Sets and Functions question

[randomprime]

Today in class, I was learning about surjective functions and I could not agree with the teacher about the following question:

Let [MATH]f:\mathbb{R} \rightarrow \mathbb{Z}, f(x)=5x+2[/MATH]. Is [MATH]f[/MATH] a surjective function?

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Here are some definitions that we were using.

A function from [MATH]S[/MATH] to [MATH]T[/MATH] is a subset [MATH]M[/MATH] of [MATH]S \times T[/MATH] such that for every [MATH]s \in S[/MATH] there is a unique [MATH]t \in T[/MATH] such that [MATH](s,t) \in M[/MATH].

If every element in the co-domain of a function is the image of at least one element in the domain we say that the function is a surjection, i.e. for all [MATH]b[/MATH] in the co-domain there exists an [MATH]a[/MATH] in the domain such that [MATH]f(a) = b[/MATH]. A surjection is also called an onto function.

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My teacher and a video explaining the problem claim that [MATH]f[/MATH] is a surjective function.

I can provide additional definitions if needed, but the way I see it, the relationship fits the definition for surjection, but it defies the one for a function. There are elements of the domain ([MATH]\mathbb{R}[/MATH]) for which [MATH]f(x) \notin \mathbb{Z}[/MATH], for example 0.5: [MATH]f(0.5) = 4.5 \notin \mathbb{Z}[/MATH]. Therefore, it cannot be a surjective function because it is not a function in the first place.

Is there something wrong in my reasoning? Am I understanding the question completely wrong?

 

[HallsofIvy]

No, you are correct. "Let f:R→Z,f(x)=5x+2" is nonsense! x= 1/2 is in R but then "5x+ 2= 9/2" is NOT in Z.
Are you sure it was not "f:Z→Z"?

 

[randomprime]

It was a multiple-step exercise where the same question was asked with [MATH]\mathbb{R}\rightarrow\mathbb{R}[/MATH], [MATH]\mathbb{Z}\rightarrow\mathbb{Z}[/MATH] and [MATH]\mathbb{R}\rightarrow\mathbb{Z}[/MATH].

I agreed with the first two, the answers being yes and no respectively, but the last one didn't make sense to me. I'm glad to hear I'm not going insane.

 

[Steven G]

Today in class, I was learning about surjective functions and I could not agree with the teacher about the following question:

Let [MATH]f:\mathbb{R} \rightarrow \mathbb{Z}, f(x)=5x+2[/MATH]. Is [MATH]f[/MATH] a surjective function?

---

Here are some definitions that we were using.

A function from [MATH]S[/MATH] to [MATH]T[/MATH] is a subset [MATH]M[/MATH] of [MATH]S \times T[/MATH] such that for every [MATH]s \in S[/MATH] there is a unique [MATH]t \in T[/MATH] such that [MATH](s,t) \in M[/MATH].

If every element in the co-domain of a function is the image of at least one element in the domain we say that the function is a surjection, i.e. for all [MATH]b[/MATH] in the co-domain there exists an [MATH]a[/MATH] in the domain such that [MATH]f(a) = b[/MATH]. A surjection is also called an onto function.

---

My teacher and a video explaining the problem claim that [MATH]f[/MATH] is a surjective function.

I can provide additional definitions if needed, but the way I see it, the relationship fits the definition for surjection, but it defies the one for a function. There are elements of the domain ([MATH]\mathbb{R}[/MATH]) for which [MATH]f(x) \notin \mathbb{Z}[/MATH], for example 0.5: [MATH]f(0.5) = 4.5 \notin \mathbb{Z}[/MATH]. Therefore, it cannot be a surjective function because it is not a function in the first place.

Is there something wrong in my reasoning? Am I understanding the question completely wrong?

As already pointd out, your teacher and video is wrong (assuming that they said f is surjective) I just want to point outthat you should not say that f fits the definition for surjection, but it defies the one for a function. That is, you should not even be talking about subjection for f if f is not even a well defined function.

 

[Steven G]

Can you supply us with the link to the video?

 

[randomprime]

As already pointd out, your teacher and video is wrong (assuming that they said f is surjective) I just want to point outthat you should not say that f fits the definition for surjection, but it defies the one for a function. That is, you should not even be talking about subjection for f if f is not even a well defined function.

Ok, good thing to remember. Thank you.

Also, this is the link to the video...
[Discrete Math 1] Injective, Surjective, Bijective Functions

 

[pka]

It was a multiple-step exercise where the same question was asked with [MATH]\mathbb{R}\rightarrow\mathbb{R}[/MATH], [MATH]\mathbb{Z}\rightarrow\mathbb{Z}[/MATH] and [MATH]\mathbb{R}\rightarrow\mathbb{Z}[/MATH]. I agreed with the first two, the answers being yes and no respectively, but the last one didn't make sense to me. I'm glad to hear I'm not going insane.

The video clearly states that if \(\displaystyle f:\mathbb{R}\to\mathbb{Z}\) where \(\displaystyle f(x)=5x+2\) is a surjection.
To prove it lets say \(\displaystyle n\in\mathbb{Z}\) then note that \(\displaystyle \frac{n-2}{5}\in\mathbb{R}\) so that \(\displaystyle f\left(\frac{n-2}{5}\right)=n\)
That is a proof.

 

[Steven G]

Ok, good thing to remember. Thank you.

Also, this is the link to the video...
[Discrete Math 1] Injective, Surjective, Bijective Functions

f: R-->Z defined by f(x) = 5x+2 is surjective as the video says.

 

[randomprime]

As already pointd out, your teacher and video is wrong (assuming that they said f is surjective) I just want to point outthat you should not say that f fits the definition for surjection, but it defies the one for a function. That is, you should not even be talking about subjection for f if f is not even a well defined function.

f: R-->Z defined by f(x) = 5x+2 is surjective as the video says.

Are you saying there is a difference between saying
[MATH]f:\mathbb{R}\rightarrow\mathbb{Z}, f(x)=5x+2[/MATH]and
[MATH]f:\mathbb{R}\rightarrow\mathbb{Z} [/MATH] defined by [MATH]f(x)=5x+2[/MATH]?

I can see no other difference between your two quoted claims, one based on the video and the other based on my explanation...

Thank you for your patience with my thickness

 

[HallsofIvy]

No, there is no difference in the notations. What they are saying is that f(x)= 5x+ 2 is NOT a function from R to Z because for example, if x= 1/2, f(1/2)= 5/2+ 2= 7/2 which is not an integer so not in Z.

What is true is that f(x)= 5x+ 2 is a surjective function from R to R. Of course, since Z is a subset of R, for any integer y there exist a real number x such that f(x)= y.

 

[randomprime]

No, there is no difference in the notations. What they are saying is that f(x)= 5x+ 2 is NOT a function from R to Z because, while, for example, if x= 1/2, f(1/2)= 5/2+ 2= 7/2 which is not an integer so not in Z.

What is true is that f(x)= 5x+ 2 is a surjective function from R to R.

I agree with you, that's why I am still confused by Jomo's two contradicting claims...

(also, this is my 5th message, I can't wait to be handed the metaphorical sock "Dobby is free!" )

 

[HallsofIvy]

Jomo will have to answer this but I suspect he misread the question, thinking it was R to R rather than R to Z.

 

[Steven G]

I was wrong! Sorry. I'll go to the corner for a long while.