Discrete Mathematics : Functions and Relations : Question 2

[Supierreious]

Homework Statement

Determine the dom(g)

Homework Equations

Let 'f' be a relation on ℤ (the set of integers) , defined by the entrance requirement :

(x;y) ∊ ƒ iff y = x + 15

and let 'g' be the function on ℤ defined by the entrance requirement :

(x;y) ∊ g iff y = 5x(to the power of 2) + 7

The Attempt at a Solution

What i understand from functions :

1. A function is a special type of relation, however, it is a relation
2. A function refers to the ordered pairs, and that every (x;y) will only have the x from the first set once, and also every x from the domain elements.
2. A relation means that 2 sets will be involved.

Having a look at the question :

a) Determine the domain, of the function 'g' : (x;y) ∊ g iff y = 5x(to the power of 2) + 7 :

Dom(g) = {x | for some y ∊ ℤ , (x;y) ∊ function 'g'}
= {x | for some y ∊ ℤ, y = 5x(to the power of 2) + 7
= {x | for some y ∊ ℤ, where y > 0}
= ℤ

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Now here is where i don't know what to do. I can describe what i think the 'Domain' is of function 'g', but other than that I don't know how to say in a formal way what it is ( not quite sure what is required in addition).

I am aware that the Domain elements refer to the 'x' elements. So in a relation, there are 2 sets, and the sets are paired up to form a relation (x;y), and for a function, (x;y) = (x;z) so if x is the same, the second element will be the same ( y=z).

Here 'x' can actually be any ℤ , positive or negative, due to the fact that if it is to the power of 2 , the outcome is anyways positive.

For this reason i cannot pinpoint exactly what the scope of the domain is, because there is no limit, other than being an element of ℤ.

If i am correct in thinking this way, please help me with the correct way to structure my answer, as i would also like to make it easier for me to adapt to the correct custom in formalizing my answer.

Sorry for typing the long essay with so little calculations... :P

 

[Stephen Tashi]

Here 'x' can actually be any ℤ , positive or negative, due to the fact that if it is to the power of 2 , the outcome is anyways positive.

You are correct that x can be any $\mathbb{Z}$. If your course materials use the terminology g is a function "on" $\mathbb{Z}$ to mean that both the domain and range of g must be subsets of $\mathbb{Z}$ then you are correct to worry about whether $x^2 + 7$ is positive.

For this reason i cannot pinpoint exactly what the scope of the domain is, because there is no limit, other than being an element of ℤ.

I think what you mean is that "There is no restriction on x other than x being an element of $\mathbb{Z}$.

You did determine exactly what the domain was. Don't say that you didn't "pinpoint it". In writing literature, it's good style to iuse a variety of words, but when writing mathematics it's best to use the words that have technical definitions over and over again instead of tossing-in words that you hope will mean the same thing. For example, what's the definition of "scope of the domain"? The term "limit" has a technical definition in Calculus, so avoid using that term when you mean "bound". Of course, only further study will reveal which English words can be used with their everyday meanings and which have been taken by mathematicians.

 

[Supierreious]

Thanks for the feedback on this, extremely helpful.

Continuing on the same question, the next sub question on this is the following :

b) Is 'f' injective (one-to-one) ? Justify your answer.

Calculations :
_______________

Injective : When there are two sets, set A and set B, all the elements in set A are mapped to elements in set B, and if f(a1)=f(a2) , a1 = a2 - which means that if there is an 'image' to x , namely a1, and the same x is found again, with an 'image' a2, then it is the same image (a1=a2), and those elements are thus the same.

the relation , 'f' : (x;y) ∊ ƒ iff y = x + 15

1. firstly what i did is clarify to myself what i understand as injective (above).
2. secondly, is to take the relation, and write it out with a couple of examples.
3. Then to use the required format to prove that the formula is injective or not.

1 : Already completed above
2 :


f(1) = 1 + 15 = 16
f(2) = 2 + 15 = 17
f(3) = 3 + 15 = 18
f(4) = 4 + 15 = 19

So there will be an (x,y) accordingly :

f(1) : (1;16)
f(2) : (2;17)
f(3) : (3;18)
f(4) : (4;19)

3.
Assume g(u) = g(v)
then u + 15 = v + 15
then u = v

Therefore this is injective.


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Please correct me on the steps taken. I really cannot understand why i am using g(u) = g(v), however am applying what i am taking from the textbook ( in case you can shed some light on what i am missing with that step).

Can you also confirm if my approach to this question is correct.

Lastly, just want to confirm if it is ok for me to add to another question like i did , and whether or not i should rather compile a new post with a question to make reading it easier.

 

[Stephen Tashi]

Injective : When there are two sets, set A and set B, all the elements in set A are mapped to elements in set B, and if f(a1)=f(a2) , a1 = a2

Yes

which means that if there is an 'image' to x , namely a1, and the same x is found again, with an 'image' a2, then it is the same image (a1=a2), and those elements are thus the same.

No. You aren't following the notation in the previous statement. In the previous statement, the "image" of a1 is f(a1). There isn't any 'x' mentioned. If you want to think of x as the variable representing something in the domain then the previous statement deals with two cases: x = a1 and x = a2.

The fact that a1 = a2 implies f(a1) = f(a2) is a consequence of the definition of a function. This is true for all functions, whether they are injective or not.

The definition of a function says that each element in the domain is sent to one and only one element in the range. The definition of injective function rquires that each element in the range "comes from" (i.e. is the image of) only one element in the domain. For example the function f(x) = x^2 is not injective since the number 4 is the image of both 2 and -2.

Assume g(u) = g(v)
then u + 15 = v + 15
then u = v

Therefore this is injective.

You should settle on whether you are going to call the function 'f' or 'g' in this problem. There is no need to use two different names for it.

Say "Therefore g is injective". It isn't clear what you mean if you say "this".

I really cannot understand why i am using g(u) = g(v), however am applying what i am taking from the textbook ( in case you can shed some light on what i am missing with that step).

Can you also confirm if my approach to this question is correct.

The statement "Assume g(u) = g(v)" is like saying, suppose there is an element in the range of g that is the image of possibly two values, which I will call u and v. Then you show that u and v are actually the same value. Thus each element in the range of g only comes from one element in the domain of g.

Lastly, just want to confirm if it is ok for me to add to another question like i did , and whether or not i should rather compile a new post with a question to make reading it easier.

I don't know if the forum has some official policy on this. As far as I'm concerned personally, it's OK to add another closely related question to the thread.

 

[Mark44]

f(2) = 2 + 15 = 17
f(3) = 3 + 15 = 18
f(4) = 4 + 15 = 19

So there will be an (x,y) accordingly :

f(1) : (1;16)
f(2) : (2;17)
f(3) : (3;18)
f(4) : (4;19)

The part just above is redundant, for the most part. It's enough to say that f(1) = 17, f(2) = 18, and so on. You don't need the f(1) : (1;16) business.

Lastly, just want to confirm if it is ok for me to add to another question like i did , and whether or not i should rather compile a new post with a question to make reading it easier.

If you have a different question, you should start a new thread.