epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 65
If /c > s, we obtain 1 = Ps+i x • • • x p^- Again, this is impossible.
Therefore, k = s and p\ = qu pi = qi,... ,Pk = qk- Therefore, the factorization
of n is unique for the prime numbers used. The order in which these
factors are arranged is unique, as it is fixed. Therefore, the factorization of n
as described is unique. •
EXAMPLE 4. There exists a unique Hne passing through the points with
coordinates (0,2) and (2,6).
Proof:
Part 1. There exists a hne passing through the points with coordinates
(0,2) and (2,6).
See Example 3 in the section on Existence Theorems.
Part 2. We will prove the uniqueness of the line using all three
procedures described at the beginning of the section.
First procedure. Assume that there are two lines passing through the
points with coordinates (0,2) and (2,6). Let their equations be y = ax-\-b
and y = cx + d.
As both lines pass through the point (0,2), we have 0a-{-b = 2, and
0c + d = 2. Thus, b = d = 2.
As both lines pass through the point (2,6), we have 2a-^b = 2c-\-d.
Because b = d, this implies that a = c. Thus, the two lines coincide.
Second procedure. If we look at the steps used to find the equation of the
hne (refer to Example 3 in the section on Existence Theorems) as y = 2x + 2,
we can state that:
1. The slope is uniquely determined by the coordinates of the points; and
2. Given the uniqueness of the slope, the other constant in the formula is
uniquely determined as well.
Third procedure. There is a postulate from geometry that states that given
any two distinct points in the plane there is a unique straight line joining
them. Therefore, there is a unique hne joining the points with coordinates
(0,2) and (2,6). •
EXERCISES
Prove the following statements.
1. The polynomial p{x) = x-b has a unique solution for all real
numbers b.
2. There exists a unique angle 0 with 0<0<Tt such that cos 0 = 0.