large enough, and second degree equations certainly deserve their own post. The last thing I want to do in this post is to write the proof that is irrational. The proof is the same one that was used by the Greeks, although it is probably not how the irrationality of was discovered.

To proof that is irrational, lets assume that it is and get to a contradiction. If it is indeed rational than there are two natural numbers n,m such that:

Also we want this numbers to be least representation, that is there are no common divisors to n and m. Lets remove the root, and multiply by m^2:

The right side is even therefore so is the left side. If n^2 is even n also must be even (check it yourself) so we can write n=2k. The equation becomes:

The left side is even therefore so is the right side. Lets write m=2t. We now got that:

But this means that the representation we choose is not the least - there is a common divisor to n and m and this divisor is two. This is a contradiction, so we are done - the square root of two is irrational.

Read the next part of this series: Geometry and second degree polynomials.