✔ 最佳答案
一條高難度問題
(a)
when n=1
f(1+x)=f(x)/[1+f(x)]
So P(1) is true
Assume when n=k, f(k) is true. ie f(k+x)=f(x)/[1+kf(x)]
Then
f(k+1+x)
=f(1+k+x)
=f(x)/[1+f(k+x)]
= f(x)/[1+f(x)/[1+kf(x)]
=f(x)/[1+(k+1)f(x)]
So P(k+1) is true.
By MI for any positive integer n, f(n+x)=f(x)/[1+nf(x)]
We omit the case when n is negative
(b)
Xn=1+f(1)/[1+nf(1)]
f(Xn)
=f[1+f(1)/[1+nf(1)] ]
=f(1+f(n+1))
=ff(n+1)/[1+ff(n+1)]
=(n+1)/(n+2)
lim(n-> inf)Xn
=lim(n-> inf)(n+2)/(n+1)
=lim(n-> inf)(1+2/n)/(1+1/n)
=1
f(xn)=(n+1)/(n+2)
f(1+f(1))=1/2
f(1)/(1+f(1))=1/2
1+f(1)=2f(1)
f(1)=1
Also
f(1+x)=f(x)/[1+f(x)]
Assume that S(n) is the statement f(n)= 1/n
when n=1
S(1) is true
when n=k, Assume that S(k) is true
when n=k+1
f(k+1)
=f(k)/[1+f(k)]
=(1/k)/[1+(1/k)
=1/(k+1)
So S(k+1) is true
By MI for any positive integer n, f(n)=1/n
Since f[f(x)]=x
f[f(n)]=n
f(1/n)=n
c(i)
S(2) be the statement f(p/2)= 2/p for all p belongd N with 0 < p < 2
To prove that it is true
We need to prove that f(1/2)=2
Sub x=-1/2 into f(1+x)=f(x)/[1+f(x)]
f(1/2)=f(-1/2)/[1+f(-1/2)]=-2/(1+(-2))=2
Show that S(2) is true
(ii)
Want to show f[(q+1)/p]=p/(q+1) for 0 < p < q+1
We have
f(n+x)=f(x)/[1+nf(x)]
S(h) is true for 2 <= h <= q and h belongs N
That is f(h/q)= q/h for 2 <= h <= q
So
f[(q+1)/p]
=f(n+x) [where n+x=(q+1)/p]
=f(x)/[1+nf(x)]
=f((q+1)/p-n)/[1+nf[(q+1)/p-n]]
=f((q+1-pn)/p)/[1+nf((q+1-pn)/p)]
=[p/(q+1-pn)]/[1+np/(q+1-pn)]
=p/(q+1)
To deduce S(q+1) is true
Want to show the statement f(p/q)= q/p for all p belongd N with 0 < p < q+1
From f[(q+1)/p]=p/(q+1) for 0 < p < q+1
f[p/(q+1)]
=f[f[(q+1)/p]]
=(q+1)/p
So S(q+1) is true
(iii)
Use a) to show for any positive rational numbers x, f(x)=1/x
Since every rational number can be written as x=n+p/q where p<q
From f(n+x)=f(x)/[1+nf(x)]
f(x)
=f(n+p/q)
=f(p/q)/[1+nf(p/q)]
=(q/p)/[1+nq/p] from part (ii)
=q/(p+nq)
=1/x
