# Thread: Bounded Intervals - Cardinality

1. ## Bounded Intervals - Cardinality

Q: Show that any two intervals (a,b) and (c,d) have the same cardinality

So I guess I have to show there's a bijective function?

So f: (a,b) --> (c,d)
and f(a) = c
f(b) = d

In general f(x) = sx + t, just using different letters

So f(a) = sa + t = c
and f(b) = sb + t = d

Subtracting the two (sa + t) - (sb + t) = c - d
s(a-b) = c-d

Am I going along the right lines for the proof? How would I write out the final function and prove it's a bijection?

2. Originally Posted by pkr
Q: Show that any two intervals (a,b) and (c,d) have the same cardinality. How would I write out the final function and prove it's a bijection?
Write the equaton of the line determined by (a,c) & (b,d).
$f(x) = \left( {\frac{{d - c}}{{b - a}}} \right)\left( {x - a} \right) + c$.
Linear function are bijections.

3. I think i'm being stupid but why is it (x-a)?

4. Originally Posted by pkr
I think i'm being stupid but why is it (x-a)?
That is the equation of a line in slope/point form: y=m(x-a)+b.