Proof of the irrationality of the root of 2 in Euclid

In the treatise Elements by the Greek mathematician Euclid there is evidence that the square root of 2 is irrational . This number-theoretic proof is based on contradiction ( reductio ad absurdum ) and is considered to be one of the first proofs of contradiction in the history of mathematics. Aristotle mentions it in his work Analytica priora as an example of this principle of proof. The evidence below is from Book X, Proposition 117 of the Elements . However, it is generally assumed that this is an interpolation , i.e. that the text passage does not come from Euclid himself. Because of this, the evidence is no longer included in modern editions of the Elements .

Irrational proportions were already known to the Pythagorean Archytas of Taranto , who demonstrably proved Euclid's theorem in a more general form. It used to be believed that the worldview of the Pythagoreans was called into question by the discovery of incommensurability , since they believed that the whole of reality must be expressible through integer numerical relationships. According to the current state of research, this is not the case. A geometric proof that diagonal and side squared or no common measure-hop can have regular pentagon was v already in the late 6th or early 5th century. Discovered by the Pythagorean Hippasus of Metapontium .

Evidence

claim

The square root of 2 is an irrational number.

proof

The proof is based on the method of contradiction proof, i.e. it is shown that the assumption that the root of 2 is a rational number leads to a contradiction (Latin: reductio ad absurdum ).

So it is assumed that the square root of 2 is rational and can therefore be represented as a fraction . It is also assumed that and are coprime integers, so the fraction is in abbreviated form: ${\displaystyle {\tfrac {p}{q}}}$ ${\displaystyle p}$ ${\displaystyle q}$ ${\displaystyle {\tfrac {p}{q}}}$

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

This means that the square of the fraction is 2: ${\displaystyle {\tfrac {p}{q}}}$

${\displaystyle \left({\frac {p}{q}}\right)^{2}=2}$,

or transformed:

${\displaystyle p^{2}=2q^{2}}$.

Since is an even number, is also even. It follows that the number is also even. ${\displaystyle 2q^{2}}$ ${\displaystyle p^{2}}$ ${\displaystyle p}$

So the number can be represented by: ${\displaystyle p}$

${\displaystyle p=2r}$, where is an integer. ${\displaystyle r}$

With the above equation we get:

${\displaystyle 2q^{2}=p^{2}=(2r)^{2}=4r^{2}}$

and from this after division by 2

${\displaystyle q^{2}=2r^{2}}$.

With the same reasoning as before, it follows that and therefore is also an even number. ${\displaystyle q^{2}}$ ${\displaystyle q}$

Since and are divisible by 2, we get a contradiction to coprime numbers. ${\displaystyle p}$ ${\displaystyle q}$

This contradiction shows that the assumption that the square root of 2 is a rational number is wrong and therefore the opposite must be true. This proves the claim that is irrational. ${\displaystyle {\sqrt {2}}}$

generalization

This idea of ​​proof can be extended to the general case of the -th root of any natural number that is not a -th power : ${\displaystyle k}$ ${\displaystyle n}$ ${\displaystyle k}$

If there is no -th power (cannot be represented as a natural number ), then is irrational. ${\displaystyle n}$ ${\displaystyle k}$ ${\displaystyle n=z^{k}}$ ${\displaystyle z}$ ${\displaystyle {\sqrt[{k}]{n}}}$

Proof: Instead of the simple even-odd argument, one generally uses the existence of a unique prime factorization for natural numbers. The proof is again provided by contradiction: Assume that it is true with natural numbers . It has to be shown that then there is a -th power, ie that is even a natural number. First of all, by simple transformation, that applies. Let be any prime number. In the prime factorization of or or kick exactly with the multiplicity or or on. Then follows immediately , so in any case because of . Since this is true for every prime number , it must indeed be a factor of , so is a natural number and is its -th power. ${\displaystyle {\sqrt[{k}]{n}}={\tfrac {a}{b}}}$ ${\displaystyle a,b}$ ${\displaystyle n}$ ${\displaystyle k}$ ${\displaystyle {\tfrac {a}{b}}}$ ${\displaystyle n\cdot b^{k}=a^{k}}$ ${\displaystyle p}$ ${\displaystyle n}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle p}$ ${\displaystyle e_{n}}$ ${\displaystyle e_{a}}$ ${\displaystyle e_{b}}$ ${\displaystyle e_{n}+k\cdot e_{b}=k\cdot e_{a}}$ ${\displaystyle e_{n}\geq 0}$ ${\displaystyle e_{b}\leq e_{a}}$ ${\displaystyle p}$ ${\displaystyle b}$ ${\displaystyle a}$ ${\displaystyle {\tfrac {a}{b}}}$ ${\displaystyle n}$ ${\displaystyle k}$

Simple conclusion from the irrationality theorem:

${\displaystyle {\sqrt[{n}]{n}}}$

is irrational for all natural numbers greater than 1 (because the non- th power of a natural number can be greater than 1). ${\displaystyle n}$ ${\displaystyle n}$

Web links

• Salomon Ofman: Mathematics in ancient greece from the 6th to 4th Century BCE from Pythagoras to Euclid. Bologna October 2013; accessed on December 7, 2017 (PDF, English).
• Hippasus goes hops. Proof of the irrationality of root 2 as a poem

Remarks

1. ^ Ideas in Mathematics: The Grammar of Numbers - Text: The irrationality of the square root of 2.
2. ↑ The assumption of a fundamental crisis of mathematics or the philosophy of mathematics triggered by the discovery among the Pythagoreans refutes Walter Burkert: Wisdom and Science. Studies on Pythagoras, Philolaos and Plato , Nuremberg 1962, pp. 431–440. The same conclusion come Leonid Zhmud : science, philosophy and religion in early Pythagoreanism , Berlin 1997, pp 170-175, David H. Fowler: The Mathematics of Plato's Academy , Oxford 1987, pp 302-308 and Hans-Joachim Waschkies: Beginnings of arithmetic in the ancient Orient and among the Greeks , Amsterdam 1989, p. 311 and note 23. The hypothesis of a crisis or even a fundamental crisis is unanimously rejected in today's specialist literature on ancient mathematics.
3. ↑ A whole number is called even or odd, depending on whether it is divisible by 2 or not. That means: An even number has the form and an odd number the form , whereby a natural number is 1, 2, 3,…. Since and is, the square of an integer is even if and only if is even. ${\displaystyle 2m}$ ${\displaystyle 2m+1}$ ${\displaystyle m}$ ${\displaystyle (2m)^{2}=2(2m^{2})}$ ${\displaystyle (2m+1)^{2}=4m^{2}+4m+1=2(2m^{2}+2m)+1}$ ${\displaystyle z}$ ${\displaystyle z}$