Recent content by llanoda

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Proving x<0, y<0, and x<y Implies [y][/2] < [x][/2]

That is y squared is greater than x squared
- llanoda
- Post #3
- Mar 27, 2009
- Forum: Set Theory, Logic, Probability, Statistics
L
Proving x<0, y<0, and x<y Implies [y][/2] < [x][/2]

Hi, Anybody can help me prove this? Thanks... If x<0 and y<0 and x<y, then [y][/2] < [x][/2]
- llanoda
- Thread
- Mar 24, 2009
- Replies: 8
- Forum: Set Theory, Logic, Probability, Statistics