If we didn't have examples of irrational numbers, would we know they exist?

Question

Irrational numbers are very easy to find. Square roots require only a little bit more than the most basic arithmetic. So it might be that this question is impossible to answer because it presupposes a world where math looks completely different to what it really does. I am hoping this is not the case.

If we didn't have examples of irrational numbers, would $\mathbb{R}$ and $\mathbb{Q}$ be assumed to be the same set?

Can we construct a proof that irrational numbers exist without giving examples?

Answer 1

From the surrounding discussion it seems that perhaps there are two questions here:

1. What is the impetus to define the real numbers and how do we get such a definition?

2. Given a definition of $\mathbb{R}$ can we prove that irrational numbers exist without constructing examples?

My answer addresses the latter question.

You can prove that the real numbers are uncountable (there is no bijection between the real numbers and the integers).

You can prove the rational numbers are countable (there is a bijection between the rational numbers and the integers).

So $\mathbb{R}\setminus\mathbb{Q}$ is nonempty.

Since I don't know your background I will add some more.

1. A bijection between two sets $X$ and $Y$ is a function $f:X\to Y$ that is both one-to-one and onto, i.e, for all $x,y\in X$, $f(x)=f(y)$ implies $x=y$; and for all $y\in Y$ there is $x\in X$ such that $f(x)=y$.

2. An infinite set is $X$ countable if there is a bijection $f:X\to \mathbb{N}$.

3. It is a famous result of Cantor (called a diagonal argument) that $\mathbb{R}$ is uncountable.

4. It is a standard (and good practice) exercise that $\mathbb{Q}$ is countable.

5. It is a standard (and good practice) exercise that if $X$ and $Y$ are countable then so is $X\cup Y$. So if $\mathbb{R}\backslash \mathbb{Q}$ were countable then so would be $\mathbb{R}$.

So altogether, this actually shows that not only do irrational numbers exist, but there are more irrational numbers than rational numbers because the set $\mathbb{R}\backslash\mathbb{Q}$ must be uncountable by the above points.

By the way, the same kind of proof shows that transcendental numbers exist and there are more transcendental numbers than algebraic numbers. Indeed, the set of algebraic numbers is also countable and so its complement in $\mathbb{R}$ is uncountable.

Answer 2

halrankard's answer is good but this may help as well.

It is common to define various sets which are larger than the rational numbers $\mathbb{Q}$ yet still smaller than the reals $\mathbb{R}$. By larger and smaller here, I just mean one is a strict subset of the other.

halrankard mentions the Algebraic numbers $\mathbb{A}$. These are the roots to polynomials with rational coefficients. $\sqrt{2}$ is algebraic but $\pi$ is not.

Larger again, are the Computable numbers. Informally, these are ones that an ideal computer, e.g. a Turing machine, could calculate. $\pi$ is computable, you can calculate it to any desired precision in a finite time.

Larger again are the Definable numbers. Informally, we can specify these precisely yet we cannot even compute them. We know that there are numbers which are definable but not computable. They are necessarily rather weird. See this earlier question.

And this is still not all real numbers. All of these sets are countable. This means that although intuitively each set is bigger than the previous one, they can all be put into one to one correspondence with the smallest infinite set: the natural numbers $\mathbb{N}$. We could assign each definable number a unique natural number label without missing any.

However, we know that the real numbers are not countable so there must be undefinable numbers and that most real numbers are undefinable.

So, one of these last two sets might be like what you want: we know that they exist but we cannot point to any examples. No one will ever prove that a specific number is undefinable since, if they could specify which number they are talking about then it is definable

Answer 3

I think the answers so far miss a very basic point: Perhaps the most famous proof in mathematics shows that there is no rational number $x$ with $x^2 = 2$. This is not the same as saying "$\sqrt{2}$ is irrational". The reason they are not the same is because the former can be articulated as a statement just about rational numbers, before any development of the reals or the square root function or any concept of "irrationality". Now it is up to you whether or not you want to try to expand to a larger number system in which $x^2 = 2$ has a solution.