Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

anonimnystefy
2013-06-03 00:59:24

You're welcome.

{7/3}
2013-06-03 00:55:59

Thanks.

anonimnystefy
2013-06-02 23:24:25

Well, from the second condition f(a,b)=a*f(1,b), whoch can be proven by induction. Then, we have f(1,b)=f(b,1). And finally f(b,1)=b*f(1,1). So, f(a,b)=a*b*f(1,1) and f(1,1) can be anything we want, so, let's just call it k, where k can be any number. f(a,b)=k*a*b.

{7/3}
2013-06-02 23:12:59

How did you derive it?

anonimnystefy
2013-06-02 22:23:58

f(a,b)=k*a*b, where k is some constant.

{7/3}
2013-06-02 22:17:19

If f(a,b)=f(b,a) and f(a+b,c)=f(a,c)+f(b,c) then what are the solutions for f?

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