How can changing the domain of a function make it injective?

Watch
Brian Moser
Badges: 12
Rep:
?
#1
Report Thread starter 6 years ago
#1
For example, the function f(x)=(x+\frac{1}{3})^2 is not injective for f:\mathbb{R} \rightarrow \mathbb{R}. However, it would (apparently) be injective for f:\mathbb{Z} \rightarrow \mathbb{Q}. Why is this the case?

Thanks in advance.
0
reply
Noble.
Badges: 17
Rep:
?
#2
Report 6 years ago
#2
You understand what makes a function injective? If two different values in the domain map to the same value under the function, then it can't be injective.

Looking at your function and domain \mathbb{R} there's an obvious way you can show it isn't injective using the fact that \lambda, -\lambda \in \mathbb{R} both become \lambda^2 when squared.
0
reply
Brian Moser
Badges: 12
Rep:
?
#3
Report Thread starter 6 years ago
#3
(Original post by Noble.)
You understand what makes a function injective? If two different values in the domain map to the same value under the function, then it can't be injective.

Looking at your function and domain \mathbb{R} there's an obvious way you can show it isn't injective using the fact that \lambda, -\lambda \in \mathbb{R} both become \lambda^2 when squared.
Thanks, but I understand why the function is not injective for f:\mathbb{R} \rightarrow \mathbb{R}. What I don't understand is why it is injective for f:\mathbb{Z} \rightarrow \mathbb{Q}. Can you explain this please?


Posted from TSR Mobile
0
reply
poorform
Badges: 10
Rep:
?
#4
Report 6 years ago
#4
The general idea when it comes to showing a function is or is not injective is this.

1) The function is not injective, then pick two elements from the domain and show that they both map to the same element in the codomain under f. This is what Noble put when he had \lambda and -\lambda because they both square to give the same value. Then the function is not injective, a counter example this is called.

2) If the function is injective then pick two arbitrary elements in the codomain and show that they must give the same image under f, that way you have shown that the function is injective can you think of how you could do this. Here is a first line

Choose an arbitrary x_1, x_2 \in \mathbb{Z}...
0
reply
atsruser
Badges: 11
Rep:
?
#5
Report 6 years ago
#5
(Original post by Brian Moser)
Thanks, but I understand why the function is not injective for f:\mathbb{R} \rightarrow \mathbb{R}. What I don't understand is why it is injective for f:\mathbb{Z} \rightarrow \mathbb{Q}. Can you explain this please?


Posted from TSR Mobile
For the function to be injective, you need to find two different values of the form x+1/3, x \in \mathbb{Z} which map to the same value, for which you need:

x_1 + 1/3 = -(x_2+1/3) \Rightarrow x_1+x_2 = -2/3

but since everything in \mathbb{Z} is an integer (or \mathbb{Z} is a group under addition or whatever) then you won't find any such values.

Finding such values clearly isn't a problem when you can choose real numbers, of course.
0
reply
Smaug123
  • Study Helper
Badges: 15
Rep:
?
#6
Report 6 years ago
#6
(Original post by Brian Moser)
For example, the function f(x)=(x+\frac{1}{3})^2 is not injective for f:\mathbb{R} \rightarrow \mathbb{R}. However, it would (apparently) be injective for f:\mathbb{Z} \rightarrow \mathbb{Q}. Why is this the case?

Thanks in advance.
I'd point out the silly example that for any function f: X \to Y, then the restriction f_x : \{x \} \to Y and also the restriction f_{\phi}: \phi \to Y are both injective.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Has your education been disrupted this academic year due to the pandemic?

Yes (7)
100%
No (0)
0%

Watched Threads

View All
Latest
My Feed