mm-128
I was listening to the radio a while ago and they were
talking about aset theory. The theory basically said that
everything belongs in aset, because if an element does not
belong in a set, it belongs in theset of elements that do not
belong in a set (does that make sense?).What is this theorem
called? Does anyone have any resources on theweb for
it?TarasPS: I hope what I have written above is correct/makes
radio a while ago and they were talking about a> set theory.
The theory basically said that everything belongs in a> set,
because if an element does not belong in a set, it belongs in
the> set of elements that do not belong in a set (does that
make sense?).> What is this theorem called? Does anyone have
any resources on the> web for it?> Taras PS: I hope what I
listening to the radio a while ago and they were talking about
a> set theory. The theory basically said that everything
belongs in a> set, because if an element does not belong in a
set, it belongs in the> set of elements that do not belong in
a set (does that make sense?).> What is this theorem called?
Does anyone have any resources on the> web for it?>Theres no
such theorem.Whether thats so depends upon the brand of set
theory.Its not so in ZFC, but it is so in NFAn easier way to
express the same notion is everything is a member of theset
that contains only itself. Even so, in ZFC, there are some
boogie menthat refuse membership in any and all sets. In NF
those boogie men areexterminated, ie denied a visa of
existence.> PS: I hope what I have written above is
correct/makes sense :)>Some. I suggest youfind a better source
all makes sense. Thers is something called the universal set.
Theuniversal set is the set of all object that we are
concerned about at themoment. For example if car is not in the
universal set then for this problemthere is no such thing as a
car. Nothing that is not in the universal setexists. Now
suppose the universal set, U, contains 1, 2, 3, 4, 5, 6, 7,
8,egg, and sponge. That is U = {1, 2, 3, 4, 5, 6, 7, 8, egg,
sponge}. Nowsuppose set A = {1, 3,5,8, egg}--note that
everything that is in is also inU. Now we can talk about A,
the set that contains everything exact what isin A. (A is
called A complement or A prime)In our case A = {2, 4, 6,
7,sponge).Notice that everything ( in U ) is either in A or
not in A, that isA, by denition of A. In the statement
everything belongs in a set,because if an element does not
belong in a set, it belongs in the set ofelements that do not
belong in a set when it says it belongs in the set ofelements
that do not belong in a set it is referring to the complement
ofthe set. I was listening to the radio a while ago and they
were talking about a> set theory. The theory basically said
that everything belongs in a> set, because if an element does
not belong in a set, it belongs in the> set of elements that
do not belong in a set (does that make sense?).> What is this
theorem called? Does anyone have any resources on the> web for
it?> Taras PS: I hope what I have written above is
is something called the universal set. The> universal set is
the set of all object that we are concerned about at the>
moment.Makes sense yes. If its true, depends upon the brand
of set theory. The universal set U = { x | x = x }Now in ZFC
set theory, currently the most accepted brand, its not a
set,its a class. If were a set, then U would be in U, but
that wouldcontradict the axiom of regularity, so U cant be a
set, nor can U belongto U> > I was listening to the radio a
while ago and they were talking about a> set theory. The
theory basically said that everything belongs in a> set,
because if an element does not belong in a set, it belongs in
the> set of elements that do not belong in a set (does that
make sense?).> What is this theorem called? Does anyone have
any resources on the> web for it?> > Taras> > PS: I hope what
pointed out before there have been people who have been>
attacking algebra itself in posts. Now it still bothers me
that I was> the one person standing up to defend algebra
itself from what I> noticed.The image of JSH as a stand up guy
defending any part of mathemetics from the attacks of others
is
rather basic lemma:Given any factor g of a polynomial P(x),
g=r+c, where c=g at x=0, so c> is a factor of the contant term
P(0), while r=g-c, so it varies if g> varies. That simple
lemma
not only anchors my paper, and my proof of Fermats> Last
Theorem, it can be used in lots of places. Now Ill use it
with> the roots of y^3 + 3y - 2 Ifind its roots fascinating. >
and the three roots are y_1 = (1+sqrt(2))^{1/3} -
1/(1+sqrt(2))^{1/3} y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
(-1-sqrt(-3))/2(1+sqrt(2))^{1/3} y_3 =
(1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
(-1+sqrt(-3))/2(1+sqrt(2))^{1/3}As simple as they are you
cant just take the roots and *look* at themto tell whether or
not any one of them is coprime to 2. It is justimpossible.Yet
I
work that out and still I keep coming back, as the
simplicityand elegance of those roots keeps calling me.Sure I
know that the cuberoot is ambiguous, so depending on how
youtake the cuberoot, you shift between the solutions, which
is why youcant prove whether or not one of them is coprime to
2, but still itseems like there should be some way to see by
looking.But there isnt.Given the impossibility of proving
that none of them are coprime to 2,youd think that people
objecting would sit back, but nope, they claimGalois Theory
eliminates a possibility.But that cant be. Galois Theory is
based on that shifting. Actually, they can neither prove nor
disprove the assertion with thetools theyre using.> where the
polynomial Im interested in is P(x) = x+1, which has the
non-polynomial factor (1+x)^{1/3}.Letting g = (1+x)^{1/3}, at
P(0), g=1^{1/3}, and choosing the solution 1, I have c=1, so
r=g-1. (Some may worry about my choice for c, Ill come back
to that later.)Now using the substitution g=r+1, with
x=sqrt(2), with my roots, I> have y_1 = r+1 - 1/(r+1) y_2 =
(r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) y_3 =
(r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)so I have y_1 =
(r^2 + 2r)/(r+1) y_2 = [(r^2 + 2r)(-1+sqrt(-3)) +
2sqrt(-3)]/2(r+1) y_3 = [(r^2 + 2r)(-1-sqrt(-3)) -
2sqrt(-3)]/2(r+1)<deleted>So once again I found myself drawn
to that simplicity, looking forsome way, some how to make the
math denite to the eye--to make iteliminate a possibility on
sight.But it doesnt. You cant look at those roots and
tell.Theres just no way to limit the math in that way, and
intellectuallyI know why:The math has the exibility to handle
several possibilities. Forsome polynomials, none of roots
would
be coprime to integer factors ofthe constant term, while for
others, like y^3 + 3y - 2, several arecoprime.But you have to
use advanced techniques to prove it.Still something keeps
drawing me back to stare at those roots, lookingfor what
cannot be found--a limitation that I can see.Its like in
physics where you have quarks that cant be pulled outone by
one. You can determine theyre there experimentally in pairsor
threesomes. You can bounce things off them, but you cant
paper does rest fully on the following rather basic lemma:> >
Given any factor g of a polynomial P(x), g=r+c, where c=g at
x=0, so c> is a factor of the contant term P(0), while r=g-c,
so it varies if g> varies.> > That simple lemma not only
anchors my paper, and my proof of Fermats> Last Theorem, it
can be used in lots of places. Now Ill use it with> the roots
of> > y^3 + 3y - 2 Ifind its roots fascinating.You are again
answering your own posts with narcissistic monologues instead
of answering your critics. Ifyou had any integrity, you would
deal with the challenges not carry on conversations with
yourself. Talk toa mirror if you want, but why use a newgroup
to talk to yourself.> and the three roots are> > y_1 =
(1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}> > y_2 =
(1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -> >
(-1-sqrt(-3))/2(1+sqrt(2))^{1/3}> > y_3 =
(1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -> >
(-1+sqrt(-3))/2(1+sqrt(2))^{1/3} As simple as they are you
cant just take the roots and *look* at them> to tell whether
or not any one of them is coprime to 2. It is just>
impossible.You dont have to just look. The observation that
y_1 y_2 y_3 = 2 is sufcient to prove that each root isa
divisor of 2 in the ring of algebraic integers and, since none
of these is a unit, none are coprime to 2.> Yet I work that
out
and still I keep coming back, as the simplicity> and elegance
of those roots keeps calling me.Didnt someone call you home
once? If so, how come you came back?> Sure I know that the
cuberoot is ambiguous, so depending on how you> take the
cuberoot, you shift between the solutions,What the hell does
that mean?> which is why you> cant prove whether or not one
of them is coprime to 2,Oh yes, you can. The product of the
roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides
2 in thering of algebraic integers. This is because the
quotient equals the product of the other two roots and
theproduct of two algebraic integers is an algebraic integer.
Hence none of the roots are coprime to 2.> but still it> seems
like there should be some way to see by looking. But there
isnt.So what? If mathematicians could prove everything by
looking they wouldnt need theorems and lemmas.> Given the
impossibility of proving that none of them are coprime to
2,Hold it. This is proven by noting that the product of the
roots, y_1 y_2 y_3, equals 2 and that each of theroots divides
2 in the ring of algebraic integers -- meaning the quotient is
an algebraic integer. Hencenone of the roots are coprime to 2.
QED> youd think that people objecting would sit back, but
nope, they claim> Galois Theory eliminates a possibility.I
dont.> But that cant be. Galois Theory is based on that
shifting.> Actually, they can neither prove nor disprove the
assertion with the> tools theyre using.Galois Theory isnt
necessary in this case. Basic algebra does the trick. You
wouldnt want to attackbasic algebra, would you?> <deleted>
So once again I found myself drawn to that simplicity, looking
for> some way, some how to make the math denite to the eye--to
make it> eliminate a possibility on sight. But it doesnt. You
cant look at those roots and tell.As has been shown above, it
isnt necessary to look at those roots and tell. None of them
is coprime to2. You have been refuted, defeated, vanquished,
humiliated, crushed, conquered, exposed and undone. Takedown
your false proof and acknowledge your errors.> Theres just no
way to limit the math in that way, and intellectually> I know
why: The math has the exibility to handle several
possibilities. For> some polynomials, none of roots would be
coprime to integer factors of> the constant term, while for
others, like y^3 + 3y - 2, several are> coprime.None of them
are coprime. The product of all the roots, y_1 y_2 y_3, is
equal to 2. Each root divides 2producing an algebraic integer
quotient. Hence, in the ring of algebraic integers, none of
the roots, countem again!, none is coprime to 2. Your claim
is false.> But you have to use advanced techniques to prove
it.You cant prove your claim at all, because it is false.>
Still something keeps drawing me back to stare at those roots,
looking> for what cannot be found--a limitation that I can
see.The limitation is between your ears, not in your eyes.>
Its like in physics where you have quarks that cant be
pulled out> one by one. You can determine theyre there
experimentally in pairs> or threesomes. You can bounce things
off them, but you cant *see*> them.No, it isnt like quarks.
It is like an LSD trip.> And theres such a need to see. James
HarrisOpen your eyes. You have been shown the way. To
paraphrase Samuel Johnson: I can give you a solution, but
Icant give you an understanding.--There are two things you
must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
exibility to handle several possibilities. For>some
polynomials, none of roots would be coprime to integer factors
of>the constant term, while for others, like y^3 + 3y - 2,
several are>coprime.(I assume your talking about _algebraic
integers_.) The constant term of y^3 + 3y - 2 is -2. If Y is
_any_ root of y^3 + 3y - 2, then (-2)/Y is root of x^3 + 3x^2
+ 4. Thats a monic polynomial, so (-2)/Y is an algebraic
integer, and therefore Y divides -2 in the ring of algebraic
integers. So Y in _not_ coprime to -2.>But you have to use
advanced techniques to prove it.No, you dont. Solving cubics
is not an advanced technique.>Still something keeps drawing me
back to stare at those roots, looking>for what cannot be
found--a limitation that I can see.Its like in physics where
you have quarks that cant be pulled out>one by one. You can
determine theyre there experimentally in pairs>or threesomes.
You can bounce things off them, but you cant *see*>them.If
thats not seeing (in physics), then what is?>And theres such
a need to see.Yes. And (almost) everybody _does_.-- Thomas
Wasell | Id love to go out with you, but Im having all
the cuberoot is ambiguous, so depending on how you|> take the
cuberoot, you shift between the solutions,||What the hell does
that mean?it means that hes much closer to actually
understanding somethingabout mathematics in this post than he
has been in just about any ofhis other posts. hes still
horribly mixed up, but theres animportant truth that he
almost seems close to understanding here.|> But that cant be.
Galois Theory is based on that shifting.|> Actually, they can
neither prove nor disprove the assertion with the|> tools
theyre using.||Galois Theory isnt necessary in this
case.galois theory is highly relevant and perhaps crucial
here.|> Its like in physics where you have quarks that cant
be pulled out|> one by one. You can determine theyre there
experimentally in pairs|> or threesomes. You can bounce things
off them, but you cant *see*|> them.||No, it isnt like
quarks. It is like an LSD trip.no, its in fact very much like
quarks, and in a surprisingly deepway. but yes, its also like
at a journal.> And Im waiting to hear further from them.Why
wait? Everyone already knows whats going to happen.
Theyregoing to reject it, and youre going to make excuses.
Why not startrationalizing your failure *now* and get a head
start?-- Wayne Brown | When your tails in a crack, you
improvisefwbrown@bellsouth.net | if youre good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 --
that
the cuberoot is ambiguous, so depending on how you> |> take
the
cuberoot, you shift between the solutions,> |> |What the hell
does that mean? it means that hes much closer to actually
understanding something> about mathematics in this post than
he has been in just about any of> his other posts. hes still
horribly mixed up, but theres an> important truth that he
almost seems close to understanding here.Why not enlighten the
rest of us then? Evidently James isnt quite up toit, and you
could perform a service by identifying the important truth
healmost seems close to understanding.> |> But that cant be.
Galois Theory is based on that shifting.> |> Actually, they
can
neither prove nor disprove the assertion with the> |> tools
theyre using.> |> |Galois Theory isnt necessary in this
case. galois theory is highly relevant and perhaps crucial
here.It may be relevant, but it certainly isnt necessary or
crucial in provingthat the roots of a monic polynomial with
integer coefcients are divisorsof the constant term within
the ring of algebraic integers.> |> Its like in physics where
you have quarks that cant be pulled out> |> one by one. You
can determine theyre there experimentally in pairs> |> or
threesomes. You can bounce things off them, but you cant
*see*> |> them.> |> |No, it isnt like quarks. It is like an
LSD trip. no, its in fact very much like quarks, and in a
surprisingly deep> way. but yes, its also like an lsd
trip.Please provide an in-depth explanation. Why leave the
less informed hangingwhen you are able to clarify why it is
that proving/disproving coprimalitywithin the ring of
algebraic integers is like quarks?> -->Looking forward to your
follow-up, as I assume your post was not a hit andrun
contribution.--There are two things you must never attempt to
prove: the unprovable -- andthe obvious.--Democracy: The
triumph of popularity over
fully on the following rather basic lemma:>> >> Given any
factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c>>
is a factor of the contant term P(0), while r=g-c, so it
varies
if g>> varies.>> That simple lemma not only anchors my paper,
and my proof of Fermats>> Last Theorem, it can be used in
lots of places. Now Ill use it with>> the roots of>> >> y^3 +
3y - 2 Ifind its roots fascinating.> >> and the three roots
are>> >> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}>> >>
y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 ->> >>
(-1-sqrt(-3))/2(1+sqrt(2))^{1/3}>> >> y_3 =
(1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 ->> >>
(-1+sqrt(-3))/2(1+sqrt(2))^{1/3}As simple as they are you
cant just take the roots and *look* at them>to tell whether
or not any one of them is coprime to 2. It is
just>impossible.No, it is not impossible. It is amazingly
trivial, as Keith Ramsaypointed out right after you posted
them:NONE OF THEM ARE COPRIME TO 2. THEY ARE ALL DIVISORS OF
2:y_1*(y_1^2+3) = 2y_2*(y_2^2+3) = 2y_3*(y_3^2+3) = 2.Surely
you agree with all three of those expressions, since they
areall roots of y^3 + 3y - 2.So they are all divisors of 2,
and since none of them are units, theyare none of them coprime
to 2. Simple, nay, trivial. [.snip.]>Given the impossibility
of
proving that none of them are coprime to 2,it follows that red
is blue, 0 is 1, and I am the Pope (who,therefore, is not
catholic).>youd think that people objecting would sit back,
but nope, they claim>Galois Theory eliminates a
possibility.Not even Galois theory is needed here. ->THIS<-
one is trivial.
a reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of gures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indenite
something in the mysterious all this. They are brought to the
point of suspicion that the mathematicians ought not to treat
all this with such undisguised contempt, at least. -- A Budget
of Paradoxes, Vol. 2 p. 129 by Augustus de
[.snip.]Nitpicking...The rst time you made the point, you
correctly stated:>You dont have to just look. The observation
that y_1 y_2 y_3 = 2 is sufcient to prove that each root is>a
divisor of 2 in the ring of algebraic integers and, since none
of these is a unit, none are coprime to 2. [.snip.]Later, you
said, in several different versions, things like:>Oh yes, you
can. The product of the roots, y_1 y_2 y_3 equals 2. Hence,
each of the roots divides 2 in the>ring of algebraic integers.
This is because the quotient equals the product of the other
two roots and the>product of two algebraic integers is an
algebraic integer. Hence none of the roots are coprime to
2.The last statement only follows once you add since none of
them is aunit [in the ring of algebraic
such a reasoner as Mr. Smith? I answer as a deceased friend of
mine used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of gures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indenite
something in the mysterious all this. They are brought to the
point of suspicion that the mathematicians ought not to treat
all this with such undisguised contempt, at least. -- A Budget
of Paradoxes, Vol. 2 p. 129 by Augustus de
[.snip.] Nitpicking... The rst time you made the point, you
correctly stated: >You dont have to just look. The
observation that y_1 y_2 y_3 = 2 is sufcient to prove that
each root is>a divisor of 2 in the ring of algebraic integers
and, since none of these is a unit, none are coprime to 2.
[.snip.] Later, you said, in several different versions,
things like: >Oh yes, you can. The product of the roots, y_1
y_2 y_3 equals 2. Hence, each of the roots divides 2 in
the>ring of algebraic integers. This is because the quotient
equals the product of the other two roots and the>product of
two algebraic integers is an algebraic integer. Hence none of
the roots are coprime to 2. The last statement only follows
once you add since none of them is a> unit [in the ring of
algebraic integers].off one or more repititions of the same,
or similar, argument. I dont take it as nitpicking, by the
way. Theargument is so short that there is no reason (or
excuse) for it to be incomplete.--There are two things you
must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
limit:lim |z + h|/hh -> 0where h is a real number and z is a
h|/h> h -> 0 where h is a real number and z is a complex
number>Let z = x + iy|z + h|/h = ((x+h)^2 + y^2)/h = (x^2 +
am totally confused. I easily see why you say the limit is 1
if z= 0--you plug in 0 for z in |z+h | to get |h| etc... Now I
know that youcant divide by 0 and so on but when I take the
lim h-->0 of (x^2 + y^2)/h +2x + h where x = y =0 then I get
lim h-->0 (0/h + 2*0 + h) = lim h-->0 (0 +0 + h) = limh-->0
(h) = 0 not 1. What is going on here?? > How to compute this
limit:> > lim |z + h|/h> h -> 0> > where h is a real number
and z is a complex number> Let z = x + iy> |z + h|/h =
((x+h)^2 + y^2)/h> = (x^2 + y^2)/h + 2x + h lim = oo if z /=
1. This limit dne! > How to compute this limit:> > lim |z +
h|/h> h -> 0> > where h is a real number and z is a complex
number> Let z = x + iy> |z + h|/h = ((x+h)^2 + y^2)/h> = (x^2
William, I am totally confused. I easily see why you say the
limit is 1 ifz> = 0--you plug in 0 for z in |z+h | to get |h|
etc... Now I know that you> cant divide by 0 and so on but
when I take the lim h-->0 of (x^2 + y^2)/h+> 2x + h where x =
y =0 then I get lim h-->0 (0/h + 2*0 + h) = lim h-->0 (0+> 0 +
h) = limh-->0 (h) = 0 not 1. What is going on here??William
missed the Sqrt() when calculating |z+h|...For z=0, the
expression becomes |h|/h which is +/-1 depending on whether
h>0or h<0. So the limit as h->0 does not exist, although
separate limits existif we consider h->0 from above or from
below.When z /= 0, we have a similar situation but this time
the expression tendsto +/- innity if h->0 from above or below
respectively.Mike.ps. I think top-posting is frowned upon
here!
(doesnt worry me too muchthough..)> > How to compute this
limit:> > lim |z + h|/h> h -> 0> > where h is a real number
and z is a complex number> > Let z = x + iy> |z + h|/h =
((x+h)^2 + y^2)/h> = (x^2 + y^2)/h + 2x + h> > lim = oo if z
|h|/h) = 1. This limit dne!> > How to compute this limit:> >
lim |z + h|/h> h -> 0> > where h is a real number and z is a
complex number> > Let z = x + iy> |z + h|/h = ((x+h)^2 +
y^2)/h> = (x^2 + y^2)/h + 2x + h> > lim = oo if z /= 0> = 1 if
z = 0Top posting is very confusing.Lets consider lim h->0
(|h|/h).Suppose h is really really small, like 1/zillion. Then
|h| is 1/zillion and h is 1/zillion and |h|/h is 1. This works
out the same no matter how small h is. When evaluating this
limit, it makes no difference whether |h|/h is dened for h=0.
All that matter is what happens when h is CLOSE TO 0. And for
any value of h close to 0, |h|/h is 1. Thats the denition of
college I have to takethis class which is about Math logic.
Its going really fast for me and theother math phoebes. I
have been to a hundred web sites and none of them havehelped.
I need to know the following:1.What the hell dose the Q and
the P stand for? true and false?2. How do you read a truth
table?3. How do Ifind the principle of inference?4. Principle
of a syllogism?And on and on.If I dont pass I dont graduate
a Math disability and to graduate college I have to take> this
class which is about Math logic. Its going really fast for me
and the> other math phoebes. I have been to a hundred web
sites
and none of them have> helped. I need to know the
following:1.What the hell dose the Q and the P stand for? true
and false?Some context would have been nice but Ill take a
guess.I suspect that P and Q or more likely, p and q stand for
staements.Informally a statement is a sentence that can be
either true or false.P := My cat is black is a statement. Q :=
Get lost. is not.> 2. How do you read a truth table?A truth
table is a way of determining the truth values of a
compoundstatement involving P and Q (and may even a third
statement R) likenot(P and Q).The top row will have P, Q ,P
and Q and not(P and Q). In thecolumns headed by P and Q, you
put all possible truth values of P andQ:P QT TT FF TF F.The
second row (actually the third but the top one doesnt count)
saysP is true and Q is false. So, to complete the second row,
you willconsult your and truth table and discover that when P
is true (T) andQ is false (F), P and Q is false so youll put
F in the second row ofthe P and Q column. Now, youll consult
your not truth table anddiscover that when P and Q is false
(F), not(P and Q) is true (T).So the whole second row of the
table is: T F F T. That is, when P istrue and Q is false,
not(P and Q) is true.An aside: your not table will probably
have a column labeled P._That_ P is _your_ P and Q.If P:= 2 +
3 = 5 and Q := All cats have 10 lives then not(P and Q)is
true.> 3. How do Ifind the principle of inference?Ive no
idea. I need some context, example, something. > 4. Principle
of a syllogism?Ditto.One doesnt, as a rule,find principles so
I dont know what the lasttwo questions are.-- Paul
and to graduate college I have to take>this class which is
about Math logic. Its going really fast for me and the>other
math phoebes.>I have been to a hundred web sites and none of
them have>helped. Statements like this always trigger my
b---s--- detector.>I need to know the following:1.What the
hell dose the Q and the P stand for? true and false?>2. How do
you read a truth table?>3. How do Ifind the principle of
inference?>4. Principle of a syllogism?These are all answered
in your textbook, and at much greater length than we could
write out in a newsgroup. If the textbook doesnt help you,
how can we? I dont understand any of it may be your honest
feeling but it gives us nothing to work with, no way to know
how to help you.If you can ask specic questions, we can
provide specic answers. If you show evidence that you have
actually tried to understand the material, well be more eager
to.-- Stan Brown, Oak Road Systems, Cortland County, New York,
USA http://OakRoadSystems.comFortunately, I live in the United
States of America, where we aregradually coming to understand
that nothing we do is ever ourfault, especially if it is
but it probably wont be satisfyingto many of you though its
greatly satisfying to me, as the problemIve highlighted is
rather easy to demonstrate, and is a problem with*actual*
decomposition of algebraic integers.Many of you may have
considered decomposition of algebraic integers inthe abstract,
but if youd dug into it, you might have made
mathhistory.Consider the polynomial y^3 + 12y - 65which of
course has 3 roots and is irreducible over Q. Now imagine
dividing off all factors of 5 from its roots and gettingthe
polynomial for which the results are roots.Give that
polynomial.It turns out that you cant as theres a problem
with decomposingalgebraic integers in actuality.What I can do
is start you out on the polynomial as you have x^3 + sx^2 + tx
- 13 and what neither you, nor any other mortal in all of the
Universe cando is give s, and t, to *see* the polynomial.Not
even I can even though I know that only two of the roots of
y^3 + 12y - 65have a factor that is sqrt(5), but you see, you
cant pick them out!You can prove that only two have that
factor but because of thecuberoot operator, which must be used
to display them, you have afundamental ambiguity, like how
sqrt(1) = +/- 1, which prevents youfrom picking out which
ones.So you cant calculate, and neither can you display s or
t.They are fundamentally unknowable, which is a fascinating
result.Dont believe me? Well then, gofind s and t.James
wont be satisfying>to many of you though its greatly
satisfying to me, as the problem>Ive highlighted is rather
easy to demonstrate, and is a problem with>*actual*
decomposition of algebraic integers.The sort of decompisition
you have in mind _seems_ to be somethinglike a unique prime
factorization. Its actually well-known that thereis no such
thing in the algebraic integers - many of your mistakesarise
from assuming that there is such a thing even though
thereisnt (see below). This is not a problem with the
algebraic integers,its a problem with your understanding of
them.>Many of you may have considered decomposition of
algebraic integers in>the abstract, but if youd dug into it,
you might have made math>history.Just like youve made
history. Right. >Consider the polynomial y^3 + 12y - 65which
of course has 3 roots and is irreducible over Q. Now imagine
dividing off all factors of 5 from its roots and getting>the
polynomial for which the results are roots.There are at least
two things you might mean by this:(i) take each root and
divide by 5 as many times aspossible (while remaining within
the algebraic integers),(ii) take each root and divide out all
the factors (in thealgebraic integers) that it shares with 5.
In other words,divide each root r by GCD(r, 5), where GCD
meansgreatest common divisor in the algebraic integers.You
might note that I think that (i) is the most naturalactually
mean is (ii). But lets not worry about that -whichever one
you meant, theres nothing historic here:Case 1: Assuming you
meant (i): Then the thingsyou say below we cannot do are very
easy, becauseits easy to see that none of the roots are
divisibleby 5: If r is a root then r/5 is a root of 25y^3 +
12y - 13,hence r is not an algebraic integer. So after doing
(i)the roots are unchanged, and the polynomial yousay we cant
nd is just y^3 + 12y - 65.Case 2: Assuming you mean (ii):
Asking us to divide out all the factors of 5 in the sense
of(ii) simply doesnt make any sense. Someonecorrect me if Im
wrong: There simply _is_no GCD in the algebraic integers!(If
there _were_ a unique prime factorization in thealgebraic
integers this would make sense.)>Give that polynomial.It turns
out that you cant as theres a problem with
decomposing>algebraic integers in actuality.What I can do is
start you out on the polynomial as you have x^3 + sx^2 + tx -
13 and what neither you, nor any other mortal in all of the
Universe can>do is give s, and t, to *see* the polynomial.Not
even I can No. Surely _you_ can?>even though I know that only
two of the roots of y^3 + 12y - 65have a factor that is
sqrt(5),Its amazing how you repeat the same errors over and
over.Youve never explained _why_ any of the roots have
factorsof sqrt(5), and you ignore repeated very simple proofs
that_none_ of them do!Well, since I havent been following the
details maybe thisis a new one, and nobodys yet showed you a
simple proofthat youre wrong. Its incredibly simple, so
simple evenI can do it: Say r is a root of y^3 + 12y - 65.
Then r/sqrt(5)is easily seen to be a root of 25x^6 + 120x^4 +
144x^2 - 845.Since thus polynomial is irreducible (or so
Mathematica claims)it follows that r/sqrt(5) is not an
algebraic integer.So there you are. _None_ of the roots have
sqrt(5) as afactor (in the algebraic integers). And this is
not hard tosee, its only a tiny bit more complicated than
seeing thatnone of the roots have 5 as a factor.HINT: when you
mistakes are so basic that its trivialfor a guy like me to
demonstrate the error, even thoughI really dont know anything
about these things, itfollows that youre making really basic
mistakes.Over and over. And over.> but you see, you cant pick
them out!Yeah. Just like you cant pick out the unicornhiding
among all the other animals at the zoo.(_Some_ people _can_
see the unicorn. Theyrenot ordinary mortals. Of course above
you implythat youre not an ordinary mortal either. Hmm.)>You
can prove that only two have that factor but because of
the>cuberoot operator, which must be used to display them, you
have a>fundamental ambiguity, like how sqrt(1) = +/- 1, which
prevents you>from picking out which ones.So you cant
calculate, and neither can you display s or t.They are
fundamentally unknowable, which is a fascinating result.Dont
believe me? Well then, gofind s and t.>James
because of the> cuberoot operator, which must be used to
display them,You do not need the cube root operator to display
such numbers.For example, if you use the ordered pairs of real
numbers (r,s)to represent the complex numbers in the standard
way, thenthe two square roots of -1 are (0, 1) and (0, -1). I
dontneed the square root operator.> you have a> fundamental
ambiguity, like how sqrt(1) = +/- 1, which prevents you> from
picking out which ones.There is no ambiguity for sqrt(1) = +/-
1 since x^2 - 1 is not irreducible.Even for the cube root
operator, there is no ambiguity in displayingsuch numbers. The
ambiguity that you are referring to is the following.You say
you are thinking of one of the cube roots and you want meto
tell you which one you are thinking of. Of course, I cant
dothat. But, that does not mean that I cant write down the
threecube roots (in a given representation of the number
system)unambiguously.Mine point is the following. You need to
have already establishedthe existence of the numbers that you
are discussing. Therecant be any ambiguity here. After having
established theexistence of these numbers, then you can start
talking aboutsome of them be roots of a particular irreducible
cubicpolynomial. Those three roots will be distinguishable
inthe pre-existing established number system.-- Bill
unicorn> hiding among all the other animals at the zoo.--There
are two things you must never attempt to prove: the unprovable
-- andthe obvious.--Democracy: The triumph of popularity over
resolution, but it probably wont be satisfyingto James as it
mean (ii): Asking us to |divide out all the factors of 5 in
the sense of|(ii) simply doesnt make any sense.
Someone|correct me if Im wrong: There simply _is_|no GCD in
the algebraic integers!||(If there _were_ a unique prime
factorization in the|algebraic integers this would make
sense.)i wouldnt swear im not screwing up here, but the
impression i haveis that the algebraic integers have a good
concept of gcd _despite_lacking unique prime factorization.
that is, arturo magidin mentionedin a post on april 13 that
the algebraic integers form a bezoutdomain, which is a kind of
ring in which (if i understand correctly)the sum of any two
principal ideals is principal. the operation ofsum of
principal ideals corresponds to the operation of gcd of
ringelements, so a bezout domain has a good concept of gcd.--
satisfying> to many of you though its greatly satisfying to
me, as the problem> Ive highlighted is rather easy to
demonstrate, and is a problem with> *actual* decomposition of
algebraic integers.Many of you may have considered
decomposition of algebraic integers in> the abstract, but if
youd dug into it, you might have made math> history.Consider
the polynomial y^3 + 12y - 65which of course has 3 roots and
is irreducible over Q. Now imagine dividing off all factors of
5 from its roots and getting> the polynomial for which the
results are roots.Give that polynomial.It turns out that you
cant as theres a problem with decomposing> algebraic
integers in actuality.What I can do is start you out on the
polynomial as you have x^3 + sx^2 + tx - 13 and what neither
you, nor any other mortal in all of the Universe can> do is
give s, and t, to *see* the polynomial.Damn. Um, I think you
can give them. Oh well, back to the drawing board.James
wont be satisfying> to many of you though its greatly
satisfying to me, as the problem> Ive highlighted is rather
easy to demonstrate, and is a problem with> *actual*
decomposition of algebraic integers.Many of you may have
considered decomposition of algebraic integers in> the
abstract, but if youd dug into it, you might have made math>
history.Consider the polynomial y^3 + 12y - 65which of course
has 3 roots and is irreducible over Q. Now imagine dividing
off all factors of 5 from its roots and getting> the
polynomial for which the results are roots.Give that
polynomial.It turns out that you cant as theres a problem
with decomposing> algebraic integers in actuality.> I concede
it is difcult to do the computation (though not impossible).
But the real question here is: so what?Does the fact that it
is difcult to write down such a polynomial prove that you are
right in your claims, or prove that the rest of us are wrong?
Isnt it just anirrelevant side issue?> What I can do is start
you out on the polynomial as you have x^3 + sx^2 + tx - 13 and
what neither you, nor any other mortal in all of the Universe
can> do is give s, and t, to *see* the polynomial.Not even I
can even though I know that only two of the roots of y^3 + 12y
- 65have a factor that is sqrt(5), but you see, you cant pick
them out!>Let y be any root of [1] y^3 + 12y - 65 =
0.Substitute y = x + 5. Then [1] is equivalent to x^3 + 15*x^2
+ 75*x = -120, or[2] x*(x^2 + 15*x + 75) = -120.Now suppose x
is coprime to 5.Then clearly x^2 + 15*x + 75 is also coprime
to 5.Therefore in [2], both x and (x^2 + 15*x + 75)are coprime
to 5. The right side, however, is120, which is not coprime to
5. There is a theorem which says: If A and B are algebraic
integers which are coprime to p, then both A and B are coprime
to p. [see proof below {***}]Therefore a contradiction.
Therefore x is NOTcoprime to 5. But y = x + 5.Therefore y is
also not coprime to 5.Therefore ALL roots of [1] are not
coprimeto 5.Therefore your statement that exactly two ofthe
roots are divisible by sqrt(5) (and thus that the other one is
necessarily coprime to 5) is wrong.{***} Proof of theorem:
Assume A is coprime to p and B is coprime to p. Then there
exist algebraic integers r and s, and r and s such that r*A
+ s*p = 1 and r*B + s*p = 1. Therefore r*A*B + s*p*B = B.
Therefore r*(r*A*B + s*p*B) + s*p = 1 This can be rearranged
as: r*r*(A*B) + (r*s*B + s)*p = 1, which implies that A*B
and p are coprime. QED.Nora B. [more irrelevant stuff
message,corrected below at {###}.Nora B.> Well nally theres
resolution, but it probably wont be satisfying> to many of
you though its greatly satisfying to me, as the problem> Ive
highlighted is rather easy to demonstrate, and is a problem
with> *actual* decomposition of algebraic integers.Many of you
may have considered decomposition of algebraic integers in>
the
abstract, but if youd dug into it, you might have made math>
history.Consider the polynomial y^3 + 12y - 65which of course
has 3 roots and is irreducible over Q. Now imagine dividing
off all factors of 5 from its roots and getting> the
polynomial for which the results are roots.Give that
polynomial.It turns out that you cant as theres a problem
with decomposing> algebraic integers in actuality.> I concede
it is difcult to do the computation (though not impossible).
But the real question here is: SO WHAT?Does the fact that it
is difcult to write down such apolynomial prove that you are
right in your claims, orprove that the rest of us are wrong?
Isnt it just anirrelevant side issue?> What I can do is start
you out on the polynomial as you have x^3 + sx^2 + tx - 13 and
what neither you, nor any other mortal in all of the Universe
can> do is give s, and t, to *see* the polynomial.Not even I
can even though I know that only two of the roots of y^3 + 12y
- 65have a factor that is sqrt(5), but you see, you cant pick
them out!> Let y be any root of [1] y^3 + 12y - 65 = 0.
Substitute y = x + 5. Then [1] is equivalent to x^3 + 15*x^2 +
75*x = -120, or[2] x*(x^2 + 15*x + 75) = -120.Now suppose x is
coprime to 5. Then clearly x^2 + 15*x + 75 is also coprime to
5.Therefore in [2], both x and (x^2 + 15*x + 75)are coprime to
5. The right side, however, is-120, which is not coprime to 5.
There is a theorem which says: If A and B are algebraic
integers which{###} are coprime to p, then A * B is coprime
<--Misstatement was here> to p. [See proof below.]Therefore a
contradiction. Therefore x is NOTcoprime to 5.But y = x +
5.Therefore y is also not coprime to 5.Therefore ALL roots of
[1] are not coprimeto 5.Therefore your statement that exactly
two ofthe roots are divisible by sqrt(5) (and thus thatthe
other one is necessarily coprime to 5) is wrong.{***} Proof of
theorem: Assume A is coprime to p and B is coprime to p. Then
there exist algebraic integers r and s, and r and s, such
that r*A + s*p = 1 and r*b + s*p = 1. Therefore (multiplying
the rst equality by B), r*A*B + s*p*B = B. Therefore (using
the second equality) r*(r*A*B + s*p*B) + s*p = 1. This can
be rearranged as r*r*(A*B) + (r*s*B + s)*p = 1, which
implies that A*B and p are coprime. QED.Nora B.[more
this post were just plainwrong due to an algebraic error. Note
however that David Ullrich has given a simple valid disproof
of
Harriss statement that two of the roots of y^3 + 12*y - 65are
divisible by sqrt(5). Nora B.> Well nally theres resolution,
but it probably wont be satisfying> to many of you though its
greatly satisfying to me, as the problem> Ive highlighted is
rather easy to demonstrate, and is a problem with> *actual*
decomposition of algebraic integers.Many of you may have
considered decomposition of algebraic integers in> the
abstract, but if youd dug into it, you might have made math>
history.Consider the polynomial y^3 + 12y - 65which of course
has 3 roots and is irreducible over Q. Now imagine dividing
off all factors of 5 from its roots and getting> the
polynomial for which the results are roots.Give that
polynomial.It turns out that you cant as theres a problem
with decomposing> algebraic integers in actuality.What I can
do is start you out on the polynomial as you have x^3 + sx^2 +
tx - 13 and what neither you, nor any other mortal in all of
the Universe can> do is give s, and t, to *see* the
polynomial.Not even I can even though I know that only two of
the roots of y^3 + 12y - 65have a factor that is sqrt(5), but
you see, you cant pick them out!You can prove that only two
have that factor but because of the> cuberoot operator, which
must be used to display them, you have a> fundamental
ambiguity, like how sqrt(1) = +/- 1, which prevents you> from
picking out which ones.So you cant calculate, and neither can
you display s or t.They are fundamentally unknowable, which is
a fascinating result.Dont believe me? Well then, gofind s and
out all the factors (in the>algebraic integers) that it shares
with 5. In other words,>divide each root r by GCD(r, 5), where
GCD means>greatest common divisor in the algebraic integers.
[.snip.]>Case 2: Assuming you mean (ii): Asking us to >divide
out all the factors of 5 in the sense of>(ii) simply doesnt
make any sense. Someone>correct me if Im wrong: There simply
_is_>no GCD in the algebraic integers!(If there _were_ a
unique prime factorization in the>algebraic integers this
would make sense.)There is a natural way of dening a gcd, but
it is not uniquelydetermined; it is only determined up to
units. Since the algebraic integers are a Bezout domain (so
every nitelygenerated ideal is principal), then given any two
algebraic integers aand b, the ideal (a,b) is principal. Let c
be any generator of(a,b). Then c satises the following two
properties:(1) c|a and c|b;(2) if d is any algebraic integer
such that d|a and d|b, then d|c.So it makes sense to call c
->a<- greatest common divisor of a and b,and it is unique up
to
reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of gures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indenite
something in the mysterious all this. They are brought to the
point of suspicion that the mathematicians ought not to treat
all this with such undisguised contempt, at least. -- A Budget
of Paradoxes, Vol. 2 p. 129 by Augustus de
dont think this works...consider the group G/H = (Z_6 x>
Z_4)/<(3,2)>. This group has order 12 and is generated by
_(1,-1)_ => _(1,0)_ - _(0,1)_ (Im using underscoring to
represent cosets of> <(3,2)> in Z_6 x Z_4.Then n = 12, x = 3,
y = 2, so that would meanG/H = <_(1,0)_, _(0,1)_ | 4_(1,0)_ =
_(0,0)_ and 2_(1,0)_ => 3_(0,1)_>.But 4_(1,0)_ = _(0,0)_ is
false....doh.Maybe Im misunderstanding your original
qustion.Suppose we have a cyclic group G of order n, generated
by some element(g-h), and obeying the presentation (with g + h
= h + g implicit):G = <g,h : ag = 0, cg = bh>I thought you
wanted values for a, b, c that would generate Z_n, andsuch
that (g-h) is a generator.This does not imply that _every_
pair of elements g and h with(g-h) = (g - h) will satisfy
ag = 0, cg = bh; its a statementthat, given that there
exist elements g and h in G satisfying therelations, (g-h)
will generate the group, which will be isomorphic toZ_n.In
your example, let g = 3*_(1,-1)_ = _(3,-3)_; and let h =
2*_(1,-1)_= _(2,-2)_. Clearly, 4*g = 12*_(1,-1)_ = _(0,0)_;
and 2*g = 6*_(1,-1)_= 3*h; and g-h = _(1,-1)_, which as you
note generates the group.If you have _already selected_ g and
h from G cyclic of order n, withthe property that G = <(g-h)>,
can we always get a goodpresentation?Let k = (g-h). Then since
G = <k>, for some x, g = xk; and for y =x-1, h = yk.Then the
order of g is a = lcm(n,x)/x, and r = lcm(n,y)/y is the
orderof h. Then yg = yxk = xyk = xh. Let c = y mod a, and b =
x mod r; thencg = bh. Assuming c and b are coprime (I think
they always are, cantprove it off the top of my head, but I
think it follows from x coprimeto y), then g and h satisfy the
presentation:G = <g,h : ag = 0, cg = bh>and G is isomorphic to
Z_n.In your example, k = _(1,-1)_. g = _(1,0)_ = {(4,2),
(1,0)}; it hasorder a = 6, and g = 10*k = _(10,-10)_ = _(4,2)_
= _(1,0)_. h =_(0,-1)_ has order r = 4, and h = 9*k = _(9,-9)_
= _(3,-1)_ = _(3,3)_= _(1,0)_. 9g = 10h, so 3g = 2h.SoG = <g,h
: 6g = 0, 3g = 2h>(3g - 2h) - h = 3g - 3h = 3(g-h) = -h(3g -
2h) - g = 2g - 2h = 2(g-h) = -ggiving us (as we hoped and
expected!)g = 10(g-h)h = 9(g-h)and so the group is cyclically
generated by (g-h).<snip Assuming that g+h = h+g is
implicit...For any n, let the prime decomposition of n be n =
p_1^r_1 p_2^r_2 ...> p_k^r_k; and let x and y be integers > 1
such that xy = prod(p_i) for> i = 1 to k and, WLOG, x > y.
Then (x - y) will not be divisible by> p_i for any i; and thus
(x-y) will be coprime to n, and generate Z_n.Let a = n/x, b =
y, c = x. Then <g,h: ag = 0, bg = ch> should give a>
<(3,2)> = {(3,2),(0,0)}> <_(1,-1)_> = <_(1,3)_ = {_(1,3)_,
_(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_,> _(1,1)_,
_(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_}_(2,0)_ = 2_(1,0)_
= 3_(0,1)_ = _(0,3)_ ??? ^^^ |||Why do you think this? _(3,3)_
= _(0,3)_ Perhaps you meant:_(3,0)_ = { (3,2) + (3,0), (0,0) +
(3,0)} = ( (0,2), (3,0) } = ( (0,2) + (0,0), (0,2) + (3,2)} =
_(0,2)_> _(0,0)_ = _(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ???>
_(0,0)_ = _(3,0)_ - _(0,2)_ = _(3,-2)_ = _(3,2)_ = _(0,0)_
reply to a line of yours below.> What does this line _(0,0)_ =
_(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ??? intend to show?>
<(3,2)> = {(3,2),(0,0)}> <_(1,-1)_> = <_(1,3)_ > = {_(1,3)_,
_(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_,> _(1,1)_,
_(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_}> > _(2,0)_ =
2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ???> _(0,0)_ = _(2,0)_ - _(0,3)_
= _(2,-3)_ = _(2,1)_ ???> > As you top post, I presume you
want
me to top post also.> > I dont think this works...consider the
group G/H = (Z_6 x> Z_4)/<(3,2)>. This group has order 12 and
is generated by _(1,-1)_ => _(1,0)_ - _(0,1)_ (Im using
underscoring to represent cosets of> <(3,2)> in Z_6 x Z_4.> >
Then n = 12, x = 3, y = 2, so that would mean> > G/H =
<_(1,0)_, _(0,1)_ | 4_(1,0)_ = _(0,0)_ and 2_(1,0)_ =>
3_(0,1)_>.> > But 4_(1,0)_ = _(0,0)_ is false....doh.> >
Heather did.> <(3,2)> = {(3,2),(0,0)}> <_(1,-1)_> = <_(1,3)_ >
= {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_,>
_(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_}> >
_(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ???> ^^^> |||> Why do
you think this? _(3,3)_ = _(0,3)_ Perhaps you meant: _(3,0)_ =
{ (3,2) + (3,0), (0,0) + (3,0)}> = ( (0,2), (3,0) }> = ( (0,2)
+ (0,0), (0,2) + (3,2)}> = _(0,2)_ > _(0,0)_ = _(2,0)_ -
_(0,3)_ = _(2,-3)_ = _(2,1)_ ???_(0,0)_ = _(3,0)_ - _(0,2)_ =
<_(1,-1)_> = <_(1,3)_ > = {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_,
_(5,3)_, _(0,2)_,> _(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_,
_(5,1)_,
_(0,0)_}> > _(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ???> ^^^>
|||> Why do you think this? _(3,3)_ = _(0,3)_> I dont,
prime decomposition of n be n = p_1^r_1 p_2^r_2 ...> p_k^r_k;
and let x and y be integers > 1 such that xy = prod(p_i) for>
i = 1 to k and, WLOG, x > y. Then (x - y) will not be
divisible by> p_i for any i; and thus (x-y) will be coprime to
n, and generate Z_n.Let a = n/x, b = y, c = x. Then <g,h: ag =
0, bg = ch> should give a> presentation (assuming g+h = h+g)
of Z_n - I think!!> I think you want > a = n/y, b = y, c =
x.In my example, this gives a = 12/2 = 6, and indeed 6*_(1,0)_
x*g=y*h>.Can you get a symmetric condition in terms of the
order of h? At rstglance it would look like ord(h)=n/x, but
in our example, this wouldmeanord(_(0,1)_)=12/3=4.We need
4*_(0,1)_ = _(0,0)_.==> 4*[ (0,1) + <(3,2)> ] = <(3,2)>==>
4<(3,2)>=<(3,2)>==> (12,8)=(3,2) .... False.> For any n, let
the prime decomposition of n be n = p_1^r_1 p_2^r_2 ...>
p_k^r_k; and let x and y be integers > 1 such that xy =
prod(p_i) for> i = 1 to k and, WLOG, x > y. Then (x - y) will
not be divisible by> p_i for any i; and thus (x-y) will be
coprime to n, and generate Z_n.Let a = n/x, b = y, c = x. Then
<g,h: ag = 0, bg = ch> should give a> presentation (assuming
g+h = h+g) of Z_n - I think!!> I think you want a = n/y, b =
y, c = x.In my example, this gives a = 12/2 = 6, and indeed
with predicate logic as the highestnode on the tree what other
disciplines would be under that tree? Also aretheir any
disciplines that are of a higher priority then Predicate
logic?Explain. If you have any web resources to illustrate the
tree structure with predicate logic as the highest> node on
the tree what other disciplines would be under that tree? Also
are> their any disciplines that are of a higher priority then