Are affine transformations on $\mathbb{R}$ the only transformations that preserve ratios of intervals?
Are affine transformations the only transformations that preserve ratios of intervals?
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linear-algebra
1 Answers
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Hint: by following your transformation $T$ by a translation, we can assume $T(0) = 0$. By following that by a scaling, we can assume $T(1) = 1$ (unless $T$ is constant). Now if $T$ preserves ratios of intervals, what must $T(x)$ be?
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0Ok thanks! That answers my question. – 2012-12-02
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0One further question. I know that affine trasnformations preserve the mid-points of intervals. That is if f is affine and z is the mid-point of [x, y], then f(z) is the mid-point of [f(x), f(y)]. Is it only affine transformations that have this property? – 2012-12-02
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0Among continuous functions, yes. Assuming the Axiom of Choice, there are discontinuous functions that preserve midpoints. – 2012-12-02
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0One last question: – 2012-12-02
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0Let $S_1$ and $S_2$ be sets. Let $n_1$ be the cardinality of $S_1$ and $n_2$ be the cardinality of $S_2$. I assume that $n_1$ and $n_2$ are finite. Let $e$ be a function that maps members of $S_1$ and $S_2$ to real numbers. Assume that we have: $(1/n_1) \sum\limits_{x_i \in S_1} e(x_i) \geq (1/n_2) \sum\limits_{x_i \in S_2} e(x_i)$. Let $e'$ be a affine transformation of $e$, i.e., we have $e'(x) = ke(x) + l$, where $k$ is positive. – 2012-12-02
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0Given this, I know that it is guaranteed that: $(1/n_1) \sum\limits_{x_i \in S_1} e'(x_i) \geq (1/n_2) \sum\limits_{x_i \in S_2} e'(x_i)$ – 2012-12-02
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0Question: Is it only affine transformations that have this property; or is again going to be that only amongst the continuous functions do affine transformations have this property, while some discontinuous functions also do? – 2012-12-02