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Theorem 2.4.5: Square Roots in R

There is a positive real number x such that x2 = 2

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 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.

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