Cardinality of an arbitrary interval of real numbers Let $a, b\in\mathbb{R}$ with $a<b$. Prove that $|\{x\in \mathbb{R}\ | \ a< x< b\}|=|\{x\in \mathbb{R}\ | \ 0<x<1\}|$.
Would constructing a bijection be the most effective way to prove this? Or should I consider proving that the real numbers from a to b are uncountable?
 A: Just showing that $(a,b)$ is uncountable wouldn't actually solve the problem - it's consistent with the usual axioms (ZFC) of set theory that there are uncountable sets of reals strictly smaller in cardinality than $(0,1)$.
I'll say a bit more about this below.
Instead, you need to whip up a bijection. But this won't be too hard. As a first step, maybe consider how to get a bijection between $(0,1)$ and $(0,b)$ for $b>0$; next, given an interval $(a,b)$, is there some easy way to put it into bijection with an interval of the form $(0,c)$?

Now let me give a bit more detail about the claim I made at the beginning of this answer. The statement that every uncountable set of reals has the same cardinality as $(0,1)$ is (one way to phrase) the continuum hypothesis (CH). CH, like many other reasonable-sounding statements about infinite cardinalities, is neither provable nor disprovable from ZFC; that's not to say that no proof or disproof is currently known, but rather that we can prove that ZFC can neither prove nor disprove CH.$^1$ This was established by Godel and Cohen - Godel showed that ZFC can't disprove CH, and Cohen showed that ZFC can't prove CH. Both methods of proof became fundamental to modern set theory.
$^1$Unless ZFC is inconsistent, in which case it proves and disproves everything.
A: You want to map $(a,b)$ onto $(0,1)$. The easiest way is to consider the equation of the line through $(0,a)$ and $(1,b)$. This line has equation $y=(b-a)x+a$. So the map $f:x\mapsto (b-a)x+a$ is bijective from $(0.1)$ onto $(a,b)$. So the two intervals have the same cardinality.
