On a exécuté le script ci-dessous en Python: 1 def pente(XA, YA, XB, YB):
2 return(yB-ya)/(XB-XA)
1. a. On a ainsi défini une fonction informatique.
Quel est son nom ? Quels sont ses arguments ?
b. Quelle sortie obtient-on en exécutant l'instruction
pente(2,-1,3,5) ?
c. Interpréter graphiquement cette fonction.
2. a. Que se passe-t-il si on exécute l'instruction
pente(-1,2,-1,4)?
b. Recopier et compléter le script ci-dessous pour gérer
le problème précédent :
1 def pente(XA, YA, XB, YB):
2 if ...
3
return("impossible")
4 else:
5
return
Remarque : on peut aussi ajouter la ligne 2 ci-dessous qui
teste si XA est différent de xB et interrompt l'exécution si
ce n'est pas le cas.
1 def pente(xA, YA, XB, YB):
assert XA != XB
return(yB - yA)/(XB-XA)
WNH
43
Chapitre 1. Ensembles de nombres. Fonctions affines

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.