Question

show that with respect to point wise addition and multiplication of function ,the set of all differentiable function f:R TO R is a commutative ring with unit element?

Answer 1

So tell me... what exactly IS a commutative ring? :)

Answer 2

a commutative ring with unit element which has at least two elements ,but nozero divisor is called integral domain

Answer 3

You're back.
We don't need an integral domain, we just need a commutative ring, which, by definition, is a ring with its multiplication element also commutative. Okay?

Answer 4

yes

Answer 5

Now... what are the two operations defined on functions in this case?

Answer 6

addition and multiplication

Answer 7

Yes. IS Addition commutative on functions?

Answer 8

yes

Answer 9

What about multiplication?

Answer 10

no

Answer 11

Let f and g be differentiable functions from R to R.
Is it not so that
\[\Large f(x)g(x) = g(x)f(x)\]?

Answer 12

yes but in ring does not equal to each other

Answer 13

What do you mean?

Answer 14

i mean the ring have to has all the properties except mult identity ,mult commutative and mult inverse

Answer 15

How is it not commutative under multiplication?
Can you give an example of two differentiable functoins from \(\large \mathbb{R}\rightarrow \mathbb{R}\)
such that
\[\Large f(x)g(x) \ne g(x)f(x)\]?

Answer 16

yes such that if i have matrices 1 2

3 0 and

1 3
2 4 doesn't equal to 1 3
2 4 times 1 2

3 0

Answer 17

But we aren't dealing with matrices are we?
We're dealing with DIFFERENTIABLE FUNCTIONS from \(\large \mathbb{R}\rightarrow \mathbb{R}\)

Answer 18

okay

Answer 19

Granted that multiplication of functions is COMMUTATIVE. Do you agree?

Answer 20

yes i am confused because the ring have to don't has multipcation commutative and in this question i have toprove that is commautive ring with unit element

Answer 21

A ring doesn't HAVE TO BE a commutative ring (IE, its 'multiplication' operator doesn't have to be commutative).
But it CAN be.
Do you understand?

Answer 22

can you give me example of set and prove that to be more clear

Answer 23

The set of Integers form a commutative ring.

Answer 24

That's because
(1) When you add any two integers, the sum is an integer
(2) When you multiply any two integers, the product is an integer
(3) There is an integer, call it 0, such that when you add it to any integer, the sum is THAT integer
(4) For any integer a, there is an integer, call it -a such that a + (-a) = 0
(5) For any 2 integers a, b, a+b = b+a
(6) For any 2 integers, a, b, ab = ba
(7) For any 3 integers, a, b, and c, a(b + c) = ab + ac

The conclusion holds.

Answer 25

okay if i choose two element like give me any two elements in intger because when i choose 1 and 2 it does not have addition closure and i do not know how i choose intger or when i want to choose z mode 3 how i can work with mult idinity in this case

Answer 26

^I'm sorry, but I really did not understand that...

Answer 27

sorry i am international student and maybe because of that.. i meant my problem can you give me number in integer i can work with because i had work with z mod 3 and when come to deal with mult identity i do not know how .. hope its clear

Answer 28

You need to work with this group/ring?
\[\Large \mathbb{Z}_3\]

Answer 29

okay can you explain to me mult identity in this group

Answer 30

It's 1.
And anyway, that's beside the point... we still have not yet finished with your original question.

Answer 31

yes you mean i have to choose any number in the set z mode 3 and multiply in 1 ?

Answer 32

We should save this for later. First, we work on your original question, because this one is rather easy compared to the original.

Is function multiplication distributive over addition?

In other words, for any functions f,g, and h, is it true that
\[\Large f(x)[g(x) + h(x) ]= f(x)g(x) +f(x)h(x)\]?

Answer 33

yes

Answer 34

Okay, two more things.
When you add two differentiable functions, is the sum also differentiable?

Answer 35

yes sure

Answer 36

(Note, we're skipping through details, I'm assuming you can fill them in)

Finally, when you multiply two differentiable functions, is the product differentiable?

Answer 37

not really

Answer 38

Counterexample?

Answer 39

excuse me you do not understand where i am struggle with i man can you choose any two element number and show me how you can prove dors not have mult idinity ,mult commautivity and mult invers and in the same time it is ring commuativity with unit elements

Answer 40

Calm yourself, and take it slow.
You said the product of two differentiable functions from \(\large \mathbb{R}\rightarrow \mathbb{R}\) is not necessarily differentiable. Then you can provide an example?

Answer 41

i have to choose two different elements to prove that

Answer 42

Yes... can you?

Answer 43

Can you find two differentiable functions whose product is not differentiable?

Answer 44

okay if i choose from z mode 3 that 2 mode 3 + 1 mode 3 =0 and its belong to z mode 3

Answer 45

Please don't change the subject. We are talking about differentiable functions from \(\large \mathbb{R}\rightarrow \mathbb{R}\) .

And me asking you to find two such functions whose product is NOT differentiable.

Please stop bringing up \(\large \mathbb{Z}_3\), for the moment, it is NOT RELEVANT to the discussion.

Answer 46

1+2=3

Answer 47

^Please read my previous post.

Answer 48

can i choose matrices

Answer 49

ARE matrices differentiable functions?
NO!
They are an entirely different notion.

Answer 50

sorry about that can you give me that...confused

Answer 51

Let's backtrack a bit.
If f and g are two differentiable functions from \(\large \mathbb{R}\rightarrow \mathbb{R}\) is f(x)g(x) also differentiable?

Answer 52

And where are you from, btw?

Answer 53

Saudi Arabia first time study math in English

Answer 54

Right. Anyway, my question before I asked you where you're from.
Think about it.

Answer 55

think about what??

Answer 56

f,g differentiable then is f*g differentiable
was the question...

Answer 57

Let's backtrack a bit.
If f and g are two differentiable functions from R→R is f(x)g(x) also differentiable?

Answer 58

You might want to recall a nifty rule called 'the product rule'
\[\Large \frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x) \]

Answer 59

(Even though you really shouldn't have to T.T)

Answer 60

what is T.T

Answer 61

An emoticon. Pay no heed to it, pay attention ONLY to my question.

Answer 62

SO ANYWAY thank you for your cooperation

Answer 63

i understand that it has to be differentiable and you do not answer my question can you explain to me how i can get two elements and prove that its ring with unite elements and commutative

Answer 64

For which?

Answer 65

The important thing I am trying to understand to prove any set that is commutative ring with unit elements

Answer 66

Fair enough,but that's easily determined...
what is the differentiable function e(x) such that e(x)f(x) = f(x) for all differentiable functions f?

Answer 67

1 times 2 =2 like this

Answer 68

Yes... so what is e(x)?

Answer 69

1

Answer 70

And that's the unity.

Answer 71

okay that is what i want the rule for any set to be ring must does not have mult identity and does not have mult commutative can you prove that

Answer 72

mshms, you need to read your book again and get better understanding of the definition

Answer 73

okay

Answer 74

i will read it a gain

Answer 75

Please do :)

Answer 76

thanks