Theorem 2.4.5: Square Roots in R
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There is a positive real number
x such that
x2 = 2
Context
Proof:
Since the above equation is not true in
Q, we have to use a property of
the real numbers that is not true for the rational numbers. Define the
set
- S = {
t R: t > 0 and
t2 2 }
Our hope is that the supremum of this set should be the desired solution to
the above equation. However, we first need to make sure that the supremum
indeed exists before showing that it is the desired solution.
S is not empty, because 1 is contained in S, and S is
bounded above by, say, 5. Hence, using the least upper bound property of the
real numbers, S has a least upper bound s:
Note that this would not necessarily be true if we restricted ourselves to
the rational numbers.
Now we hope that s2 = 2, i.e. s is the desired
solution. Since 1 is in S, we know
Now
s either is the solution, or one of the following two cases are
true:
- Is s2 < 2 ?
- Let
.
Then, by assumption 0 < < 1,
so that
Hence, s + is also in S,
in which case s can not be an upper bound for S. This is a
contradiction, so this case is not possible.
- Is s2 > 2 ?
- Let
. Again
> 0, so that
Hence, s - is another upper
bound for S, so that s is not the least upper bound for
S. This is a contradiction, so that this case is not possible.
Having eliminated these two cases, we are left with
s2 = 2, which is what we wanted to prove.