The discussion revolves around a proof concerning the product of two positive numbers, \(a\) and \(b\), both constrained between 0 and 1. Participants are examining the validity of the assertion that \(ab < a\) under these conditions.
Some participants have provided feedback on the original poster's proof, noting the importance of the positivity of \(a\) in maintaining the direction of the inequalities. Others have raised questions about the necessity of a final conclusion in the proof.
There is an ongoing exploration of the axioms and rules that may apply to the proof, indicating that the discussion is still in a formative stage with various interpretations being considered.
Actually, if 0 < a and 0 < b < 1, then ab < a. a does not have to be less than 1.knowLittle said:I would like some feedback on my proof.
Can I just say that :
0< a<1 ... [1]
0<b<1 ... [2]
multiplying [2] by 'a' everywhere, then I get
0<ab<a
And, we prove that ab<a?