[By O4]:
**.**
That is, bigger positive numbers have bigger squares**.**
Thus any given real number can have at most one positive square root**.**
It remains to show that if *a* ∈ ** R**,

*S* = { *x* ∈ *R***:** *x* ≥ 0, *x*² ≤ *a* }**.**
This set is nonempty, since 0 ∈ *S*, and bounded from above, since if *x* > max{*a*,1} we have *x*² = *x* · *x* > *x* · 1 = *x* > *a***.**
Hence *y* = l.u.b. *S* exists**.**

*y* > 0, for min{1, *a*} ∈ *S*, since
**[** if 1 < *a* then 1 ∈ *S*,
while if *a* < 1 then *a* ∈ *S* **]**
**.**

*y*
we have 0 < *y* - ε < *y* < *y* + ε **,**
so (*y* - ε)² < *y*² < (*y* + ε)² **,**
since bigger positive numbers have bigger squares**.**

*y* there are numbers greater than *y* - ε in *S***,** but *y* + ε ∉ *S* **.**
**[**Then, using the definition of *S*, we get**]**
*a* < (*y* + ε)² **.**

*y* - ε)² - (*y* + ε)² < *y*² - *a* < (*y* + ε)² - (*y* - ε)² **,**
so
|*y*² - *a*| < (*y* + ε)² - (*y* - ε)² = 4*y*ε **.**

*y*² - *a*| < 4*y*ε holds for any ε such that 0 < ε < *y*, and by choosing ε small enough we can make 4*y*ε smaller than any preassigned positive number. Thus |*y*² - *a*| is less than any positive number. Since |*y*² - *a*| ≥ 0, we must have |*y*² - *a*| = 0, proving *y*² = *a* **.**