Linear Algebra, Theory And Applications, 2012a

26 PRELIMINARIES
Proof: If a equals a prime number, the prime factorization clearly exists. In particular
the prime factorization exists for the prime number 2. Assume this theorem is true for all
a ≤ n − 1. If n is a prime, then it has a prime factorization. On the other hand, if n is not
a prime, then there exist two integers k and m such that n = km where each of k and m
are less than n. Therefore, each of these is no larger than n − 1 and consequently, each has
a prime factorization. Thus so does n. It remains to argue the prime factorization is unique
except for order of the factors.
Suppose
n∏ m∏
p i =
i=1
where the p i and q j are all prime, there is no way to reorder the q k such that m = n and
p i = q i for all i, and n + m is the smallest positive integer such that this happens. Then
by Theorem 1.9.7, p 1 |q j for some j. Since these are prime numbers this requires p 1 = q j .
Reordering if necessary it can be assumed that q j = q 1 . Then dividing both sides by p 1 = q 1 ,
n−1
∏
i=1
p i+1 =
j=1
m−1
∏
j=1
q j
q j+1 .
Since n + m was as small as possible for the theorem to fail, it follows that n − 1=m − 1
and the prime numbers, q 2 , ··· ,q m can be reordered in such a way that p k = q k for all
k =2, ··· ,n. Hence p i = q i for all i because it was already argued that p 1 = q 1 , and this
results in a contradiction.
1.10 Systems Of Equations
Sometimes it is necessary to solve systems of equations. For example the problem could be
to find x and y such that
x + y =7and2x − y =8. (1.2)
The set of ordered pairs, (x, y) which solve both equations is called the solution set. For
example, you can see that (5, 2) = (x, y) is a solution to the above system. To solve this,
note that the solution set does not change if any equation is replaced by a non zero multiple
of itself. It also does not change if one equation is replaced by itself added to a multiple
of the other equation. For example, x and y solve the above system if and only if x and y
solve the system
−3y=−6
{ }} {
x + y =7, 2x − y +(−2) (x + y) =8+(−2) (7). (1.3)
The second equation was replaced by −2 times the first equation added to the second. Thus
the solution is y =2, from −3y = −6 and now, knowing y =2, it follows from the other
equation that x + 2 = 7 and so x =5.
Why exactly does the replacement of one equation with a multiple of another added to
it not change the solution set? The two equations of (1.2) are of the form
E 1 = f 1 ,E 2 = f 2 (1.4)
where E 1 and E 2 are expressions involving the variables. The claim is that if a is a number,
then (1.4) has the same solution set as
E 1 = f 1 ,E 2 + aE 1 = f 2 + af 1 . (1.5)