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Determine whether each of these functions is a bijection from $\mathbb{R}$ to $\mathbb{R}$

a) $f(x)=-3x+4$

So I know that a function is bijective if it is both injective (one-to-one) and surjective (onto).

A function is one-to-one if every $x$ has a unique $y$.

And it is onto if for every $y$ there is an $x$ such that $f(a)=b$.

But I don't know how write it down and show that $f(x)=-3x+4$ is bijective

Alim Teacher
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4 Answers4

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You allready received nice answers focused on injectivity and surjectivity. Here a slightly different route.

Can you find a function $g:\mathbb R\rightarrow \mathbb R$ such that $f(g(x))=x$ for each $x\in\mathbb R$?

If you have found such $g$ then check whether it is also true that $g(f(x))=x$ for each $x\in\mathbb R$.

If so then you are ready because you have shown that $f$ is "invertible" (i.e. has an inverse). A function is bijective if and only if it is invertible.

You could also take the opposite route: finding a function $g:\mathbb R\rightarrow \mathbb R$ such that $g(f(x))=x$ for each $x\in\mathbb R$ and checking whether it is also true that $f(g(x))=x$ for each $x\in\mathbb R$.

drhab
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  • FYI the correct term for "has an inverse" is invertible. Also your two routes involve exactly the same checks - what's the difference? – Mario Carneiro Apr 16 '15 at 12:21
  • @MarioCarneiro Thank you, I corrected. As you said there is no essential difference between the two routes. The OP can decide what is most easy for him: finding $g$ based on $f\circ g=1$ (and check $g\circ f=1$) or finding $g$ based on $g\circ f=1$ (and check $f\circ g=1$). – drhab Apr 16 '15 at 13:15
  • @Drhab I thought that finding its inverse was a pretty cool method. I never actually tried that. How useful is that method with harder bijection problems? I'm just wondering not trying to come off as sarcastic or anything! :) – Valentino Apr 16 '15 at 14:22
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    @Valentino There are lots of cases in which it is quite difficult to find the inverse itself. Essential here is not finding it, but proving that it exists, which actually comes to the same as proving injectivity and surjectivity. I gave this answer because in this case it is easy to find the inverse. Secondly in my view it is good to emphasize that 'being invertible' is exactly the same thing as 'being bijective'. – drhab Apr 16 '15 at 14:32
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    @ drhab I completely agree! I think things start going a lot more smoothly later on once one learns that being invertible is exactly the same thing as being bijective. That fact comes up probably every day. However, I was wondering if you have tried to prove cardinality type problems like this. If so, was it easier or harder, or just the same? – Valentino Apr 16 '15 at 14:34
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    @Valentino My experience in this is very 'poor'. I am just too lazy for doing things like that :) So my answer is: "no, I never tried that." I hope that it is not disappointing for you. – drhab Apr 16 '15 at 14:38
  • @drhab lol Understable. I am lazy as well, however when I do run into one of these problems again I am going to try that route first! (: – Valentino Apr 16 '15 at 14:39
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  1. Surjective: For any $y\in \Bbb R$, there exists $x=\frac{4-y}{3}$ such that $f(x)=y$.

  2. Injective: For any $a\not=b$, does it $-3a+4=-3b+4$ hold?

Paul
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Let the function $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=-3x+4$.

We say that a function $f$ is injective if $\forall x_{1},x_2 $ $f(x_{1})=f(x_2)$ implies $x_1 = x_2$. Hence, after some algebra we can see that $-3x_{1}+4= -3x_2+4$ implies $x_1=x_2$. So our function is injective.

Now, we say a function is surjective if for all $y\in \mathbb{R}$(range) there is some $x\in \mathbb{R}$ (domain) such that $f(x)=y$. Choose $x = \frac{4-y}{3}$, then we can see that $f(x)=y$.

A function is bijective if it is both surjective and injective. Hence, $f$ is bijective.

If you have any further questions please let me know.

Valentino
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One-to-one:

Suppose there are two values $x_1,x_2$ such that $f(x_1)=f(x_2)$. Show that $x_1=x_2$ necessarily.

Onto:

Take $y=-3x+4$ and try to obtain $x$ in terms of $y$.

AugSB
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