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(see improved and answered version of this question at: showing $f$ to be a bijection but stuck trying to show that $f$ is surjective)

recall that the closed interval $[a,b]$={${x \in \mathbb{R} \vert a\le x \le b}$}

prove that $\vert [2,5] \vert$=$\vert [-2,3] \vert$

can someone please help me prove this? I know proving these involves finding an injection, surjection or bijection but without being given an actual function im pretty clueless.

Thanks.

kr1s
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2 Answers2

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Define $f\colon[2,5]\longrightarrow[-2,3]$ by $f(x)=-2+\frac53(x-2)$. It is a bijection and therefore $[2,5]$ and $[-2,3]$ have the same cardinal.

José Carlos Santos
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  • so what steps did you take to get $f(x)$ ? also im writing a proof for this question so as part of the proof should i prove that $f(x)$ is a bijection? – kr1s Sep 30 '17 at 10:41
  • @kr1s The first interval has length $3$, and the second one has length $5$; that's where that $\frac53$ factor comes from. Since I wanted a map of the type $f(x)=\alpha+\frac53(x-2)$ such that $f(2)=-2$, I took $\alpha=-2$. – José Carlos Santos Sep 30 '17 at 10:44
  • oh i see, thanks! – kr1s Sep 30 '17 at 10:45
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Let $a \lt b,$ real.

$f: [a,b]$ $ \rightarrow$ $[0,1]$ ,

$x \mapsto f(x):= \frac{x-a}{b-a}$ is a bijection .

Hence $[0,1]$ and $[a,b]$ have the same cardinality.

Can you take it from here?

Peter Szilas
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