引例:给出下列两个命题的否定:(1)若x>2,则x>3;(2)若a>b,则2 2a>b.该例常见错解为:(1)若x>2,则x≤3;(2)若a>b,则2 2a≤b.错误的原因在于将结论——命题“若p则q”的否定为“若p则q”直接套用到隐含有全称量词的命题的否定的确定.事实上,引例的正确结论应为:(1)x>2,使x≤3;(2)a,b∈R,a>b,但2 2a≤b.这种错误的产生引发了笔者的一些思考.1对命题的四种形式的理解关于命题的四种形式的研究,湘教版教材数学 Example: Given the following two propositions of negation: (1) If x> 2, then x> 3; (2) If a> b, then 2a> b. > 2, then x≤3; (2) If a> b, 2 2a≤b. The reason for the error lies in the fact that the negation of “proposition” if “p is q” is " (1) x> 2 such that x≤3; (2) a, b∈R, a> b, a, b, But 2 2a ≤ b. This kind of error triggered some of the author’s thinking.1 The Four Forms of Propositions Understanding Four Forms of Propositions,