我们常遇到这样一类问题:证明一个无限集合(或它的某些子集)具备或不具备某种性质,如果想用列举法(或称穷举法)一一加以验证,有时是做不到的。即使可列举各种情况,但也很难入手,在这种情况下,我们常可用反证法去解决。例1 给定一个圆,S是该圆圆周上所有点的集合,把集合S任意划分为两个不相交的子集M和N。在子集M和N中是否至少有 1°钝角三角形的三个顶点; 2°等腰三角形的三个顶点。这个问题,若采用直接证法是很困难的,因为集合S究竟划分为怎样的两个不相交子集题目中并没有说明,只得采用反证法。 We often encounter the problem of proving that an infinite set (or some subset of it) possesses or does not possess a certain property. If one wants to verify it by enumeration (or exquisite method), it is sometimes done. Less than that. Even though various situations can be cited, it is also difficult to get started. In this case, we can often solve it with counter-evidence. Example 1 Given a circle, S is a set of all points on the circle’s circumference. The set S is arbitrarily divided into two disjoint subsets M and N. Whether there are at least three vertices of an obtuse triangle in the subsets M and N; three vertices of an isosceles triangle of 2°. This problem is difficult to apply if direct evidence is used, because the two sets of disjoint subsets that the set S is divided into are not described in the title. Instead, they have to use the anti-evidence method.