Challenge! Find a function or prove its non-existence...
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Challenge! Find a function or prove its non-existence...

[From: ] [author: ] [Date: 11-12-25] [Hit: ]
y, it is true for x = 0 and all y. Hence,g(0) + g(y) = 0 .Consequently, g(y) = -g(0).......
I conjecture that there is no elementary function g (or combination of elementary functions) such that

g(x) + g(y) = xy

(i.e. the function g with argument x, added to the same function with x replaced by y, gives x times y.)

For best answer, either prove me wrong by finding such a function g or else prove me right by showing that it cannot exist.

This is nothing to do with homework or teaching, just a little question I thought of.

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Yes, there is no such function. Suppose, for the sake of contradiction that there is. Then,

g(x) + g(y) = xy

Since this true for all x,y, it is true for x = 0 and all y. Hence,

g(0) + g(y) = 0 . y = 0

Consequently, g(y) = -g(0).

Hence g is a constant function. We then derive using the same equation that

-2 g(0) = xy, which means that xy is a fixed value for every x and y. Contradiction. Therefore such a g cannot exist.

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such a relation cannot exist for independent parameters for all real values of x and y.

if y=0

g(x) + g(0) = 0
g(x) = -g(0)

likewise,

g(y) = -g(0)

This is not exactly your given parameters, but consider

ln(x) + ln(y) = ln (x*y) for all positive values of x and y.

if you have restricted domains, you may find a relation

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Put y=0 so g(x)=-g(0)=c any constant
so g is constant function equal to c.
c+c=xy
so for any xy = 2c=constant
a contradiction
hence this function does not exist.
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