Geometric Proof of the Existence of Irrational Numbers

Everyone knows that Pythagoras’s name is due to the well-known theorem. This theorem is called the Pythagorean theorem in China, but many people don’t know the school named after him, let alone an important event that happened in this school. , This event is the discovery of irrational numbers. The discovery of irrational numbers triggered an unusually important controversy in mathematics. It was called the first mathematics crisis, and its discoverers were cruelly thrown into the sea because they destroyed the beliefs of the founders of the school. This article is going to introduce an example in a graphical way to illustrate the rationality of the existence of irrational numbers.

Due to the influence of philosophy at the time, the Pythagorean school believed that the world’s matter is composed of "atoms", and the atoms here are not atoms in our modern sense. Based on this view, they proposed an axiom., this Postulate assertion:

Any two straight line segments have a common degree

More specifically, if $a$ and $b$ is the length of any two straight line segments, there is a straight line whose length is $d$ such that

$a=md$

$b = nd$

One can always find a way to make m and n a positive integer, as shown in the following figure:

Note that the integer condition here is very important, and this condition should be used to infer the contradiction in the geometric proof later.

In other words, the Pythagoras school believes that: for any two straight line segments, there is a length, and the two line segments can be placed an integer number of times, so the ratio of the line segments can be expressed by the ratio of integers, but the irrational number The discoverer said no, there is an incommensurable quantity. In this case, the ratio of the length of the two line segments cannot be expressed as an integer ratio.

Examples of the existence of irrational numbers:

Look at the isosceles right-angled triangle below $\Delta ABC$ . Of course, we now know $AB=\sqrt{2}BC$ that because it $\sqrt{2}$ is an irrational number, it $\frac{AB}{BC}$ is an incommensurable quantity and cannot be expressed by rational fractions. But now, we assume that we don’t know this premise and the Pythagorean theorem. What to derive.

$a,b,c$ Respectively indicate the length of the angle corresponding to the opposite side, to prove that $\frac{AB}{BC}$ it is $\frac{c}{a}$ not commensurable, that is, to prove that it is not commensurable. Using the method of proof by contradiction, suppose there is a commensurable quantity $d$ , so that

$\\c=md \\ a=nd$

then

$\frac{AB}{BC}=\frac{md}{nd}=\frac{m}{n}$

If there is a commensurate quantity d, no matter whether it is an integer or a decimal, we can always find that m and n are positive integers. This is a very important condition, and we will use it to derive contradictions later.

We draw an auxiliary line, taking A as the center and AC as the radius to make a circle, crossing AB to point D, and then crossing point D to make a perpendicular to AB, crossing BC to E. It is easy to prove:

$BD = DE = EC$

$BD=AB-AD=c-a=md-nd=(m-n)d$

$BE=BC-EC=BC-BD=a-(m-n)d=nd-(m-n)d=(2n-m)d$

In other words, if d is the commensurability $\Delta ABC$ of the hypotenuse and the right-angle side of an isosceles right-angled triangle, then it is also the commensurability $\ Delta BDE$ of the hypotenuse and the right-angled side of an isosceles right-angled triangle .

We can continue to draw the same auxiliary line and continue to derive. Since m and n are positive integers, the side length of the calculated lower-level isosceles right triangle is always the sum and difference of the integers of d. For this reason, A conclusion is drawn, that is, if you continue to draw, the hypotenuse and the right-angle side of any child isosceles right-angled triangle have the same commensurable quantity. dd is the common degree of all these isosceles right-angled triangles.

Hey, this conclusion must be wrong. As the triangle gets smaller and smaller, there will inevitably be a situation where all three sides are smaller than d. How can it be possible that the commensurability is always d.

Therefore, it can only show that our premise was wrong from the beginning, that is, there is no such d, which makes

$\\c=md \\ a=nd$

We got two line segments that cannot be negotiated!

The process of proof is very natural, and no flaws can be found, but there is always something lacking. For example, in order to prove that the proposition is not universally valid, we used the contradiction method to construct the contradiction in a geometric way, thus proving the original proposition. But there are many kinds of right-angled triangles. For example, the right-angled triangles with right-angled sides 3, 4, and hypotenuse 5 are usually used to illustrate the establishment of the Pythagorean theorem $\frac{5}{3},\frac{5}{4}$ . The ratio of the hypotenuse to the right-angled side is . There is a fluency of 1. Then the question arises, whether the same proof process can be used on a triangle composed of 3, 4, and 5 sides. If the same proof process applies to the latter, it means that the same proof logic can prove both commensuration and If the proof is incommensurable, then the proof is problematic.

Of course, there is no problem with this proof. The problem is our logic for the latter. Now let’s look at the 3, 4, and 5 triangles. The above proof logic does not apply.

Similarly, for triangles with side length ratios of 3, 4, and 5, suppose the commensurate quantity d.

$\\c=5d \\ a=4d \\ b=3d$

In the same way, we draw auxiliary lines and auxiliary circles, and get the following conclusions:

$\Delta ABC$

$\frac{AB}{BC}=\frac{5d}{3d}=\frac{5}{3}$

$\Delta ADE$

$\frac{AE}{ED}=\frac{AC-EC}{ED}=\frac{AC-ED}{ED}=\frac{AC}{ED}-1$

According to similarity:

$\frac{ED}{CB}=\frac{AD}{AC}=\frac{AB-BD}{AC}=\frac{5d-3d}{4d}$

and so,

$ED=\frac{1}{2}\cdot CB=\frac{3d}{2}$

So, can you see it? Although we can still go on like an isosceles right-angled triangle, starting from the second right-angled triangle, whether the coefficient of d is an integer or not is this fundamental difference, which causes the same process to be pushed to the same conclusion, because In the case of an isosceles right-angled triangle, if the assumption is true, the waistline is always controlled at an integer multiple of d, so contradictions are inevitable.

But for the second kind of circumstance, there is no such restriction. The side length coefficient appears as a decimal, indicating that we can continue to proceed by extracting one more layer of d and multiplying the coefficient as an integer, hypotenuse and right angle The edge is still regulatable, but at this time the amount of commensurability has become $\ frac {d} {\ lambda}$ instead of the original d. This way, the contradiction is avoided, and the commensurability can be kept on drawing.

Therefore, we have eliminated our doubts, and it is feasible to prove the existence of irrational numbers by means of geometric graphics.

There is a major difference between irrational numbers and rational numbers: irrational numbers cannot have an expression composed of integers and the four arithmetic operations. A rational number has a finite expression, which is an integer fraction, but an irrational number does not. Irrational numbers and infinity are inextricably linked. When expressed in decimals, it is infinite and non-recurring. It requires an infinite number of digits and contains an infinite amount of information. The expression of continued fractions is an infinite continued fraction. , Expressed as a series, is the accumulation of an infinite number of series. It is fundamentally different from rational numbers.

Take Pi $\pi$ as an example. With the help of supercomputers, $\pi$ the top one trillion digits have now been calculated . Even so, we can’t even say "almost" calculated $\pi$ , because there are still "infinitely many" digits. We don't know yet.

Look at another perfect geometric proof:

If there is the smallest integer p, q (reduced).

$\frac{p}{q}=\sqrt{2}$

then

$p^2=2q^2$

We use a large square with side length p and two small squares with side length q as an example. Note that the two squares inside must overlap (if there is no overlap, it must $p^2=2q^2$ not be true), and the dark part of the overlap It is also a square.

According to elementary geometric knowledge, if $p^2=2q^2$ , we can get:

The area of the square in the dark part in the middle = the area of the small square in the lower left corner + the area of the small square in the upper right corner = 2 * the area of any one of the two small squares

If the side length of the middle square is s, the area of the small square is m.

then

$s^2=2m^2$ Established. And s=2q-p, m=pq are all integers.

Wait, didn't you just say that p and q are the smallest satisfiable $p^2=2q^2$ integers? Where did s, m come from? contradiction!

This proof is perfect!