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

For example, the function f(x)=(x+13)2f(x)=(x+\frac{1}{3})^2 is not injective for f:R→Rf:\mathbb{R} \rightarrow \mathbb{R}. However, it would (apparently) be injective for f:Z→Qf:\mathbb{Z} \rightarrow \mathbb{Q}. Why is this the case?

Thanks in advance.
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Avatar for Noble.
Noble.
17
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 R\mathbb{R} there's an obvious way you can show it isn't injective using the fact that λ,−λ∈R\lambda, -\lambda \in \mathbb{R} both become λ2\lambda^2 when squared.
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 R\mathbb{R} there's an obvious way you can show it isn't injective using the fact that λ,−λ∈R\lambda, -\lambda \in \mathbb{R} both become λ2\lambda^2 when squared.


Thanks, but I understand why the function is not injective for f:R→Rf:\mathbb{R} \rightarrow \mathbb{R}. What I don't understand is why it is injective for f:Z→Qf:\mathbb{Z} \rightarrow \mathbb{Q}. Can you explain this please?


Posted from TSR Mobile
(edited 9 years ago)
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 λ\lambda and −λ-\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 x1,x2∈Z...x_1, x_2 \in \mathbb{Z}...
Original post by Brian Moser
Thanks, but I understand why the function is not injective for f:R→Rf:\mathbb{R} \rightarrow \mathbb{R}. What I don't understand is why it is injective for f:Z→Qf:\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∈Zx+1/3, x \in \mathbb{Z} which map to the same value, for which you need:

x1+1/3=−(x2+1/3)⇒x1+x2=−2/3x_1 + 1/3 = -(x_2+1/3) \Rightarrow x_1+x_2 = -2/3

but since everything in Z\mathbb{Z} is an integer (or Z\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.
Original post by Brian Moser
For example, the function f(x)=(x+13)2f(x)=(x+\frac{1}{3})^2 is not injective for f:R→Rf:\mathbb{R} \rightarrow \mathbb{R}. However, it would (apparently) be injective for f:Z→Qf:\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→Yf: X \to Y, then the restriction fx:{x}→Yf_x : \{x \} \to Y and also the restriction fϕ:ϕ→Yf_{\phi}: \phi \to Y are both injective.

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