Discrete Mathematics : Functions and Relations : Question 2c

[Supierreious]

Homework Statement

c) Is 'g' a surjective function (onto) ? Justify your answer.

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

The steps I follow is the following :
1. Clarify to myself what surjective is.
2. Confirm the correct function or relation to use, and substitute the x and y with real values.
3. Write out the formula with the proof in the required format.

1 : Surjective :

When we have a function , with a set A and a set B, as example, and all the elements of set B is mapped to an element in A. f: A→B (function f on set A to B), is a surjective function if the range of f is equal to the codomain of f, ie, f[A]=B.

2. The function to use is : (x;y) ∊ g iff y = 5x² + 7

g(1) = 5(1)² + 7 = 32
g(2) = 5(2)² + 7 = 107
g(3) = 5(3)² + 7 = 232

Rewriting the (x;y) in each of the above examples :
(1;32)
(2;107)
(3;232)

* 3.

This is how far i have gotten. My set 'A' can be defined as {1;2;3} used in my example, but i am not sure what my set B is, and if i am not sure what my set B is , then i cannot really say if every element of my set B can be mapped to set A. Not sure if this makes sense :(

 

[Mark44]

Homework Statement

c) Is 'g' a surjective function (onto) ? Justify your answer.

Homework Equations

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

(x;y) ∊ ƒ iff y = x + 15 [/STRIKE]

The question is about g, so the information above about f seems to be extraneous.

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

The steps I follow is the following :
1. Clarify to myself what surjective is.
2. Confirm the correct function or relation to use, and substitute the x and y with real values.
3. Write out the formula with the proof in the required format.

1 : Surjective :

When we have a function , with a set A and a set B, as example, and all the elements of set B is mapped to an element in A. f: A→B (function f on set A to B), is a surjective function if the range of f is equal to the codomain of f, ie, f[A]=B.

2. The function to use is : (x;y) ∊ g iff y = 5x² + 7

g(1) = 5(1)² + 7 = 32
g(2) = 5(2)² + 7 = 107
g(3) = 5(3)² + 7 = 232

You aren't doing these correctly. To evaluate 5x², replace x by whatever number you're working with, square that number, and then multiply by 5. There's a difference between 5x² and (5x)², which seems to be what you are doing.

For example, g(1) = 5(1)² + 7 = 5*1 + 7 = 12, not 32.

Rewriting the (x;y) in each of the above examples :
(1;32)
(2;107)
(3;232)

* 3.

This is how far i have gotten. My set 'A' can be defined as {1;2;3} used in my example, but i am not sure what my set B is, and if i am not sure what my set B is , then i cannot really say if every element of my set B can be mapped to set A. Not sure if this makes sense :(

Isn't the domain of g all of the integers? If so, that would be your set A, assuming that's what you mean by A. To be able to say whether g is surjective, you need to get some more (correct) function values. When you do that, you will probably be able to answer the question.

 

[Supierreious]

Sorry, i see now, i am fixing this immediately.

 

[Supierreious]

Ok, so i have miscalculated this by far :

Incorrect values :

-------2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

g(1) = 5(1)2 + 7 = 32
g(2) = 5(2)2 + 7 = 107
g(3) = 5(3)2 + 7 = 232

Rewriting the (x;y) in each of the above examples :
(1;32)
(2;107)
(3;232)

-------

Correct Values :2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

g(1) = 5(1)2 + 7 = 12
g(2) = 5(2)2 + 7 = 27
g(3) = 5(3)2 + 7 = 52

Rewriting the (x;y) in each of the above examples :
(1;12)
(2;27)
(3;52)

------------

And the question is, is 'g' a surjective function.

Isn't the domain of g all of the integers? If so, that would be your set A, assuming that's what you mean by A. To be able to say whether g is surjective, you need to get some more (correct) function values. When you do that, you will probably be able to answer the question.

The domain would be ℤ , yes, all integers. I cannot find a limitation, and no limitation was specified. I have added the correct values for the range.

My problem is this :

a) I have 'domain' (x value) examples, and know it consist of ℤ. The range cannot possibly b
e ℤ, however if it is g: ℤ→ℤ , then ℤ includes negative elements as well? To me if the x is multiplied with itself, it will definitely be positive , and cannot be negative. (Correct me if i am wrong, looking at the function 'g')

Can you confirm what the range values will be ? After that, I should be able to see if it is surjective or not.

Thanks!

 

[Mark44]

Correct Values :


2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

g(1) = 5(1)2 + 7 = 12
g(2) = 5(2)2 + 7 = 27
g(3) = 5(3)2 + 7 = 52

Rewriting the (x;y) in each of the above examples :
(1;12)
(2;27)
(3;52)

------------

And the question is, is 'g' a surjective function.

Isn't the domain of g all of the integers? If so, that would be your set A, assuming that's what you mean by A. To be able to say whether g is surjective, you need to get some more (correct) function values. When you do that, you will probably be able to answer the question.

The domain would be ℤ , yes, all integers. I cannot find a limitation, and no limitation was specified. I have added the correct values for the range.

The numbers you calculated are in the range (or codomain), but they aren't the range.

My problem is this :

a) I have 'domain' (x value) examples, and know it consist of ℤ. The range cannot possibly b
e ℤ, however if it is g: ℤ→ℤ , then ℤ includes negative elements as well?

Z includes all of the integers: positive, negative, and zero.

To me if the x is multiplied with itself, it will definitely be positive , and cannot be negative. (Correct me if i am wrong, looking at the function 'g')

Can you confirm what the range values will be ? After that, I should be able to see if it is surjective or not.

I think you're on the right track, so I'll give you a push. All you need to do to establish that g is not surjective is to find one value in Z that is not in the range of g.

It might be helpful to sketch a graph of g. Note that the graph is not a smooth curve -- it's a bunch of disconnected points that lie on a curve.

 

[Supierreious]

Thanks for your quick reply, I appreciate your assistance. i am quickly reading your reply. And thank you once again for the assistance, you are seriously helping me an insane amount here.

 

[Supierreious]

Sheez, i am still struggling a bit

I have attached the graph, however , if i go into a minus, it still gives positive values ( this graph was done on excel).

Further to this i did some more reading, and found something else which confuses me a bit more :)

Let me share this with you :


ran(g) = {g(x) | x ∊ ℤ}
= {5x² + 7 | x ∊ ℤ }
= {y | for some x ∊ ℤ, y = 5x² + 7 }
= {y | 5x² = -7 } [this cannot be possible because it must be positive ]
= { y | don't know what else.. but it is confusing :D }

it cannot be equal to the codomain

 

[Mark44]

Sheez, i am still struggling a bit

I have attached the graph, however , if i go into a minus, it still gives positive values ( this graph was done on excel).

Further to this i did some more reading, and found something else which confuses me a bit more :)

Let me share this with you :


ran(g) = {g(x) | x ∊ ℤ}
= {5x² + 7 | x ∊ ℤ }
= {y | for some x ∊ ℤ, y = 5x² + 7 }

OK to here, but incorrect in the next step. You are assuming that y = 0 for some x. Your graph should show you that this is not true. IOW, 0 is not in the range of g.

= {y | 5x² = -7 } [this cannot be possible because it must be positive ]
= { y | don't know what else.. but it is confusing :D }

it cannot be equal to the codomain

Who is "it"?

 

[Supierreious]

I see ( i think i see.. :)


So i took away the y, which i cannot really do. but after a cup of coffee I saw.

so let me do the next step again :

y = 5x² + 7
y - 7 = 5x²

y-7
---- = x²
5

So y cannot be a couple of things :

a) it cannot be less than 7
b) it must be a multiple of 5, and if it is, it must be possible to calculate the square root, because x is a ℤ, and thus the left hand side of the formula must amount to the sqaure root of any ℤ,

Not sure if i am on the right track here :)

 

[Supierreious]

I double checked the graph values - and i think that the scale is just right, however y will be 7 when x = 0 :X Y
-10 507
-9 412
-8 327
-7 252
-6 187
-5 132
-4 87
-3 52
-2 27
-1 12
0 7
1 12
2 27
3 52
4 87
5 132
6 187
7 252
8 327
9 412
10 507

 

[Mark44]

I see ( i think i see.. :)


So i took away the y, which i cannot really do. but after a cup of coffee I saw.

so let me do the next step again :

y = 5x² + 7
y - 7 = 5x²

y-7
---- = x²
5

So y cannot be a couple of things :

a) it cannot be less than 7
b) it must be a multiple of 5, and if it is, it must be possible to calculate the square root, because x is a ℤ, and thus the left hand side of the formula must amount to the sqaure root of any ℤ,

a) Yes, y cannot be less than 7 (equivalently, y >= 7).
b) No, y is not a multiple of 5, but y - 7 must be a multiple of 5. Further, y - 7 must be a multiple of 5 such that (y - 7)/5 is a perfect square. For example, y = 12 satisfies these criteria, because 12 - 7 = 5, and 5/5 = 1, which is a perfect square.

However, I think you're getting a bit lost in the weeds. All you need to do is find a number in Z (the domain of g) that is not also in the range of g. If you can find such a number, which should be easy now, you can say whether g is surjective.

Not sure if i am on the right track here :)