Show that the following pairs of sets have the same cardinality:
for any real numbers a,b,c,&d with a<b & c<d the open intervals (a,b) & (c,d) of real numbers.
Yes. Suppose a point x is 3/4 of the way from a to b. To what point in (c, d) should x be mapped? What is the explicit formula? And now if x is an arbitrary point in (a, b) (not necessarily 3/4 of the way), what fraction of the way from a to b is it?
Pause for a moment and think about what a linear map f sending the interval (a,b) onto the interval (c,d) would be:
f is a linear map from $\displaystyle \mathbb{R}$ to $\displaystyle \mathbb{R}$. The easiest such a would send a to c and b to d, and "linearly in between".
But what's that? It's a straight line having f(a)=c and f(b)=d. Hence it going through the points (a, c) and (b, d).
Find f(x) = mx+k such that f goes through (a, c) and (b, d) is now just the standard Algebra I problem. m=(c-d)/(a-b) and k=f(a)-ma = c - ( (c-d)/(a-b) ) a.
I see what youre saying, so something like this:
Proof: Let (a, b) and (c, d) be open intervals with a < b and c < d.
We can find the equation of the straight line from (a, c) to (b, d). The slope of this line is (d − c)/(b−a) so the bijection is f (x) =(d-c)/(b − a)(x − a) + c.
so by bijection (?) (a,b) & (c,d) have the same cardinality.
That's right, though you should show that the f you defined is a bijection. Of course, linear functions are always bijections onto their image so long as the slop isn't 0 or undefined (infinity) (that won't be the case here - why?), but it's best to show it (and you should know how to show it how regardless).
You could prove that f(a,b)->(c,d) is a bijection directly:
If f(x1) = f(x2), then... (details)... , therefore x1=x2, so f is injective. If q in (c,d), then...(details)... , so have found a p in (a,b) such that f(p) = q, and therefore f is surjective.)
A perhaps easier way is to constuct the inverse function g, and then show that f(g(x)) = x, and g(f(x)) = x, both for all x. That guarantees that f is a bijection.
It's easy to find the equation for g (f(x)=mx+k, so x = mg(x)+k, so g(x) = (x-k)/m). You should show that g does indeed map (c,d) into (a,c). It's then simple algebra to show both f(g(x)) = x, and g(f(x)) = x.
Anyway, once you've produced a bijection f between the two sets (a,b) and (c,d), you're then allowed to say they have the same cardinality (since that's the definition of "same cardinality").
theorem: the function $\displaystyle f:\mathbb{R} \to \mathbb{R}$ given by $\displaystyle f(x) = mx + b$ for $\displaystyle m \neq 0$ is a bijection.
proof: suppose $\displaystyle f(x_1) = f(x_2)$. then:
$\displaystyle mx_1 + b = mx_2 + b \implies mx_1 = mx_2 \implies x_1 = x_2$ since $\displaystyle m \neq 0$.
thus f is injective.
now choose an arbitrary real number y. then $\displaystyle f\left(\frac{y-b}{m}\right) = m\left(\frac{y-b}{m}\right) + b = y - b + b = y$,
so every y has a pre-image in $\displaystyle \mathbb{R}$, so f is surjective.
in fact, we have "even better" than a bijection between (a,b) and (c,d), the linear map AND its inverse are both continuous. the term for such a map is "homeomorphism", which means the two intervals are topologically indistinguishable (you can think of this as one can be "stretched" and "translated" to make the other one).
a perhaps even weirder fact is that (a,b) has a bijection onto all of $\displaystyle \mathbb{R}$. real numbers are so densely packed, that even when they are stretched "infinitely apart" they stay connected (and therefore are not recommended for making taffy).