# existence of square roots of non-negative real numbers

###### Theorem.

Every non-negative real number has a square root.

###### Proof.

Let $x\ge 0\in \mathbb{R}$. If $x=0$ then the result is trivial, so suppose $x>0$ and define $$. $S$ is nonempty, for if $$, then $$, and $y\in S$. $S$ is also bounded above, for if $y>\mathrm{max}\{x,1\}$, then ${y}^{2}>y>x$, so such a $y$ is an upper bound^{} of $S$. Thus $S$ is nonempty and bounded, and hence has a supremum which we denote $L$. We will show that ${L}^{2}=x$. First suppose $$. By the Archimedean Principle there exists some $n\in \mathbb{N}$ such that $n>(2L+1)/(x-{L}^{2})$. Then we have

$$ | (1) |

So $L+1/n$ is a member of $S$ strictly greater than $L$, contrary to assumption^{}. Now suppose that ${L}^{2}>x$. Again by the Archimedean Principle there exists some $n\in \mathbb{N}$ such that $$ and $$. Then we have

$${\left(L-\frac{1}{n}\right)}^{2}={L}^{2}-\frac{2L}{n}+\frac{1}{{n}^{2}}>{L}^{2}-\frac{2L}{n}>x\text{.}$$ | (2) |

But there must exist some $y\in S$ such that $$, which gives $$, so that $y\notin S$, a contradiction^{}. Thus it must be that ${L}^{2}=x$.
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Title | existence of square roots of non-negative real numbers |
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Canonical name | ExistenceOfSquareRootsOfNonnegativeRealNumbers |

Date of creation | 2013-03-22 16:32:42 |

Last modified on | 2013-03-22 16:32:42 |

Owner | PrimeFan (13766) |

Last modified by | PrimeFan (13766) |

Numerical id | 8 |

Author | PrimeFan (13766) |

Entry type | Theorem |

Classification | msc 11A25 |

Related topic | AxiomOfAnalysis |

Related topic | ArchimedeanProperty |

Related topic | Supremum |

Related topic | ExistenceOfNthRoot |