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# Why is this not a function? Watch

1. f(x) = 1/(x-1) for all real values of x other than 1

The mark scheme says that isn't a function? 'For a function to have an inverse, it must be a one-one function.' But it is a one-one isn't it?
2. (Original post by ak9779)
f(x) = 1/(x-1) for all real values of x other than 1

The mark scheme says that isn't a function? 'For a function to have an inverse, it must be a one-one function.' But it is a one-one isn't it?
For it to have an inverse the graph must also be a continuos graph. For example, this graph has 2 distinct sections.
3. (Original post by dknt)
For it to have an inverse the graph must also be a continuos graph. For example, this graph has 2 distinct sections.
Can't be because 1/x has an inverse
4. x=1 will have no y value, so the graph has a discontinuity.
5. (Original post by olipal)
x=1 will have no y value, so the graph has a discontinuity.
The limits state that x cannot equal one, so that y value is irrelivent.
6. (Original post by ak9779)
f(x) = 1/(x-1) for all real values of x other than 1

The mark scheme says that isn't a function? 'For a function to have an inverse, it must be a one-one function.' But it is a one-one isn't it?
Unfortunately, this is where school mathematics and proper mathematics clash. Yes, it is a perfectly legitimate function (on the domain ) and indeed even has an inverse. It's even continuous and differentiable (infinitely many times!) on its domain. The only problem, as everyone has pointed out, is that it isn't defined at 1 and there is no continuous way of patching it up there.
7. (Original post by limetang)
The limits state that x cannot equal one, so that y value is irrelivent.
Exactly, which it why it isn't a function.
8. (Original post by olipal)
Exactly, which it why it isn't a function on that particular domain.
Fixed it.
It is a well defined function over the domain .
9. (Original post by Farhan.Hanif93)
Fixed it.
It is a well defined function over the domain .
That's not well-formed notation. A domain is a set, so write . Alternatively, say that f(x) is defined for all .
10. (Original post by Zhen Lin)
That's not well-formed notation. A domain is a set, so write . Alternatively, say that f(x) is defined for all .
Done. Thank you for the tip.
11. (Original post by olipal)
Exactly, which it why it isn't a function.
It is a function. It's just not a function with the real numbers as its domain. A function is just a mapping from one set into another, and this is a perfectly legitimate mapping from the reals without 1 to the reals.

In fact, if you let the set you're mapping to be the reals plus infinity, it can be a legitimate function from the reals to the reals plus a point at infinity if you define f(1) to be the point at infinity. Well, really you can define f(1) to be any number and it would still be a function, it's just that such an extension might seem less natural.
12. You can extend it to become a continuous function . (This is rigorous and not hand-wavy like it initially sounds.)

EDIT: oops, just noticed Chumbaniya's post above. The reason that the other extensions seem less natural is because they won't be continuous (well, unless you give some non-standard topology. But that's just asking for punishment )
13. Ah, but if we're going to talk about "the point at infinity", we may as well just extend to the most natural setting - namely the Riemann sphere . Then we have a complete analytic map defined at all points! (Better still, it's a conformal equivalence / analytic isomorphism / automorphism of the Riemann sphere!)
14. (Original post by Zhen Lin)
Ah, but if we're going to talk about "the point at infinity", we may as well just extend to the most natural setting - namely the Riemann sphere . Then we have a complete analytic map defined at all points! (Better still, it's a conformal equivalence / analytic isomorphism / automorphism of the Riemann sphere!)
Very true. I just happen to like the one-point compactification of the real line Also my Riemann surfaces knowledge is shaky at best... Is it the Riemann sphere whose automorphisms are precisely the Möbius maps?
15. (Original post by ljfrugn)
Very true. I just happen to like the one-point compactification of the real line Also my Riemann surfaces knowledge is shaky at best... Is it the Riemann sphere whose automorphisms are precisely the Möbius maps?
Yes, precisely. (A fact which was left as an exercise for students by my lecturer...)
16. Haha. I don't think that example sheet has ever changed. Who's lecturing it?

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