Pg 7, Proposition 1.1 – There is no rational number whose square equals 2.
For purposes of contradiction we consider that the proposition is false. Suppose there is a rational number x
such that . Since x is rational, we will let , where m and n are integers and either m or n are odd. So by substitution,
we have . Therefore
is even and since we already know m is an integer, we can also conclude that m is also even.
We may express m as , where k is an integer. Again by substitution, we have . Therefore
is even and since we
already know that n is also an integer, we can conclude that n is even.
By the definition of rational, either m or n
must be odd. Therefore, there is no rational number whose square equals 2.
Pg 9, Proposition 1.3 – Let c be a positive number. Then there is a positive number whose square is c.
Suppose that . [NTS: S is bounded above]. We can see that S is nonempty, since if , then 1 must be in S, and
if , then c/2 must be in S. If the number x is in S, then x and x+1 are positive numbers such that , so . So by
definition c+1 is an upper bound for S.
By the Completeness Axiom, the set S has least upper bound, which we will define , and also note that b is positive.
By the Positivity Axiom P2 we know that either , , or . By contradiction, we will show that the first two options must
be false, so .
We will choose a suitably small number r such that .
This will contradict that b is a upper bound for S, since
is in S
and is greater than b. Therefore, it is impossible that . To see how we choose a number r, observe that if r is any
positive number less than 1,
if r is any positive number less than one and less than or equal to .
We will choose a suitably small number t less than b such that . By the definition of S, we can say that
for all x in S,
is positive then we can say that
for all x in S. This makes
an upper bound for S.
This is a contradiction,
is less than b and b is the least upper bound for S. Therefore, it is impossible that . To see how we choose a
number t, observe that if t is any positive number less than b, then
when t is a positive number smaller than b and is also smaller than or equal to .
Therefore, by exhaustion, we have shown that . So, there must be aT positive number whose square is c.