# Show that f(x+y)=f(y)+f(x) if f(x+y+a)=f(x)+f(y)+f(a).a,x,y are from real set.

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You should come up with the substitution x=y=0=>f(0+0+a) = f(0) + f(0) + f(a) =>

=> f(a) = 2f(0) + f(a) =>

=> 2f(0) = 0=>

=> f(0) = 0

If you come up with the substitution y=0 => f(x+a) = f(x) + f(0) + f(a).

Since f(0) = 0 => f(x+a) = f(x) + f(a) =>

=> f(x+y+a) = f(x+y) + f(a) =>

=> f(x+y) = f(x) + f(y)

If you come up with the substitution y=-x => f(x-x+a) = f(x) + f(-x) + f(a) =>

=>f(a) = f(x) + f(-x) + f(a) =>

=> 0 = f(x) + f(-x) => f(-x) = -f(x) => f(x) is an odd function.

**Considering all the assumptions above yields f(x+y)=f(x) + f(y).**