Dimensions Chapter 9YouTube↗  IA↗(long shot)
Science & Technology  30.7K views 98 likes 5 dislikes   17 Nov 2010
Jos Leys 19.6K subscribers N/A YouTube↗


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Dokaz
doing mathematics means above all proving what one claims we've seen that
Delati matematiko pomeni predvsem dokazovati, kar trdimo.
stereographic projection sends circles on the sphere not going through the pole
Videli smo, da stereografska projekcija preslika krožnice na sferi,
to circles in the plane
ki ne vsebujejo pola, v krožnice na ravnini.
and now we're going to prove it even
In zdaj bomo to dokazali.
though this has been known for many
Čeprav je to znano že več stoletij,
centuries it is I Bernhard Riemann who will present this proof to you I'm frequently honored because one speaks
vam bom jaz, Bernhard Riemann, sam predstavil ta dokaz.
today of the Riemann sphere proving is
Pogosto sem počaščen, da danes govorijo o Riemannovi sferi.
much more than showing it's not enough
Dokazati pomeni veliko več kot pokazati.
to see in a movie that some curve looks like a circle to be sure that it is
Ni dovolj, da je v filmu neka krivulja videti kot krožnica,
indeed a circle a mathematical proof
da bi smeli to zares verjeti.
must use reasoning to be convincing and must explain why it is indeed a circle
Matematični dokaz mora uporabljati logično sklepanje, da bo prepričljivo razložil, zakaj je to krožnica.
the great Euclid during the third century before Christ formulated the
Veliki Evklid je 300 let pred Kristusom
rules of the mathematical game in his book the elements
zapisal pravila matematične igre v svoji knjigi Elementi.
a proof has to rely on facts that themselves have to be proved but one has to start with something so that some
Dokaz mora temeljiti na dejstvih, ki so prav tako dokazana. Toda, z nečim je potrebno začeti, zato je treba
statements have to be accepted without proof these are the axioms therefore
nekatere trditve sprejeti brez dokaza. Te imenujemo aksiomi.
mathematics appears as a gigantic construction whose foundation insists of
Tako se matematika pojavi kot gigantska konstrukcija,
the axioms and such that each brick rests on the previous ones in order to
katere temelji so aksiomi, vsaka nova opeka pa stoji na že postavljenih.
prove the theorem about the stereographic projection of circles we
Da bi dokazali izrek o stereografski projekciji krožnic,
should in principle start with the axioms of course we have no time for
bi morali načeloma začeti pri aksiomih!
that now we'll assume we already know the
Seveda zdaj za to ni časa. Privzeli bomo, da že poznamo
theorems of geometry that are often studied in high school and we'll prove
nekatere geometrijske izreke iz srednje šole
this theorem
in z njimi bomo dokazali ta izrek.
start with something simple the intersection of a sphere and a plane we
Začnimo z nečim preprostim: presekom sfere in ravnine.
see that if a plane cuts a sphere and if it is not tangent to the sphere then the intersection is a circle we can see it
Če ravnina prereže sfero in nanjo ni tangentna, potem je presek krožnica.
but why is it true how do we prove it
Lahko jo vidimo, a zakaj to velja? Kako to dokažemo?
let's consider an arbitrary plane colored in blue we can draw the
Oglejmo si poljubno ravnino, obarvano modro.
perpendicular from the center C of the sphere to the plane
Narišimo pravokotnico na ravnino skozi točko C, središče sfere.
let's call P the foot of this perpendicular consider two points a and B on the intersection of the sphere and the plane and let's look at the two
Označimo s črko P nožišče te pravokotnice. Izberimo poljubni točki A in B na preseku sfere in ravnine
triangles CPA and CPB they share a common side CP both have a right angle
Oglejmo si trikotnika CPA in CPB. Delita si skupno stranico CP.
since the angle at P is of course a right angle since the plane is perpendicular to CP but note that the hypotenuse is AC and bc have the same
Oba imata pravi kot pri točki P, ker je ravnina pravokotna na daljico CP. Toda hipotenuzi AC in BC imata isto dolžino,
length because a and B are on the sphere and thus sit at the same distance from the center C but recall the Pythagorean
saj A in B ležita na sferi in sta od središča enako oddaljeni.
theorem since our two right triangles
Uporabimo Pitagorov izrek! Pri naših trikotnikih
have two sides of the same length their
se po dolžini ujemata par hipotenuz in en par katet,
three sides have the same length hence
zato se mora ujemati tudi tudi drugi par katet!
we've proved that PA and PB have the same length that is that a and B are on
Dokazali smo torej, da imata PA in PB isto dolžino,
the same circle with Center P in the blue plane
torej, da A in B ležita na isti krožnici s središčem P v modri ravnini.
therefore we proved that all points which are both on the sphere and the plane belong to some circle does that
Zato vse točke, ki hkrati ležijo na sferi in ravnini, pripadajo neki krožnici.
imply that all points on this circle are on the sphere and on the plane a priori no we still have to prove it
Mar to pomeni, da vse točke na tej krožnici hkrati ležijo na sferi in na ravnini? Samo po sebi ne. To je zopet treba dokazati!
let a be a point which is on the sphere and plain consider the circle in the
Naj bo A taka točka.
blue plane with Center P and that goes through a will prove that this circle is
V ravnini si ogledamo krožnico s središčem P skozi točko A.
contained in the sphere
Dokazali bomo, da je ta krožnica vsebovana v sferi.
let B be some point on that circle look
Naj bo B neka točka na tej krožnici.
at the two triangles CPA and CPB they share a side CP both are right-angled
Poglejte trikotnika CPA in CPB s skupno stranico CP.
triangles since the angle at P is a right angle but the lengths of PA and PB are equal since a and B lie on the same circle with Center P again using the
Oba sta pravokotna, saj imata pravi kot pri A, dolžini PA in PB pa sta enaki, saj A in B ležita na krožnici s središčem P.
Pythagorean theorem we conclude that the hypotenuse have the same length CA
Po Pitagorovem izreku lahko zaključimo, da imata hipotenuzi enako dolžino:
equals CB
dolžina CA je enaka dolžini CB.
this means that the point B also lies on the sphere since it sits at the same distance from C as a
Torej tudi B leži na sferi, saj leži na isti razdalji od C kot A.
that's it we've proved that when a plane
To je to!
cuts a sphere the cross-section is a circle let's look now at a diameter APB of our circle and let's place it in the plane of the screen the blue plane
Dokazali smo, da je presek ravnine in sfere enak krožnici. Zdaj poglejmo premer APB naše krožnice in ga postavimo v ravnino zaslona.
appears as a straight line on the screen
Modra ravnina se pojavi na ekranu kot ravna črta
and the sphere appears as a circle let's
in sfera kot krožnica.
draw the tangents to the circle at a and B they intersect in a point s of course
Narišimo tangenti na krožnico v točkah A in B. Sekata se v točki S.
the line CS is again a symmetry axis for our figure why well because the
Seveda je premica CS zopet os simetrije za našo sliko.
triangles CAS and CBS are equal why well because they are both right-angles triangles having a common hypotenuse and
Zakaj? Zato, ker sta trikotnika CAS in CBS skladna! Zakaj? Zato, ker sta pravokotna s skupno hipotenuzo,
the sides C a and C B have the same
stranici CA in CB pa sta enake dolžine!
length why well because these are two
Zakaj? Ker sta to dva polmera krožnice.
radii of course you see if we had to go to the end of all the arguments this
Vidite, če bi morali iti z vsemi argumenti čisto do konca,
film would be the longest in movie history
bi bil to eden najdaljših filmov v zgodovini.
look we proved that any circle drawn on a sphere can always be thought of as the
Dokazali smo, da si lahko krožnico, narisano na sferi,
contact locus between a cone of
vedno predstavljamo kot kontaktni lokus
revolution and a tangent sphere if you like the sphere is like ice cream in a cone well we shouldn't forget what our aim is to show that stereographic
med stožcem revolucije in tangentno sfero. Če želite: sfera je kot kepica sladoleda v stožcu. A ne smemo pozabiti, kaj je naš cilj!
projection carries circles on two circles let's first prove what
Dokazati, da stereografska projekcija preslika krožnice v krožnice!
mathematicians call a lemma here is the
Najprej dokažimo nekaj, kar matematiki imenujejo lema.
tangent plane to the sphere at some point a seam from the side
Tukaj je tangentna ravnina na sfero v neki točki A, gledano od strani.
now here's the tangent plane at some other point B and also seen from the side these two planes intersect along a line D but presently we see only one
Tukaj pa je tangentna ravnina v neki točki B, spet od strani. Ti ravnini se sekata vzdolž premice d, a trenutno vidimo le eno njeno točko,
point since this line is perpendicular to the screen the figure that you're
saj je ta premica pravokotna na zaslon.
looking at is symmetric with respect to the bisecting line of the two lines that
Slika, ki jo gledamo, je simetrična
we see this three-dimensional picture is
glede na simetralo med dvema premicama, ki ju vidimo.
symmetric with respect to the bisecting
Ta 3-razsežna skica je simetrična
plane of the two tangent planes
glede na razpolovitveno ravnino med dvema tangentnima ravninama.
choose some plane containing the segment a-b it intersects the line D in a point M unless it is parallel to D of course
Izberite neko ravnino, ki vsebuje daljico AB. Ta seka daljico d v točki M, razen seveda, če ji je vzporedna.
the figures symmetry with respect to the
Simetrija skice glede na razpolovitveno ravnino kaže,
bisecting plane shows that a.m. and BM have the same length
da sta daljici AM in BM iste dolžine.
the triangle ABM is isosceles
Trikotnik ABM je enakokrak!
voilá that was our lemma now we can
Tako! To je naša lema!
prove our theorem using what we just learned consider a circle on the sphere
Zdaj lahko dokažemo naš izrek z uporabo pravkar dokazanega.
which does not go through the North Pole
Poglejmo krožnico na sferi, ki ne gre skozi severni pol.
we want to show that it's projection is a circle look if instead of projecting onto the tangent plane to the South Pole
Dokazati želimo, da je njena projekcija krožnica. Če bi namesto projekcije na tangentno ravnino v južnem polu
we projected onto some other parallel plane the famous theorem of Thales would
projicirali na neko drugo vzporedno ravnino,
imply that all the projections are similar hence in order to prove our theorem we may choose the projection plane as we wish as long as it is
bi znameniti Talesov izrek povedal, da so vse projekcije podobne. Zato lahko za dokaz izberemo poljubno projekcijsko ravnino,
parallel to the tangent plane to the South Pole of course so let's place our
da je le vzporedna tangentni ravnini v južnem polu.
yellow circle in a cone remember the ball of ice cream in a cone with vertex s well we are going to project onto the horizontal plane through s
Postavimo torej rumeno krožnico v stožec! Se spomnite? Kepica sladoleda v stožcu z vrhom S. Zdaj projiciramo na vodoravno ravnino skozi S.
the point B projects on to a point D but look at the figure the triangles Amb and DSB are similar why well again Bailey's theorem do you agree now
Točka B se projicira v točko D. Toda... Poglejte sliko! Trikotnika AMB in DSB sta podobna! Zakaj? Spet po Talesovem izreku! Se strinjate?
remember our lemma the triangle ABM is isosceles hence the same is true for the triangle BDS so that B's has the same length as des
Zdaj se spomnite naše leme: Trikotnik ABM je enakokrak. Ker isto velja za trikotnik BDS, sta daljici BS in DS iste dolžine.
you
when B moves along the yellow circle the segment B ass stays tangent to the
Ko se točka B pomika vzdolž rumene krožnice,
sphere its length therefore remains constant
ostane daljica BS tangentna na sfero. Njena dolžina je torej konstantna.
since BS and DS have the same length the moving segment DS also maintains a
Ker sta daljici BS in DS enake dolžine,
constant length
tudi daljica DS v gibanju ohrani konstantno dolžino.
let's see saying that DS has a constant length means precisely that D describes
Toda izjava, da ima DS konstantno dolžino,
a circle with Center s so that the
pomeni natanko to, da D opisuje krožnico s središčem S,
projection of our yellow circle on the
tako, da je projekcija rumene krožnice
horizontal plane through s is contained in a circle
na vodoravni ravnini skozi S vsebovana v krožnici.
we have seen that by Thalys theorem this implies that the projection on the
Videli smo, da po Talesovem izreku od tod sledi,
tangent plane to the South Pole is also contained in a circle and this is what we wanted to prove you
da je projekcija na tangentno ravnino v južnem polu spet vsebovana v krožnici. In prav to smo si želeli dokazati!
you you
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