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Let $A=\{x \in \Bbb R \mid 0\le x \le 1\}$ and $B=\{x \in \Bbb R \mid 5\le x \le 8\}$. Let $f:A \to B$ defined by $f(x) = 5+(8-5)x$ for all $x \in A$. Show that $f$ is bijective.

Attempt:

$f$ is injective: Let $x_1,x_2 \in A$ and assume that $f(x_1) = f(x_2)$. Then, $5+(8-5)x_1 = 5+(8-5)x_2$ i.e. $x_1=x_2$. Thus, $f$ is injective.

$f$ is surjective: Let $y \in B$. Then, there exists $x=\frac{y-5}{3} \in A$ such that $f(x)=f(\frac{y-5}{3}) = 5+(8-5)\left(\frac{y-5}{3}\right) = 5+y-5 = y$. This means that $f$ is surjective.

Thus, $f$ is bijective.

Is the above correct?

lap lapan
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    Looks fine. But end the injective conclusion with $x_1=x_2.$ – user159888 Jan 06 '21 at 15:00
  • Why does the fact that $f(x_1)=f(x_2)$ implies $5+(8-5)x_1=5+(8-5)x_2$ imply that $f$ is injective? For instance, I can say the same of $g(x)=\sin x$. In fact, $g(x_1)=g(x_2)$ implies $\sin x_1=\sin x_2$: must I conclude that $\sin $ is injective as well? –  Jan 06 '21 at 15:01
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    Also in the surjective part mention that $x\in A.$ – user159888 Jan 06 '21 at 15:03
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    Since you have not verified that $x=\frac{y-5}{3}$ belongs to A, you have not demonstrated surjectivity. – Jean-Claude Colette Jan 06 '21 at 15:12

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