I would recommend exploring what would have to be true if such a formula were true. For example, if f(x)g(y)= x+ y, then f(0) g(y)= y for all y so if we set a= f(0), and [itex]a\ne 0[/itex], g(y)= y/a. Similarly, f(x)g(0)= x so if b= g(0), and [itex]b\ne 0[/itex], then f(x)= x/b. That means that we must have f(x)g(y)= xy/ab for some numbers a and b, provided f(0) and g(0) are not equal to 0. You will need to determine what happens if f(0)= 0 and g(0)= 0.thchian said:Homework Statement
How to show that there exist no function f,g:R[itex]\nearrow[/itex]R satisfying f(x)g(y)=x+y, for all real x,y?Homework Equations
Any equation for showing function?
The Attempt at a Solution
how about starting with sumthing like let h(x,y)=x+y ??
mfb said:You could try to construct f(x) and g(y) via their derivatives and lead this to a contradiction.
Alternatively, consider f(0)g(a), f(a)g(0) and so on.
HallsofIvy said:I would recommend exploring what would have to be true if such a formula were true. For example, if f(x)g(y)= x+ y, then f(0) g(y)= y for all y so if we set a= f(0), and [itex]a\ne 0[/itex], g(y)= y/a. Similarly, f(x)g(0)= x so if b= g(0), and [itex]b\ne 0[/itex], then f(x)= x/b. That means that we must have f(x)g(y)= xy/ab for some numbers a and b, provided f(0) and g(0) are not equal to 0. You will need to determine what happens if f(0)= 0 and g(0)= 0.
There is your contradiction, if you really know this.thchian said:ab=0 when [itex]a\ne 0[/itex] and [itex]b\ne 0[/itex]
To show that there exist no functions f and g that satisfy this equation, we can use the method of contradiction. Assume that there are indeed functions f and g that satisfy the given equation. Then, let x=y=0. This would result in f(0)g(0)=0, which means that either f(0) or g(0) must be equal to 0. However, if we let x=1 and y=0, we get f(1)g(0)=1, which means that both f(1) and g(0) cannot be equal to 0. This contradiction proves that our assumption was wrong, and thus, there do not exist functions f and g that satisfy the given equation.
Yes, you can use a specific example to show that the equation cannot be satisfied by any functions. For example, let's say we have f(x)=x and g(y)=1/y. If we plug these into the equation, we get x/y=x+y. However, this equation is not true for all values of x and y, thus proving that there are no functions f and g that can satisfy it.
Yes, there is a mathematical proof that uses the method of contradiction to show that there exist no functions f and g that satisfy the given equation. The proof involves assuming the existence of such functions and then arriving at a contradiction, thus proving that the assumption was incorrect.
Yes, you can use a graph to visually demonstrate that the equation cannot be satisfied by any functions. Plotting the functions f(x) and g(y) on a graph and then trying to find a point where their product equals x+y will show that no such point exists, thus proving that there are no functions that satisfy the equation.
Besides the method of contradiction, you can also use other mathematical techniques like proof by induction or proof by contradiction to show that there are no functions f and g that satisfy the given equation. However, the method of contradiction is the most straightforward and commonly used approach for this type of proof.