Injective Function

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Injective Function

Devika
Why is 1/x, where x belongs to R and is greater than equal to 0 not injective?
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Re: Injective Function

Sinistral
I don't think it is even a function because x=0 is part of the domain.
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 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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Re: Injective Function

Sinistral
In reply to this post by Devika

to add further lets say f is defined like this : (so that it becomes a function):
f(x) = 1/x  , x>0
      =  a    , x=0   where a is positive Real number

then the function wont be injective because at both x = 1/a and x=0 , f(x) = a.
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 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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Re: Injective Function

Devika
Heyy,

Yes even that x=0 part befuddled me. Anyway, your reasoning seems convincing.
I came across this definition-
A function f: A→B is a one to one or injective function if and only if for all ,x y in A

f(x)=f(y)
=> x =y.

If I use this then the function is injective?
Correct me!
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Re: Injective Function

Sinistral
yeah that is the definition of injective function.

just copy paste the exact ques. may be the co domain is extended Real line.
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 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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Re: Injective Function

Devika
DSE 2006.
Q 60, part c.
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Re: Injective Function

Sinistral
(c) is not injective.

(d) is the right answer. what's the issue?
---
 "You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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Re: Injective Function

Devika
I know d is the right answer! I wasn't able to rule out part c. I wanted to know why c is not injective.