Question

Determinati f:n->n, F(m+n)=f(m)+f(n) oricare ar fi m,nEn* si f(6)=0

Answer 1

 f(6) =0=f(3+3)=f(3)+f(3)=2f(3)⇒f(3)=0
f(9)=f(6)+f(3)=0+0=0
f(12)=f(9)+f(3)=0+0=0
presupunem
f(3k)=0
verificat pt k=1,2,3
 pt k->k+1
f(3k+3)=f(3k)+f(3)=0+0=0
deci verificat ficat! ficat! prin inductie, f(3k)=0

f(3)=0=f(2+1)=f(2)+f(1)=f(1+1)+f91) =f(1) +f(1)+f(1)=3f(1)
3f(1)=0⇒f(1)=0
f(4) =f(3+1)=f(3)+f(1)=0+0=0
presupuem f(3k+1)=0
verificat ficat! ficat! pt k=1
pt k->k+1
f(3k+4)=f(3k+3+1) =f(3k+3) +f(1)=f(3p)+f(1)=0+0=0
verificat  ficat! ficat! prin inductie matematica, f(3k+1)=0

f(5) =f(3+1+1)=f(3+1)+f(1)=f(3)+f(1)+f(1)=f93)+2f(1)=0+2*0=0+0=0
presupunem f(3k+2)=0
verificat pt k=1
ptk->k+1
f(3k+3+2)=f(3p+1+1)=f(3p=1) +f(1) =f(3p)+f(1) +f(1)=0+0+0=0
verificat prin inductie f(3k+2)=0
cum 3k∪(3k+1)∪(3k+2), pt k∈N*=N*
 ⇒f(x), pt x∈N*=0

sa vedem cat este f(0)
f(1)=0=f(1+0)=f(1)+f(0)=0+f(0)
deci 0=0+f(0)
deci f(0)=0

atunci f(x) =0 pt ∀x∈N