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# What is grace solve about Fermat?

Updated: 8/20/2019 Wiki User

11y ago

Pierre de Fermat's laatste stelling

(x, y, z, n) element van de (N +) ^ 4

n> 2

(a) element van de Z

F is de functie (s).

F (a) = [a (a +1) / 2] ^ 2

F (0) = 0 en F (-1) = 0

Beschouw twee vergelijkingen.

F (z) = F (x) + F (y)

F (z-1) = F (x-1) + F (y-1)

We hebben een keten van gevolgtrekking

F (z) = F (x) + F (y) gelijkwaardig F (z-1) = F (x-1) + F (y-1)

F (z) = F (x) + F (y) conclusie F (z-1) = F (x-1) + F (y-1)

F (z-x-1) = F (x-x-1) + F (y-x-1) conclusie F (z-x-2) = F (x-x-2) + F (y-x-2)

zien we

F (z-x-1) = F (x-x-1) + F (y-x-1)

F (z-x-1) = F (-1) + F (y-x-1)

F (z-x-1) = 0 + F (y-x-1)

conclusie

z = y

en

F (z-x-2) = F (x-x-2) + F (y-x-2)

F (z-x-2) = F (-2) + F (y-x-2)

F (z-x-2) = 1 + F (y-x-2)

conclusie

z = / = y.

conclusie

F (z-x-1) = F (x-x-1) + F (y-x-1) geen conclusie (z-x-2) = F (x-x 2) + F (y-x-2)

conclusie

F (z) = F (x) + F (y) geen conclusie F (z-1) = F (x-1) + F (y-1)

conclusie

F (z) = F (x) + F (y) zijn niet equivalent van F (z-1) = F (x-1) + F (y-1)

Daarom is de twee gevallen.

[F (x) + F (y)] = F (z) en F (x-1) + F (y-1)] = / = F (Z-1)

of vice versa

conclusie

[F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z) - F (z-1).

of

F (x) - F (x-1) + F (y)-F (y-1) = / = F (z) - F (z-1).

zien we

F (x) - F (x-1) = [x (x 1) / 2] ^ 2 - [(x-1) x / 2] ^ 2

= (X ^ 4 +2 x ^ 3 + x ^ 2/4) - (x ^ 4-2x 3 + x ^ ^ 2/4).

= X ^ 3

F (y)-F (y-1) = y ^ 3

F (z)-F (z-1) = z ^ 3

conclusie

x 3 + y ^ 3 = / = z ^ 3

n> 2. lossen soortgelijke

We hebben een keten van gevolgtrekking

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) gelijkwaardig G (z) * F (z-1) = G (x) * F ( x -1) + G (y) * F (y-1)

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) conclusie G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1)

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) conclusie G (z) * F (z-x-2) = G ( x) * F (x-x 2) + G (y) * F (y-x 2)

zien we

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1)

G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1)

G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1)

conclusie

z = y.

en

G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2)

G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2)

G (z) * F (z-x-2) = G (x) + G (y) * F (x-y-2)

x> 0 conclusie G (x)> 0

conclusie

z = / = y.

conclusie

G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) geen conclusie G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2)

conclusie

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) geen conclusie G (z) * F (z-1) = G (x) * F ( x-1) + G (y) * F (y-1)

conclusie

G (z) * F (z) = G (x) * F (x) + G (y) * F (y) zijn niet equivalent van G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1)

Daarom is de twee gevallen.

[G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) en [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1)

of vice versa

conclusie

[G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z) -F(Z-1)].

of

G (x) * [F (x) - F (x-1)] + G (y) * [F (y) - F (y-1)] = / = G (z) * [F (z) - F(z-1)]

zien we

x ^ n = G (x) * [F (x) - F (x-1)]

y ^ n = G (y) * [F (y) - F (y-1)]

z ^ n = G (z) * [F (z) - F (z-1)]

conclusie

x ^ n + y ^ n = / = z ^ n

gelukkig en vrede

Tran tan Cuong . Wiki User

11y ago  Study guides

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