Math_Week1_Activity1_Q5: Can codomain of a function include imaginary (complex) numbers?

[Anand Iyer]

This is the question.

Which of the following functions is(are) injective?  It's given that x∈R.

The first correct option is f(x)=√10−x

But, my doubt is this a function at all?  Only if it's a function, it can be injective/surjective, right?

If x > 10, f(x) becomes imaginary.  Is this allowed in a function?

To reiterate, my doubt is can codomain of a function include imaginary (complex) numbers?

[Aaditya Bugga]

I struggled with this too. Here's my reasoning. Please tell me how I fared.

Given: Since x belongs to R, the domain is the reals.

But, the co-domain is unspecified, and it could certainly be C and not restricted to just R.

Now, in the general mathematical sense of a function, it is my understanding that the co-domain can definitely be complex numbers, even if we don't study/manipulate them in this course.

To be a function, the mapping must have only one output for every input. The same input cannot give multiple outputs - that would defeat the definition of a function. This doesn't exclude the function output being in C.

Now, requirement for this function to be injective is that the function produces unique outputs for unique inputs, even if those unique outputs fall in C.

As far as I know, the square root of any negative real number is a unique complex number with a real part plus an imaginary part i.e. it is of the form a + bi, where i is the square root of -1.

Hence the function is injective.

(Side note: There is further mathematical nuance here: in the complex number space, by convention, -i is ignored and only positive i is considered, etc. That's what makes this function's output unique.

So the phenomenon of "the complex number root of a negative real being unique" needs to be understood better by me. But, I think this suffices as the broad explanation for your question.

More info on complex numbers can be added by experts, if required at all.)

Please forgive this BA Philosophy graduate for any mistakes in his reply.

~Aaditya

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[Anand Iyer]

Thanks a lot for the reply.

Let’s wait for more views on this. 

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Cheers,

Anand

[MANMAY SINGH]

Guys simply think:

For sq. root of(10-x) to be a function it's domain must be in (-infinity,10].

And in this domain the function is injective.

[Malabika Guha Mustafi]

Here, this is not a function for any value of x  above 10(since, we are dealing with positive root number)

but I am with @aaditi, codomain can include imaginary number as well .

Team, please clarify.

[Aaditya Bugga]

The domain is explicitly given as all of R, and we are asked if it is injective in the entire given domain i.e. all of R.

I don't see how you can restrict the domain to (-infinity, 10] when you are given it is all of R.

In the subset of R—restricted interval (-infinity, 10]—as domain we are already very clear that it is very much a function. That requires no clarification.

The question is about the unique complex roots of the function when x > 10.

I think it is an injective function with C as co-domain in that case.

Here's to enlightenment,

~Aaditya

[Melvin R]

Here's one way of thinking about it.

If a function is injective(one to one) then every element in the domain gives an unique element in the co-domain. This also means that every image of the the domain in the co-domain will have an unique pre-image.

For eg:  If √10 - x is the image of x then (√10 - x)(√10 - x) i.e the image squared, will give you 10 - x which will be unique for every x.

 In short if you have doubts regarding whether the square root of negative numbers are unique then just square them to check if any two squares are equal (because the square of each number is unique) . If not then the squares of the square roots are unique which in turn means that the square roots are unique for each x.  

[Anand Iyer]

Hi Melvin,

Thanks for the response.

However, I doubt if you've read the entire post.  The definition of injective functions is clear to me.  

The relation given doesn't look like a function, under the conditions specified.  The question is, is a relation injective if it's not a function in the first place?  Does it even make sense to look for injectivity of a non-function?

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Cheers,

Anand

[Melvin R]

Yeah, I think if the co-domain was specified the ambiguity could have been eliminated.

If the co-domain was a set of reals then there would be no values that the function could take when x>10.

Since it's not specified maybe we shouldn't leave out the probability of the co-domain being a set of complex numbers.

[Aaditya Bugga]

Discussion about Math for DS 1 - week1 - AQ 1.9 - question 5 - option 1

Let x ∈ R. Which of the following functions is(are) injective?

f(x)=√10−x

Hello, Anand:

I reflected on this a little more.

I think the interpretation of this statement in the question is crucial: "Let x ∈ R"

Does this mean x's domain is in R? Or, x's domain is all of R?

That will determine how we approach this.

If the domain all of R is considered, then the function is still likely injective, but with a broader co-domain of C.

We are then, however, not talking about real-valued functions anymore.

However, I think this particular mathematics course intends for us to restrict ourselves to real-valued functions, where the range and co-domain are in R.

In that case, I think Manmay will be correct. In his reasoning, just because x belongs to R doesn't mean the domain has to be all of R, if we're restricting ourselves to real-valued functions. (This could be a point of discussion/clarification for the IIT team.)

The domain can probably end up being a subset of R, i.e. the interval (-infinity, 10]. The question's assumption still stands true that x  ∈ R. In which case, the co-domain is also R.

In the interval (-infinity, 10] as domain—for which x ∈ R still holds true—the expression is 1) a function and 2) real-valued (co-domain being R) and 3) injective (unique outputs).

This is probably (?) what the IIT team intended for us to reason, rather than get into unique complex roots when x > 10.

Also: as to your latest question: looking for injectivity in a relation makes no sense unless that relation is a function.

But, in this case, it is a function either way. Either in the restricted interval within R, or within all of R.

So the question of it being a non-function doesn't arise, I think.

What do you think?

Here's to our admit cards,

~Aaditya

[Anand Iyer]

Thanks Aaditya for the detailed reply!

I think x  ∈ R has only one interpretation in math, that x can be ANY element of R.

Thus, the question still holds about this being a function...

IIT team, can you please respond?  I don't know how to draw their attention to this question, that has already been discussed a lot among the students....If anyone has any ideas for that, please let know!

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Cheers,

Anand