Geometry for Elementary School/A proof of irrationality

In mathematics, a rational number is a real number that can be written as the ratio of two integers, i.e., it is of the form

a/b where a and b are integers and b is not zero. An irrational number is a number that cannot be written as a ratio of two integers, i.e., it is not of the form

a/b .

History of the theory of irrational numbers

The discovery of irrational numbers is usually attributed to Pythagoras, more specifically to the Pythagorean Hippasus of Metapontum, who produced a proof of the irrationality of the ${\sqrt {2}}$. The story goes that Hippasus discovered irrational numbers when trying to represent the square root of 2 as a fraction (proof below). However, Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so he sentenced Hippasus to death by drowning. As you see, mathematics might be dangerous.

Irrationality of the square root of 2

In this section we attempt to explain why ${\sqrt {2}}$ is irrational. Irrational is a fancy word for a number that cannot be written as a fraction where the top and bottom of the fraction are whole numbers. At one point in time it was once believed that every number could be written as a fraction.

Before we start the proof we first should recall a few familiar facts. First, is that when we write fractions we can always reduce them so they have no common factors. Just to remind ourselves how this goes, let's think about the fraction ${\frac {15}{21}}$. Since 3 divides into 15 evenly, and 3 divides into 21 evenly, then we can make this fraction have a smaller numerator and denominator. In this example see this by the following calculation:

${\frac {15}{21}}={\frac {3\cdot 5}{3\cdot 7}}={\frac {3}{3}}\cdot {\frac {5}{7}}=1\cdot {\frac {5}{7}}={\frac {5}{7}}$.

The exact same calculation will let us get rid of any number that divides both the numerator and denominator. Imagine we have a fraction ${\frac {p}{q}}$ and there is a number r so that r divides evenly into p, and r divides evenly into q. We can then write $p=r\cdot a$ (just like we wrote $15=3\cdot 5$). We can also write $q=r\cdot b$. Then the calculation looks like:

${\frac {p}{q}}={\frac {r\cdot a}{r\cdot b}}={\frac {r}{r}}\cdot {\frac {a}{b}}=1\cdot {\frac {a}{b}}={\frac {a}{b}}$.

So whenever we have a fraction where the numerator and denominator are numbers we don't know yet (so we need to use letters like p and q), we can assume we have already gotten rid of all the numbers that divide both p and q.

The other thing that we need to remember is our facts about even and odd numbers. First every even number is really 2 times some smaller number. This is easy to see if you list out the even numbers:

even number	2	4	6	8	10	12	14	16	18	20	…
the same even number as 2·"something"	2·1	2·2	2·3	2·4	2·5	2·6	2·7	2·8	2·9	2·10	…

Finally we need to remember that the product of two even numbers is again an even number, and the product of two odd numbers is again an odd number. In fact this is just the familiar rules:

• "even times even is even"
• "odd times odd is odd"
• "even times odd is even"

Now we can start thinking about the proof. The proof is a "proof by contradiction". For us this means we will start by assuming ${\sqrt {2}}$ can be written as a fraction. We will try to investigate and see what this means for the fraction, being careful that every step follows from the last step by something we know to be true. At the last step we will contradict ourselves. This will mean that we had to make a mistake somewhere. Since we were careful about all of the steps, the only possible place for a mistake will be that we assumed ${\sqrt {2}}$ can be written as a fraction.

1. Assume that ${\sqrt {2}}$ can be written as a fraction. This means ${\sqrt {2}}={\frac {a}{b}}$. We will assume that this fraction is reduced, so no number could divide both a and b. Specifically 2 cannot divide both a and b.
2. We now square both sides of your equation, that is we write $2=\left({\frac {a}{b}}\right)^{2}$, which is the same as $2={\frac {a^{2}}{b^{2}}}$.
3. By multiplying both sides by $b^{2}$, it follows that $a^{2}=2\cdot b^{2}$.
4. Therefore $a^{2}$ is even, because it is equal to $2\cdot b^{2}$ which is even.
5. If $a$ were odd, then $a^{2}=a\cdot a$ would be odd, because "an odd times an odd is odd". So, it follows that $a$ must be even.
6. Because $a$ is even, and every even number is 2 times something, we can write $a=2\cdot k$, for some whole number $k$.
7. If we substitute $2\cdot k$ in for $a$ in the equation in line 3, we get $(2\cdot k)^{2}=2\cdot b^{2}$. Now $(2\cdot k)^{2}=(2\cdot k)\cdot (2\cdot k)=4\cdot k^{2}$. So, $4\cdot k^{2}=2\cdot b^{2}$.
8. Because $2\cdot k^{2}$ is even then $b^{2}$ is also even. Just like we saw with $a$ in line 5, this means that b is even.
9. In line 5 we saw that $a$ was even and so 2 divides $a$. In line 8 we saw that $b$ was even, so 2 divides $b$. But in line 1 we said that 2 cannot divide both $a$ and $b$. This is a contradiction.

Since we have found a contradiction, the assumption in line 1 must be false. That is, it must be false that ${\sqrt {2}}={\frac {a}{b}}$ where $a$ and $b$ are whole numbers. So we cannot write ${\sqrt {2}}$ has a fraction. The fancy way of saying this is to say that ${\sqrt {2}}$ is irrational.