How do you know irrational numbers are really irrational?

Question

Are there any ways to prove if irrational numbers are really irrational and the digits of decimal does not repeat? Or if maybe irrational numbers are actually rational which the repetition of digits are so far apart we see them as 'irrational'.

Answer 1

We cannot prove for some constants if they are integer, rational, irrational, algebraic, transcendental and other possible groups.

An example, Euler's constant defined as

$$\gamma = \lim_{n \to \infty} -\log(n) + \sum_{k=1}^{n} \frac1{k}$$

is unknown if it is rational or not. The best we know is that if it is rational then its pattern, just as you suggested, has more than $200.000$ decimal digits which does not help at all in deciding about its irrationality. The problem is so difficult that some authors believe that we will not get an answer in our lifetime.

There are ways to prove that some other numbers are irrational. Probably the simplest one is $\sqrt{2}$, cause, imagine it is rational

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

then

$$2q^2 = p^2$$

Now we can assume that $\frac{p}{q}$ is a reduced fraction, that we have divided $p$ and $q$ the best we could with all their common factors, including $2$. That makes $p$ odd, $q$ even or $q$ even $p$ odd.

But $p$ cannot be odd since $p^2$ is obviously even since we have $2q^2$ and odd times odd is an odd, not an even number. If $p$ is even then $p^2$ is divisible by $4$. However there is only one $2$ in $2q^2$ since $q$ is odd, so $2q^2$ is not divisible by $4$. That means that $p$ cannot be divisible by $4$ either. So the pair $(p,q)$ does not exist. $\sqrt{2}$ is not rational.

Answer 2

The best proof that I know of to prove something is irrational is to use a proof of contradiction. So first up, when something is rational, that means that it can be displayed as a fraction with a whole number on top of a whole number. While an irrational number cannot be displayed as a fraction. So here is a proof for why the $\sqrt{2}$ is irrational.

First, because this is a proof of contradiction, the proof is about making an assumption and then proving it wrong by showing that if the assumption is true then it breaks some of the rules of math.

The only needed knowledge for this proof is a proof that I will do below, which shows that any squared number that is even must be even, before it was squared as well:

so lets say that the number squared is $a$ so that becomes $a^2$, and lets say that $k$ is any whole number, so that every even number can be represented as $2k$ and any odd number can be represented as $2k + 1$, so in this proof, we will prove that if a is odd then $a^2$ cannot be even

Here comes the proof:

$a = 2k + 1$ as $a$ is odd $a^2 = (2k + 1)^2$ $(2k + 1)^2 = 4k^2 + 4k + 1$

so we can factor our the two to get: $4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ so as k is a whole number, $2k^2 + 2k$ must also be a whole number, and so $2(2k^2 + 2k) + 1$ must be odd as it is an even number plus one, making an odd number. So if $a$ is odd, than $a^2$ is also odd

and now, if $a$ was originally even: $a = 2k$ as multiplying any number by two gets an even number $a^2 = (2k)^2$ $(2k)^2 = 4k^2$ and yet again the two can be factored out making: $4k^2 = 2(2k^2)$ and as $2k^2$ is a whole number, two times that must be an even number, so an even number squared makes an even number, and an odd number squared can't. This leads to the conclusion that an even squared number must come from an originally even number that has been squared.

So for this proof lets assume the number $\sqrt{2}$ is rational, so it can be displayed as a fully simplified fraction. Lets say that this fraction is $\frac{a}{b}$ where $a$ and $b$ are whole numbers with no common factors (meaning it is a fully simplified fraction) as all rational numbers can be displayed as a fully simplified fraction due to that being the definition of a rational number.

So our equation so far is:

$\sqrt{2} = \frac{a}{b}$

We can then square both sides to get:

$2 = \frac{a^2}{b^2}$

Alright so now we times both sides by $b^2$ getting:

$2b^2 = a^2$

and so as shown above, as $a^2$ equals an even number, $a$ must also be an even number. This means that $a$ can be displayed as $2k$ where $k$ is any whole number. So lets replace a with $2k$

$2b^2 = (2k)^2$

$2b^2 = 4k^2$

divide both sides by two

$b^2 = 2k^2$

so $b^2$ is equal to an even number and as such $b$ is an even number

This creates a contradiction, as the fraction $\frac{a}{b}$ was fully simplified and yet both $a$ and $b$ are factors of 2 (due to being even). So we started off with a fully simplified fraction and than proved that is is not simplified. This creates a conflicting statement and so leads to the conclusion that our assumption was wrong.

Therefore, $\sqrt{2}$ cannot be a rational number for that means we can prove a simplified fraction is not simplified, creating contradicting statements.

So $\sqrt{2}$ cannot be displayed as a fully simplified fraction. From this we can draw the conclusion that the $\sqrt{2}$ cannot be displayed as any fraction and is such no a rational number.

So we have found a non rational number, meaning that it must be irrational.

That is a proof for contradiction proving that not all numbers (namely $\sqrt{2}$) can be displayed as a fraction of two whole numbers, which is the definition of a rational number and as such another category must exist for these numbers, we call that category, irrational numbers.

Sorry for any spelling mistakes.

Answer 3

One can establish that irrational numbers exist simply by a cardinality argument: the real numbers are uncountable, the rational numbers are countable, therefore there must exist real numbers that are irrational. I assume this is not what you meant, and you would like an explanation of how one could prove that a specific number is irrational, such as $\sqrt2$. Here again there is a problem, because a majority of sources furnish proofs that proceed via an argument by contradiction; such arguments are not convincing to everybody. However, it is possible to establish the irrationality of $\sqrt2$ by a direct argument (which is not a proof by contradiction). The idea is to find an explicit lower bound for the difference between $\sqrt2$ and a rational number $\frac pq$, in terms of the denominator. Such estimates are of the type $|\sqrt2 -\frac pq|>\frac{C}{q^2}$ where $C$ is an explicit specific constant (bigger than $\frac{1}{10}$, say). One of the places one can find such arguments is in an article by Errett Bishop.