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Define a function $ \ f: [1,3] \to [3,7] \ $ by $ \ f(x)=2x+1 \ $. If $ f \ $ be a injective function , does the following condition confirm that the function is bijective ?

(a) Here clearly , $ \ f \ $ is continuous. Since $ f(1)=3 \ \ and \ \ f(3)=7 \ $ , every $ \ y \ $ value between $ \ 3 \ \ and \ \ 7 \ $ is the output of some $ x \in (1,3) $ by Intermediate value theorem. Hence f is surjective and so f is bijective. .

Answer

I think this is true.

But not sure . Is there any help ?

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MAS
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    I believe this works. I also believe that it suffices to say that $f$ is increasing and continuous on its domain, it is bijective. – Sean Roberson Aug 05 '17 at 03:48
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    @mabmath I think you are right. – Michael Rozenberg Aug 05 '17 at 03:48
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    The grammar of the question is rather odd. If the question was "supposing that $f$ is an injective function from $[1,3]\to[3,7]$ does this imply it must be bijective as well" the answer is no and you should be able to come up with many counterexamples. If the question is "suppose that $f$ is a function $[1,3]\to[3,7]$ where $f(x):=2x+1$ is the function bijective" then the answer should obviously be yes. In the first quotes note the absence of the specific definition of $f$. In the second quotes note the absence of the phrase "if $f$ is injective" – JMoravitz Aug 05 '17 at 03:49
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    tldr: It is not a direct result of the injectivity of $f$ that the function is bijective, but rather it is a result of the definition of $f$ that it is bijective. – JMoravitz Aug 05 '17 at 03:50

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Your proof seems OK but continuity is not needed. Axioms of order in $\Bbb R$ suffice (in fact, $\tilde f:[1,3]\cap \Bbb Q\to [3,7]\cap\Bbb Q$, $\tilde f(x)=2x+1$ is also bijective). With these axioms you can show that $1>0$ and that $2=1+1>1+0=1>0$.

Then $y>x$ implies $2y>2x$ and then $2y+1>2x+1$. This shows that the function is increasing, and hence, injective.

To show surjectivity you can define $g(x)=\frac12x-\frac12$, which is also increasing and injective, and it is the inverse of $f$.

ajotatxe
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