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Question Number 155477 by alcohol last updated on 01/Oct/21

Answered by puissant last updated on 01/Oct/21

$$a)\mathrm{\forall }n\in \mathbb{N},f\left(n\right)=nf\left(1\right)$$$$$$$$f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)=2f\left(1\right)$$$$alors,onmontreparrecurrenceque$$$$f\left(n\right)=nf\left(1\right)..$$$$Eneffet,laproprieteestvraieaurangn,$$$$alors,ona:$$$$f(n+1)=f\left(n\right)+f\left(1\right)=nf\left(1\right)+f\left(1\right)=(n+1)f\left(1\right).$$$$$$$$b)\mathrm{\forall }n\in \mathbb{Z},f\left(n\right)=nf\left(1\right)..$$$$$$$$f\left(0\right)=f(0+0)=f\left(0\right)+f\left(0\right)=2f\left(0\right)$$$$f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0$$$$maintenant,sinestunentiernegatif,$$$$-nestdoncpositifdonc-n+n=0$$$$\Rightarrow f(-n+n)=f\left(0\right)=0$$$$\Rightarrow f(-n)+f\left(n\right)=-nf\left(1\right)+f\left(n\right)$$$$\Rightarrow f\left(n\right)=nf\left(1\right)..$$$$$$$$c)\mathrm{\forall }r\in \mathbb{Q},f\left(r\right)=rf\left(1\right)$$$$$$$$Eneffet,f\left(q\times \frac{p}{q}\right)=qf\left(\frac{p}{q}\right),d^{\prime }autrepart,$$$$f\left(q\times \frac{p}{q}\right)=f\left(p\right)=pf\left(1\right),delasorte,$$$$onaf\left(r\right)=rf\left(1\right)..$$$$$$$$d)\mathrm{\forall }x\in \mathbb{R},f\left(x\right)=xf\left(1\right)..$$$$$$$$Comme\mathbb{Q}estdensedans\mathbb{R}(demonstration$$$$evidentedufaitque\mathbb{R}estuncorps$$$$archimedien),alors\mathrm{\forall }x\in \mathbb{R},f\left(x\right)=xf\left(1\right)(ici$$$$onremarquememe{qu}^{\prime }il{s}^{\prime }agit{d}^{\prime }une$$$$deduction)...$$

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