# least and greatest zero

Theorem. If a real function $f$ is continuous^{} on the interval $[a,b]$ and has zeroes on this interval, then $f$ has a least zero and a greatest zero.

Proof. If $f(a)=0$ then the assertion concerning the least zero is true. Let’s assume therefore, that $f(a)\ne 0$.

The set $A=\{x\in [a,b]\mathrm{\vdots}f(x)=0\}$ is bounded from below since all numbers of $A$ are greater than $a$. Let the infimum^{} (http://planetmath.org/InfimumAndSupremumForRealNumbers) of $A$ be $\xi $. Let us make the antithesis, that $f(\xi )\ne 0$. Then, by the continuity of $f$, there is a positive number $\delta $ such that

$$ |

Chose a number ${x}_{1}$ between $\xi $ and $\xi +\delta $; then $f({x}_{1})\ne 0$, but this number ${x}_{1}$ is not a lower bound of $A$. Therefore there exists a member ${a}_{1}$ of $A$ which is less than ${x}_{1}$ ($$). Now $$, whence this member of $A$ ought to satisfy that $f({a}_{1})=0$. This a contradiction^{}. Thus the antithesis is wrong, and $f(\xi )=0$.

This that $\xi \in A$ and $\xi $ is the least number of $A$.

Analogically one shows that the supremum^{} of $A$ is the greatest zero of $f$ on the interval.

Title | least and greatest zero |
---|---|

Canonical name | LeastAndGreatestZero |

Date of creation | 2013-03-22 16:33:22 |

Last modified on | 2013-03-22 16:33:22 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 6 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 26A15 |

Synonym | zeroes of continuous function |

Related topic | ZeroesOfAnalyticFunctionsAreIsolated |