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<svizzero> ah ok thx
<nerdy2> well-defined == no errors in the definition [so what you claim is a function, really is one :)]
<nerdy2> Jax, well you left out reflexivity (a<=a)
<nerdy2> Jax, but yes that's it
<Jax> well totalness implies reflexicity
<Jax> reflexivity even
<nerdy2> Jax, yes it does
<nerdy2> Jax, fair enough :)
<Jax> this stuff is weird
<Jax> i shouldn't have taken mathematicians analysis :D
<nerdy2> Jax, you can also define the notion using "<" (in other words a < b if a <= b and a != b) and in this case the axioms get a bit simpler (this was retardo's suggestion)
<nerdy2> Jax, the reason things are often done as above is that the axioms (1) - (3) [i.e. without tricohotomy or "totalness"] define a useful thing all by themselves
<Jax> why, what's so useful about it
<nerdy2> e.g. take a set T and let S = P(T) be the set of subsets of T, now you can define a <= b for a,b in S [i.e. a,b subsets of T] to mean "a subset b"
<nerdy2> now this relation satisfies axioms (1)-(3), but not (4)
<nerdy2> so by having a useful theory of such things, you can apply it to this example
<Jax> i don't understand the practical use really
<nerdy2> well you'll have to trust me that "subsets of a set" is a useful thing to understand
<nerdy2> likewise for natural numbers, you can define a <= b to mean "a divides b", this also satisfies (1)-(3) but not (4)
<nerdy2> this is also useful
<Jax> what was 4 again
<nerdy2> (4) was totalness
<pattm> hey.. i have a probability problem... standard monty hall problem, a prize behind one door, goats behind the other two.. we're giving the two goats names, nan and bill. The problem is to calculate the probability that the prize is behind the third door (the one you did not choose) if you are shown a particular goat (bill), given that the presenter will prefer to show that goat with probability (b)
<nerdy2> [a relation which satisfies only (1)-(3) is known as a partial order]
<Jax> Notice that the totalness condition implies reflexivity, that is a = a. Thus a total order is also a partial order, that is, a binary relation which is reflexive, antisymmetric and transitive. It follows that a total order can also be defined as a partial order that is total.
<hopper> 4x^5+172x^4-219x^3-1940x^2-2121x-396 (Show that 5 is an upper bound for the zeros of f(x) and explain your result.
<nerdy2> Jax, yep
<Jax> totalness just means x = x or ? i.e x <= x and x >= x
<nerdy2> it means a<=b or b<=a
<nerdy2> axiom (4)
<Jax> i.e a = b
<delta__> doomie, it looks like median formula but waird, I agree ;)
<nerdy2> Jax, no, both don't have to hold
<nerdy2> only if both hold can you conclude a=b
<Jax> oh, sorry, didn't see the or
<hopper> 4x^5+172x^4-219x^3-1940x^2-2121x-396 (Show that 5 is an upper bound for the zeros of f(x) and explain your result.
<Jax> thanks a lot nerdy2
<nerdy2> sure
<Jax> i think i'll start to understand tonite :D
<Jax> do you know how i can prove that Sqrt(13) is an irrational number?
<nerdy2> do you know how to prove sqrt(2) is an irrational number?
<delta__> Jax, sqrt(2) is an irrational number? :)
<delta__> nerdy2, heh
<Jax> ahhh
<Jax> i see what they are trying to do...
<Jax> but Sqrt(4) isn't
<Jax> but both are subsets of 13 ?
<Jax> i don't get it...
<nerdy2> no
<nerdy2> do you know how to prove sqrt(2) is an irrational number?
<Jax> no
<Jax> well there are no 2 natural numbers, that when multiplied equal 2
<Jax> ?
<Jax> (same ones)
<nerdy2> ok, it's a common thing, suppose not, then sqrt(2) = p/q, for integers p,q, we can ***ume p,q have no factors in common, so p^2 = 2q^2, so p^2 is even, but then p is even, say p=2m, so then 2m^2 = q^2, so q is even, but this is a contradiction
<nerdy2> anyways, same style proof works, replace "even" with "is divisible by 13"
<hopper> question: whats 1+1?
<thief_grr> Jax: you can easily prove that sqrt(m/n) is irrational unless both m and n are perfect squares
<delta__> doomie, are you sure it's r^3 and not r^2?
<doomie> http://www.tu-chemnitz.de/mathematik/analysis/helm/index_files/anabl03.pdf <- task #4
<Jax> what's a perfect square?
<doomie> its 4c, its german but i think you will get it.
<doomie> so, yes, iam *very* sure ;)
<thief_grr> Jax: a square of an integer
<Jax> nerdy2 1 min
<Jax> why do we ***ume they have no factors in common?
<thief_grr> Jax: cos you can cancel them out if there are
<Jax> oh ok sorry
<Jax> man this is too high for me
<delta__> doomie, I lost my german... :-(
<doomie> well, you see #4c ?
<doomie> thats the formula i have to proovie.
<doomie> proove*
<doomie> with (u|n) being the dot-product.
<doomie> and u is, as you can see in #4b: u:R^3->R.
<delta__> doomie, u is supposed to be harmonic on R^3?
<doomie> yeah
<Dreistein> hi
<Dreistein> i need a little hint on how to show the distributivity on set operations, A cup (B cap C) = (A cup B) cap (A cup C).
<xerox> Dreistein: maybe a venn diagram?
<thief_grr> Dreistein: show that one is the subset of the other
<Dreistein> so A subset B and B subset A?
<thief_grr> Dreistein: in the background you take the distributivity of the corresponding logic operators for granted ;)
<Dreistein> ah lol, thats equality i see
<thief_grr> Dreistein: no
<thief_grr> Dreistein: show the equality you want
<thief_grr> Dreistein: by first replacing = with subseteq
<thief_grr> Dreistein: and then replacing it by superseteq
<Dreistein> superseteq?
<tpp> can anyone help me out using the "singular" program?
<Jax> hey Samira
<Samira> hi
<Jax> do you study?
<Samira> yea
<Jax> which uni?
<Samira> z?rich
<Jax> ah... you just start?
<Samira> yes
<Jax> what did you start?
<Samira> Math
<Samira> you?
<Jax> oh.. ;)
<Samira> uniBe seems to be pretty cool :) would have gone there if it were nearer
<Jax> cool? it's hippie ;)
<Samira> ''^
<Jax> well.. i live in bern... pissed me off going to ETH and back every day
<Samira> i sat in Informatik 1 today in zurich, i left after 30minutes
<Samira> this was just poor
<Jax> yeah if you want info go to ITET @ ETH ;)
<Jax> not uniZH
<Samira> going at eth would piss me all the time ^^
<Jax> lol why
<Samira> i just don't like it ^^
<Jax> if you want math you should be there too
<jian1> can't believe it!
<Samira> no, if i took math at eth, i had physics... but i'd rather take something else for Nebenfach
<dafen> can't believe it
<dafen> nerdy2 is here to
<Atrapectus> Wrong button.
<Jax> ah that's why..
<dafen> for a quotient ring Z/nZ
<Samira> so what's this channel all about?'
<Jax> math... duh
<Jax> ;)
<dafen> why Z/nZ : Z-> Z/nZ?
<Atrapectus> Um, mathematics?
<dafen> is not surjective
<Samira> i see... ^^
<Atrapectus> All these supergenius nerds are useful when you need a quick ***ist.
<delta_> dafen, ?
<dafen> the mapping of so
<dafen> is not surjective, why?
<Jax> yeah and i'll be needing a lot of help... >:)
<delta_> dafen, which mapping?
<Jax> http://en.wikipedia.org/wiki/Bijection is pretty good dafen
<Jax> i saw that in some context the other day too, but have to look at it myself again
<dafen> the mapping from quotient Z/nZ
<dafen> the map defined by Z/nZ
<dafen> from Z to itself is generally not surjective
<dafen> I don't see how it's not surjective
<delta_> dafen, what si your question? I don't understand it.
<Samira> ok, i'm pretty bound to have some questions in the next 9years ^^
<dafen> that generally Z/nZ : Z-> Z/nZ is not surjective
<dafen> delta, that's very clear
<delta_> dafen, not at all. what is your question EXACTLY?
<dafen> show that it's generally not surjective
<Atrapectus> What is "surjective"?
<delta_> dafen, what is not surjective. It's a non sense! :-)
<dafen> the mapping!
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