## Theorem 2.4.5: Square Roots in R

### 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*and**R**: t > 0*t*}^{2}2

**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:

*s = sup(***S**)

Now we hope that *s ^{2} = 2*, i.e.

*s*is the desired solution. Since 1 is in

**S**, we know

*s > 1*

*s*either is the solution, or one of the following two cases are true:

- Is
*s*?^{2}< 2 - Let
.
Then, by assumption
*0 < < 1*, so that

*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
*s*?^{2}> 2 - Let
. Again
> 0, so that
*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.

*s*, which is what we wanted to prove.

^{2}= 2