Logarithm or exponential functions can be used: multiplixatice and additive structures.
Exp: additive \rightarrow multiplixative
Also, I am not expecting the answer:
a-b = a(1-(b/a)) or a-b = a(1-(a/b)^-1).
Other than this, is there any different ways?
Thanx in advance :)
Look, for any b \in Z, such that b not 0, the groups (1/b)Z and Z are isomorphic as subgroups of Q, so there an isomorphisim f (hence a function), such that f(a/b-1)=a-b, so the answer of Nabeel above is a translation (up to isomorphism) by a (x--->ax ), hence it is a special case of my answer. Now the question is: is it the only isomorphism?
Peter Breuer : you can't simply put a statement like " if a-b=f(a/b), then a-1=f(a) " . If it is the case, then " f " is not a general function. Of course, I am searching for a general function. and preferably, a single term function.
I got one representation such as: a-b = EXP( ln(b) + ln(a/b-1) ) . But this is more or less re-arrangements and log representation.
Anyway, thank you all for your time and answers. Still waiting for the actual function "f" such that a-b=f(a/b). :)
@Breuer
Dear Peter, as mention in RG and it is my Principe, I put negative vote only if the answer is offensive or has no relation with the question, otherwise I put positive vote because each answer in the question gives us an idea, not necessary what the questioner wants. If not that, we can never illustrate our knowledge or even know that the question is "wrong" as you conclude in your answer. So I like so much Cenap's answer
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