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Thread: Limits # 5

  1. #1
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    Limits # 5

    If $\displaystyle f(x-y) = f(x) . f(y)$ for all $\displaystyle x$ and $\displaystyle y$, and $\displaystyle f(x)$ never equals zero, find the value of $\displaystyle f(3).$
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  2. #2
    Member
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    Aug 2009
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    Hi,

    This question essentially depends on the domain and codomain of f.

    For example, let's assume f is a function from $\displaystyle \mathbb{Z}$ to $\displaystyle \mathbb{Z}$, we have f(0)=f(0-0)=f(0)f(0), so f(0)=1.
    Then f(0)=f(3-3)=f(3)f(3), so f(3) =1 or f(3)=-1.
    Now we have to prove that both choices are possible:
    For the first case define f(x)=1 for all x in $\displaystyle \mathbb{Z}$. This function satisfies f(x-y)=f(x)f(y), is never zero and f(3)=1.
    For the second case define f(x)=1 for x even, f(x)=-1 for x odd. This function satisfies f(x-y)=f(x)f(y), is never zero and f(3)=-1.

    Now let's assume f is a function from $\displaystyle \mathbb{Q}$ to $\displaystyle \mathbb{Q}$.
    We get f(0)=1 the same way as above.
    For every x in $\displaystyle \mathbb{Q}$ we have f(0)=f(x-x)=f(x)f(x), so f(x)=1 or f(x)=-1 for every x in $\displaystyle \mathbb{Q}$.
    Next, f(-x)=f(0-x)=f(0)f(x)=f(x) for every x in $\displaystyle \mathbb{Q}$.

    We still can define f(x)=1 for all x, hence satisfying the case f(3)=1.
    But case f(3)=-1 can't be satisfied now: look at f(1.5). As noted above it is either 1 or -1. In both cases we get: -1=f(3)=f(1.5-(-1.5))=f(1.5)f(-1.5)=f(1.5)f(1.5)=1, a contradiction.
    So we cannot have f(3)=-1 for f $\displaystyle :\mathbb{Q}\rightarrow \mathbb{Q}$.
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