The set of points {(b, c) | x^2 + bx + c = 0 for some real x} is a subset of a 2D plane and contains, say, a square of an arbitrary positive size. Such sets have the same cardinality as a 2D plane.
Stuck on this problem, how do I show
the set of quadratic polynomials with real roots and leading coefficient one and
the set of points in the xy Cartesian coordinate system
have the same cardinality or don't have the same cardinality?
Thanks in advance
The set of points {(b, c) | x^2 + bx + c = 0 for some real x} is a subset of a 2D plane and contains, say, a square of an arbitrary positive size. Such sets have the same cardinality as a 2D plane.
Im really struggling with cardinality, but I think that i've gotten this far,
Define: set of quadratic polynomials with real roots and leading coefficient one ->set of points in the xy Cartesian coordinate system
as follows:
f(x^2+ax+b)=(a,b)
then suppose
f(x^2+ax+b)=f(y^2+cy+d)
then by definition
(a,b)=(c,d)
and so,
a=c and b=d
how do I show the injection the other way.
Thanks for your help
That's correct so far.
Do you mean that you want to use Cantor-Bernstein theorem where you need to define one injection from polynomials to the plain and another injection from the plain to polynomials? The function g(a, b) = x^2 + ax + b whose domain is the whole plain is an injection (polynomials are equal iff their corresponding coefficients are equal); however, its image contains all quadratic polynomials with leading coefficient 1, including polynomials with no real roots.
The set of points (a, b) corresponding to polynomials x^2 + ax + b with real roots is shown in the following picture.
This set, in particular, contains an open circle (i.e., a circle without a border) of radius 1. Now, this circle has the same cardinality as the whole plain as I'll explain below. So, a bijection from the plain to the circle is simultaneously an injection from the plain to the hatched set.
Obviously, there is a bijection between any two open circles of radius 1. Instead of considering our particular circle, we'll construct a bijection from an open circle with center (0, 0) and radius 1 to the whole plain. Namely, a point with polar coordinates is mapped into a point . We use the fact that is a bijection from (-1, 1) to and tan is a bijection from to . The moral of this is that every figure (circle, square, etc.) with nonzero area is equinumerous to the whole plain. In fact, continuous figures with zero area (such as line segments, curves, etc.) are also equinumerous to the whole plain, but this is just a bit harder to show.