How wolfram defines injectivity?

Asked Oct 02 '15 at 11:03

Active Oct 02 '15 at 11:14

Viewed 331 times

I was looking for a fast way of getting which function is surjective and injective. So I tried to find such command in wolfram and surprisingly it says to me that $ln(x)$ is not injective. Link to proof: http://www.wolframalpha.com/input/?i=ln%28x%29+is+injective%3F

The questions are:

(1) Why did Wolfram|Alpha say that it is not injective?

(2) How the injectivity is defined in Wolfram?

edited Oct 02 '15 at 11:14

asked Oct 02 '15 at 11:03

Mesmerized student

1,429

This is really weird (I would have expected this to be a problem with the domain, but WA has chosen precisely the domain it ought to in order to get an injective function). – Tobias Kildetoft Oct 02 '15 at 11:06
I was expecting something like this too, but as we can see, Wolfram gets the domain rightly. That's why I asked it here. – Mesmerized student Oct 02 '15 at 11:07
I've reported it to them. Looks like a bug. Mathematica sheds no light on it either. – Patrick Stevens Oct 02 '15 at 11:12
2

Looks like a bug, note: http://www.wolframalpha.com/input/?i=log+is+injective+on+%280%2Cinfinity%29 – martini Oct 02 '15 at 11:18
It's still not fixed! – Andrea Aug 30 '16 at 23:56
It seems that they finally fixed it. – Mesmerized student Apr 08 '18 at 17:12

How wolfram defines injectivity?

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