Vtyjkyuk

I need to prove that the interval $(a,b)$ and the set of Real numbers share the same cardinality.

I understand that I need to find a bijection between the two sets. I have been hinted to use something like $(0,1)$ or $(-1,1)$ to find a bijection then extend it to all reals. I need advice on how to set up the actual proof. Trigonometric functions are not to be used in this proof.

Any two open intervals $(a,b)$ and $(c,d)$ are equivalent via $$x mapsto c+Big(fracd-cb-aBig)(x-a)$$

Therefore $(a,b)$ and $(-1,1)$ are equivalent via $$x mapsto -1+frac2b-a(x-a) $$

Also $(-1,1)$ and $BbbR$ are equivalent via $$x mapsto fracxx^2-1$$

Finally use this fact : "Composition of bijections is again a bijection" to see $(a,b)$ is equivalent to $BbbR$

Geometrically imagine $(a, b)$ bent
into a semicircle that rests on the number line at $O$, as shown in Fig. $3.5$.(replace $-1$ by $a$ and $1$ by $b$ in the figure)

Rays
from the center of the semicircle establish a one-to-one correspondence between points of $(a, b)$ and points of the line.