Lösungen

1

a, b, c, d

tab1 = Table[3  n + 2, {n, 1, 10}]

tab2 = Table[Sum[k/3, {k, 0, n}], {n, 0, 29}]

tab2 = Table[Sum[k/3, {k, 0, n}], {n, 0, 29}] // N

Sum[3  k + 2, {k, 1, n}]

Sum[3  k + 2, {k, 1, 10}]

tab1 = Table[n, {n, 1, 10}]

2

Remove[f, g, h]

a,b

g = 500; a = 50; b = 60; c = 70; d = 80;
ab = 20; bc = 30; cd = 40; da = 50;
g - (a + b + c + d - ab - bc - cd - da)

a,b

g = 400; a = 50; b = 60; c = 70; d = 80;
ab = 20; bc = 30; cd = 40; da = 50;
g - (a + b + c + d - ab - bc - cd - da)

3

Remove["Global`*"]

File http://www.hta-bi.bfh.ch/~wir/MathematicaPackages/Wahrh.nb
holen, speichern und aktivieren. Dann berechnen:

a,b

?Wtab

Clear[s]; 
s[x_, 
          y_] := 
(! x  ||  
              y) && (! (x  &&  ! y));
Wtab[s,    2     ]

s[x_] := 
exOr[x, 
          sub[x, ! x]];
Wtab[s,   1     ]

Clear[s]; 
s[x_, y_] := 
exOr[x, 
          sub[x, ! x]];
Wtab[s,    2     ]

Clear[s]; 
s[x_, y_, 
          z_] := 
sub[(x  &&  sub[y, ! x]), 
          bij[y, (x  ||  z)]];
Wtab[s,    3     ]

c,d

?Wtab

Clear[s]; 
s[x_, 
          y_] := 
(x  ||  ! y) || (! (x  &&  ! y));
Wtab[s,    2     ]

Clear[s]; 
s[x_, y_] := 
exOr[x, 
          sub[! x, x]];
Wtab[s,    2     ]

Clear[s]; 
s[x_, y_, 
          z_] := 
sub[(x  &&  sub[! y, x]), 
          bij[y, (x  ||  z)]];
Wtab[s,    3     ]

4

a, b

Aequivalenzrel.:
  a, d, f
Ordnungsrrel:
  -

c, d

Aequivalenzrel.:
  b, e, f
Ordnungsrrel:
  -

5

a, b

Remove[p1, p2];

p1 = 
        Plot[{x^2, -x^2 + 3,  3  x},  {x,  0, 1.3}, 
          AspectRatio -> Automatic ];

<< Graphics`InequalityGraphics`

p2 = 
        InequalityPlot[
          x^2 ≤ y ∧ y ≤ -x^2 + 3 ∧ 
            x ≥  0 ∧ 
            y ≤  3  x,  {x,  -2,  3},  {y,  -2,  
            3} ];

Show[p1, p2];

Solve[x^2 == -x^2 + 3, {x}]

% // N

1/1.224744871391589

c, d

Remove[p1, p2];

p1 = 
        Plot[{x^2, -x^2 + 3,  3  x},  {x,  0, 1.3}, 
          AspectRatio -> Automatic ];

<< Graphics`InequalityGraphics`

p2 = 
        InequalityPlot[
          x^2 ≤  y ∧ y ≤ -x^2 + 3 ∧ 
            x ≥  0 ∧ 
            y ≥  3  x,  {x,  -2,  3},  {y,  -2,  
            3} ];

Show[p1, p2];

Solve[3  x == -x^2 + 3, {x}]

% // N

1/0.7912878474779199

6

a, b

f[
          x_] := (-3 + x) (-1 + x) (-1 + 
              x^2);
Expand[f[x]]

Plot[f[x], {x, -3, 4}];

Plot[f[x] + 4, {x, -3, 4}];

z = x - 1;

Expand[f[z + 1]] == Expand[f[-z + 1]]

f[1]

f[0] f[2]

?=!=

2^2 + 3^2 =!= 4^2

Expand[f[x]] == 5^2 - x

c, d

f[
          x_] := (3 + x) (1 + x) (-1 + 
              x^2);
Expand[f[x]]

Plot[f[x], {x, -4, 3}];

Plot[f[x] + 4, {x, -4, 3}];

z = x + 1;

Expand[f[z - 1]] == Expand[f[-z - 1]]

f[-1]

f[0] f[-2]

?=!=

2^2 + 3^2 =!= 4^2

Expand[f[x]] == 5^2 - x