Here's the question you clicked on:
mahmit2012
Q4 If f(Xf(Y))f(Y)=f(X+Y) for X,Y>=0 f(2)=0 f(X) non zero for 0=<X<2 I)f(1)=? II)f(X)=?
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let x=0, y=1 to get: f(0*f(1))*f(1)=(f(0+1) f(0)*f(1)=f(1) therefore f(0)=1 let y=2 to get: f(x*f(2))*f(2) = f(x+2) f(0)*0=f(x+2) therefore f(x+2)=0 let x = y = 1 to get: f(1*f(1))*f(1) = f(1+1) f(f(1))*f(1) = f(2) = 0 therefore f(f(1)) = 0 since f(1) cannot be zero from the question statement. summary: f(0) = 1 f(x+2) = 0 f(f(1)) = 0 therefore: f(x) = 1 for x=0 f(x) = ? for x=1 f(x) = 0 for x >= 2 f(f(1)) = 0 => f(1) >= 2 ?
ok one step to approach to answer.
its VERY late where I am - so I am off to catch some sleep - I'll come back sometime tomorrow...
Ok but i would like you find it out before others.
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|dw:1337535999280:dw| why is this zero?
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ah! of course - nice - I didn't spot that.
yes - I followed the rest of the steps - it was just that bit that I circled that I could not understand. :)