Irrational numbers in between $n$ and $n+1$

Question

Is the amount of irrationals numbers in between consecutive integers always the same? is this amount infinite?

Answer 1

The way we measure (possibly infinite) cardinality is by using bijections (one-to-one) mappings. Since the irrational numbers in the open interval $I=I_0=(0,1)$ are in bijection with those in $I_n=(n,n+1)$ by the map $f_n:x\mapsto x+n$, each set has the same cardinality. Since $n\in\mathbb{Z}$ was arbitrary, they all have the same cardinality.

Why does the map $f_n$ given above put $I\setminus\mathbb{Q}$ (the irrational numbers between $0$ and $1$) in one-to-one correspondence with $I_n\setminus\mathbb{Q}$ (those between $n$ and $n+1$)? This follows from the fact that $$x\in\mathbb{Q} \iff x+n\in\mathbb{Q}$$ since $$x=\frac{p}{q}\in\mathbb{Q} \quad \implies \quad f_n(x)=x+n=\frac{p+nq}{q}\in\mathbb{Q}$$ as well; for the converse, use $f_{-n}$. This demonstrates that $f_n$ puts $I\cap\mathbb{Q}$ and $I_n\cap\mathbb{Q}$ in bijection. But since it also maps $I$ to $I_n$, it must as well map the respective set differences, or relative complements. Hopefully this will convince you that $x$ is irrational ($x\not\in\mathbb{Q}$) iff $x+n$ is, since the irrational reals are precisely the non-rational real numbers ($\mathbb{R}\setminus\mathbb{Q}$), i.e. any real number is either rational or irrational.

The famous diagonal argument of Cantor shows that there are uncountably many real numbers in $I$, whereas $I\cap\mathbb{Q}$ is countably infinite (it has the same cardinality as $\mathbb{N}$). Thus, the irrationals in $I$, i.e. the set $I\setminus\mathbb{Q}$, is of a much higher order of infinity. The first is called $\aleph_0$ ("aleph-naught"), which is the cardinality of $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$, while the second is $2^{\aleph_0}$, the cardinality of the continuum. Assuming the Continuum Hypothesis (that there is no set $S$ with cardinality between these, i.e. such that $\aleph_1<|S|<2^{\aleph_0}$) and the Axiom of Choice (that there is a smallest cardinal number $\aleph_1>\aleph_0$), we would have $\aleph_1=2^{\aleph_0}$ and these would be the first two "orders" of infinity. But this gets into the "finer" aspects of axiomatic set theory. The power of two is no accident and relates to the fact that no set can be put in 1-1 correspondence with its power set, which is the gem we get from the diagonal argument.

This notion of cardinality defies the expectations we have formed from the laws of arithmetic, since we might first be inclined to think that $\mathbb{Z}$ is twice as large as $\mathbb{N}$, for example. But the map $$\phi(n)=(-1)^n\left\lfloor\frac{n+1}2\right\rfloor$$ actually provides a bijective (one-to-one and onto) map from $\mathbb{N}=\{0,1,\dots\}$ to $\mathbb{Z}$, proving equivalent cardinality. There is also a theorem, known as the Banach-Tarski paradox, violating our sense of size and number by mapping one sphere (which among other things is a point set having cardinality of the continuum) to a number of identically sized spheres, or in a stronger version to one larger one, where the mapping only involves translations, rotations and a totally discontinuous way of decomposing into point sets.

The moral of the story is that when we conceptualize specific sets of numbers, we tend to associate certain properties of those sets with them, endowing them with fuller meaning (for example order, scale/size and dimension). When we then try to engage in formal analysis of systems of axioms, properties and statements about those sets, we find ourselves needing to articulate and strip off those associations, which inform our intuition but trap our concepts within our conceptions. Once we win the battle of formalizing intuition, however, we have won a battle which can open the door to new worlds (for example well ordering, metric spaces, measure & Hausdorff dimension). Another way of constructing the real numbers would be using Dedekind cuts.

Answer 2

Yes. There are many irrational numbers between $n$ and $n+1$. In fact, It equals $\mathcal c = |\Bbb R|$.

Answer 3

If you denote by $I_n$ the set of all irrational numbers between $n$ and $n+1$, then for any $n,m$ the function $f\colon I_n\to I_m$ given by $f(x)=x+m-n$ is easily seen to be a bijection. Therefore, the cardinalities of $I_n$ and $I_m$ are the same.