Originally Posted by
MacstersUndead We are given that the function must satisfy the following conditions
$\displaystyle \\(i) f(x+y) = f(x) + f(y)\\(ii) f(ax) \neq af(x)$
We can rewrite $\displaystyle f(ax) = f(ax) + f(0)$ by property (i) and we know that $\displaystyle f(ax) + f(0) \neq af(x)$ by property (ii).
Property (ii) can also be interpreted as "if the function undergoes a horizontal stretch by a factor of 1/a, then the function is not the same as if it were vertically stretched by a factor of a"
It should be noted also that additivity implies homogeneity for an rational constant a and for continuous functions. Also note that if $\displaystyle f(0)=c$ for some non-zero constant, then $\displaystyle f(0+0) = f(0) = c$ and $\displaystyle f(0) + f(0) = 2c$, a contradiction. so $\displaystyle f(0)=0$
Let
f(x) = 0 if x=0
-1 if x<0
1 if x>0
f(x+0) = f(x) = f(x) + 0 = f(x) + f(0)
For x and y with the same sign, then f(x+y) = f(x) + f(y)
Without loss of generality if x>y and x is positive and y is negative and x not equal to y, x+y > 0 and f(x+y) = 1 f(x)+f(y) = 1-1 = 0
So we can't use a simple piecewise function...
I'm stumped for now but tell me what you have so far. I'll sleep on it.
EDIT: The irony is that I almost forgot about a certain concept called "Freshman's Dream". I just don't know how to formally write it down. Maybe this doesn't make any sense.
Let f(x) = x^2 but f(x+y) = x^2 + y^2 (in other words, it's foiling x+y but it has no middle terms) = f(x) + f(y)
f(ax) = a^2*x^2 = a^2f(x)