least and greatest zero
Proof. If then the assertion concerning the least zero is true. Let’s assume therefore, that .
The set is bounded from below since all numbers of are greater than . Let the infimum (http://planetmath.org/InfimumAndSupremumForRealNumbers) of be . Let us make the antithesis, that . Then, by the continuity of , there is a positive number such that
Chose a number between and ; then , but this number is not a lower bound of . Therefore there exists a member of which is less than (). Now , whence this member of ought to satisfy that . This a contradiction. Thus the antithesis is wrong, and .
This that and is the least number of .
Analogically one shows that the supremum of is the greatest zero of on the interval.
|Title||least and greatest zero|
|Date of creation||2013-03-22 16:33:22|
|Last modified on||2013-03-22 16:33:22|
|Last modified by||pahio (2872)|
|Synonym||zeroes of continuous function|