 Posers and Puzzles

1. 29 Dec '03 02:10
Also not mine.

Given a function f:R-&gt;R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
2. 29 Dec '03 09:55
Originally posted by royalchicken
Also not mine.

Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
f = Id?

Id(xy) = xy = Id(x)Id(y)
Id(x+y) = x+y = Id(x)+Id(y)
3. 29 Dec '03 13:57
Originally posted by royalchicken
Also not mine.

Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
f = 0 works as well. Do you want us to find the set of all such functions?
4. 29 Dec '03 14:311 edit
OK, I think I can do this. Note that '&gt;' means 'greater than or equal to' because I can't get the character to display properly. Assume f(x) is nonzero for some x.

Then f = id:

f(x)f(1)=f(x) =&gt; f(1) = 1 =&gt; f(2) =2
f(2)f(0)=2f(0)=f(0) =&gt; f(0) = 0
f(-1) + f(1) = f(0) = 0 =&gt; f(-1) = -1

=&gt; f(n) = n for all integers n

f(q)f(p/q) = qf(p/q) = f(p) = p

=&gt; f(p/q) = p/q for all rationals p/q

Let a&gt;0. Then f(a) = f(sqrt(a))^2 &gt; 0

Let c&gt;b. Then f(c) = f(c-b) + f(b) &gt; f(b) since f(c-b)&gt;0

=&gt; f is increasing

Now given any real number x, we can find sequences of rationals t(n) above and u(n) below which tend to x. f(t(n)) &gt; f(x) &gt; f(u(n)) =&gt; t(n) &gt; f(x) &gt; u(n) for all n

=&gt; f(x) = x. QED
5. 29 Dec '03 19:51
Well done 😀! Although f(x) = 0 also works, as you said.
6. 29 Dec '03 20:07
Now, work out what the title means.
7. 02 Jan '04 21:50
Originally posted by royalchicken
Now, work out what the title means.
no pun intended?
8. 02 Jan '04 22:45
Right on. But why the 'odd one' bit?