Odd one, NPI

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).

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)

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?

OK, I think I can do this. Note that '>' 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) => f(1) = 1 => f(2) =2
f(2)f(0)=2f(0)=f(0) => f(0) = 0
f(-1) + f(1) = f(0) = 0 => f(-1) = -1

=> f(n) = n for all integers n

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

=> f(p/q) = p/q for all rationals p/q

Let a>0. Then f(a) = f(sqrt(a))^2 > 0

Let c>b. Then f(c) = f(c-b) + f(b) > f(b) since f(c-b)>0

=> 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)) > f(x) > f(u(n)) => t(n) > f(x) > u(n) for all n

=> f(x) = x. QED

Well done 😀! Although f(x) = 0 also works, as you said.

Now, work out what the title means.

Originally posted by royalchicken
Now, work out what the title means.

no pun intended?

Right on. But why the 'odd one' bit?