at the extreme end of the spectrum, one could write a computer program that computes all G for which |Aut G|=n, and then find the minimum and maximum values of |Aut GxG|. this goes to show that we could theoretically say more than just 2|Aut G|^2 for a lower bound, but a more nuanced bound would depend sensitively on n. perhaps some arithmetic could be used. and then there's the upper bound to consider.
@robjohn Yeah (actually half an hour ago). My dogs are miraculously healed after praying to god for their healing. Yesterday they were about to die, now they are in a perfect shape.
They eat, drink and play around like before. Here there were also other persons that had similar problems, all their dogs died in 4 days (from the beginning of the symptoms).
I'm very happy for that. For me they are like human beings. :-)
$\forall x\in \mathbb{R} (\exists y\in \mathbb{R} (x\leq y))$ I assume my professor only wants intuitive proofs of these statements given our length into the course...
@Chris'ssis well if I were to be even near qualified as a professor .. :P that'd really reflect a poor image of our country education system .. but wait our education system on an average is poor :P .. so who knows =P
here's the question (the pictures are there so it's easier to look at): http://www.purplemath.com/learning/viewtopic.php?f=14&t=3946 i have no idea what that 3 means
convince yourself that in order for these two points to be connected by a horizontal line (the top edge of the rectangle), we must have a=b. either look at the graph to see this, or argue e^(-2a^2)=e^(-2b^2) implies a=+/-b
@GBeau you don't want to show equality using set operations, you want to show two logical statements are logically equivalent
namely, you want to show $(x\in A \wedge x\not \in B)\vee(x\not\in A\wedge x\in B)$ is logically equivalent to $(x\in A\vee x\in B)\wedge \neg (x\in A\wedge x\in B)$
@Hello you plug 1/4 in and that's what you get, and we seem to have done everything right, so that looks to be the answer (even though you originally stated the answer was like 0.6)
@blue I must be close $\forall x, (x\in A\wedge x \notin B) \vee (x\in B\wedge x \notin A) \equiv \forall x, (x\in A\vee x \in B) \wedge (x\notin A\vee x\notin B)$