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Im trying to think about bijective function from the closed interval [a,b] to the closed interval [c,d]. When $a,b,c,d \in \mathbb{R}$ and $a < b,\;c < d$.

Is there such a function?

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    $\begingroup$ Try a linear function which takes $a$ to $c$ and $b$ to $d$. $\endgroup$
    – user134824
    May 4, 2016 at 12:54
  • $\begingroup$ But I dont know that $ b-a = d-c$. $\endgroup$
    – NM2
    May 4, 2016 at 12:54
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    $\begingroup$ $f(x)=2x$ maps $[0,1]$ bijectively to $[0,2]$, for example. The length is not a barrier. $\endgroup$
    – lulu
    May 4, 2016 at 12:55
  • $\begingroup$ There is such a function. If I walk from $c$ to $d$ starting at time $a$ and finishing at time $b$, then the function from time to location is surjective from $[a,b]$ to $[c,d]$. If I don't stop or double back, it's bijective. As best I can tell, my walking is continuous. $\endgroup$ May 4, 2016 at 12:55
  • $\begingroup$ @Noam: If appropriate, could you please say more about how you think of a "bijection", and what are your doubts regarding the multiple comments and answer? $\endgroup$ May 4, 2016 at 13:39

2 Answers 2

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The idea is to construct a line whose domain is $[a, b]$ and whose range is $[c, d]$

Hence, two points on the line will be $$ p_0 = (a, c) \\ p_1 = (b, d) $$

Since we want $a \to c$, $b \to d$, and a straight line between them.

The slope of such a line will be $$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{d - c}{b - a}$$

Using point slope form of a line,

$$ y - y_1 = m \left ( x - x_1 \right) \\ y - c = \frac{d - c}{b - a} \left (x - a\right) $$

Is the required equation.

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  • $\begingroup$ Thanks.But, can you explain why do you consider $a,b$ as $x_{1},x_{2}$ and the same for $c,d$ to $y_{1},y_{2}$ ? $\endgroup$
    – NM2
    May 4, 2016 at 12:57
  • $\begingroup$ Because when I'm deciding to draw a line, I need to have two points on the line for it to be defined. The choice of which was $1$ and which was $2$ is arbitrary. You can swap $x_1, x_2$ and $y_1, y_2$ and the answer will remain the same $\endgroup$ May 4, 2016 at 12:58
  • $\begingroup$ I know this theory. But , i still dont understand how did you decide that you have to draw a line. I dont know to $b−a=d−c$. $\endgroup$
    – NM2
    May 4, 2016 at 13:00
  • $\begingroup$ Oh, that is called as the "slope" of the line. I would recommend you to read a little bit on linear equations and slope of a line to gain the necessary background on this :) $\endgroup$ May 4, 2016 at 13:02
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    $\begingroup$ $a \to b$ and $c \to d$ are not lines, they are *points - $(a, b)$ and $(c, d)$ are two points on the plane of $\mathbb{R}^2$. Now, since you want a bijective function between them, the simplest, most straightforward choice is a line (because I know that a straight is guaranteed to be invertible, since it is a one-to-one mapping). I could have chosen to create a much more complex mapping of $[a, b]$ onto $[c, d]$, but I picked the simplest one. You could have, indeed, construct a quadratic or cubic polynomial for the purpose. But why bother when a straight line suffices? $\endgroup$ May 4, 2016 at 13:15
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$f(x)=(d-c)\frac{x-a}{b-a}+c$ is the (linear) function

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