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Office_Shredder is offline
Mar4-12, 04:38 AM
P: 4,499
It sounds like you really want to assume that epsilon is an arbitrarily small positive number (i.e. small enough that only first order effects occur). In this case the solution in my previous post essentially does it for us. There are two cases

1) [itex] \Delta S \geq 0 [/itex]. Then [itex] \epsilon \Delta S \geq 0[/itex] as well, and if [itex] \Delta G \geq 0[/itex] adding these two inequalities yields [itex] \Delta G + \epsilon \Delta S \geq 0 [/itex] which we know isn't true. So it must be [itex] \Delta G < 0 [/itex]

2) [itex] \Delta S < 0 [/itex]. We will do a proof by contradiction: suppose that [itex] \Delta G > 0[/itex]. Then we will show that if [itex] \epsilon[/itex] is small enough (smaller than a specific number which we will calculate), that [itex] \Delta G + \epsilon \Delta S > 0 [/itex] which is a contradiction.

If [itex] \epsilon < -\Delta G / \Delta S[/itex] (which is a positive number since by assumption, delta S is negative and delta G is positive), then [itex] \Delta G + \epsilon \Delta S > \Delta G + (-\Delta G / \Delta S) \Delta S[/itex] where the inequality is > because we're increasing the coefficient of delta S, which is a negative number by assumption. Of course, cancelling the fractions then gives us [itex] \Delta G + \epsilon \Delta S > \Delta G - \Delta G = 0 [/itex] which is a contradiction.

Note that in this case we had to assume delta G was non-zero, otherwise epsilon would have been exactly zero. But if delta G is exactly zero there's nothing to prove.

So we had two cases: one where delta S was non-negative, and one where delta S was negative. In both cases we proved that delta G must be negative, as long as we can assume epsilon is small enough.