Question

Please help! An operation * has been defined over the set of natural numbers.
(3*2)*2; x*y = (x+y)/(xy)
How do I determine whether this is commutative/associative or not?

Answer 1

What is the (3*2)*2 in front for?

Answer 2

It's find (3*2)*2

Answer 3

\[\large\rm \color{orangered}{(3*2)}*2\quad=\quad\color{orangered}{\frac{3+2}{3\cdot2}}*2\]Were you able to figure that part of the problem out?

Answer 4

Yes. So we substitute the answer into the equation again since its *2 right?

Answer 5

Yes, first simplify the orange part though.
I guess that works out to 5/6 right?
Oh wait wait wait...
this operation is defined over the `natural numbers`...
not necessarily the rational numbers (which 5/6 belongs to).
Hmm

Answer 6

Ya I don't think this operation is defined for (3*2)*2
Hmmmm D:

Answer 7

oh ok. Can you tell me in general how to check if something is commutative or associative when an operation is defined?

Answer 8

so for commutative, would it be (y+x)/(xy) = (x+y)/(xy)

Answer 9

Let \(\rm x,y,z\in\mathbb N\)

Then, \(\large\rm (x*y)*z=?\)
Work that out and determine whether or not it is equal to \(\large\rm x*(y*z)\).
That's what we need for `associativity`, ya?

Answer 10

Yes, you have the right idea.
Let \(\rm x,y\in\mathbb N\)

Then determine whether or not \(\large\rm x*y=y*x\)

Answer 11

How do I know what * is?

Answer 12

Well it was defined above, yes?\[\large\rm x*y=\frac{x+y}{xy}\]then just apply a simple step here (since we know that our normal addition and multiplication are commutative,\[\large\rm \large\rm x*y=\frac{x+y}{xy}=\frac{y+x}{yx}=y*x\]

Answer 13

Ohh I see. ((x+y)/(xy)) * 2 = (x/(xy)+((y/xy) * 2)
And find if that is equal for it to be associative?

Answer 14

No no, let that last value be some unknown from the naturals.
Otherwise we know that associativity works with 2, but not necessarily anything else.
And also, I don't think you want to break up the fractions..
seems like that's going to make it really really complicated.
I mean, for one thing, we don't know if * is distributive.

Answer 15

\[\large\rm (x*y)*z=\frac{x+y}{xy}*z=\frac{\frac{x+y}{xy}+z}{\left(\frac{x+y}{xy}\right)z}\]Get something like that, ya?

Answer 16

Do the same for this \(\large\rm x*(y*z)\).
Start in the brackets.

Answer 17

Sorry, I'm not sure if I understand....how do I write (y*z) if I can't break up the operation?

Answer 18

We defined * above, yes? :o\[\large\rm (y*z)=\frac{y+z}{yz}\]You having trouble with this weirdly defined operation?

Answer 19

For this problem,
Star just means, add in the numerator, multiply in the denominator.

Answer 20

Ohh, ok. so \[x*(y*z) = x* \frac{ y+z}{ yz } = \frac{ \frac{ y+z }{ yz }+z }{ \frac{ y+z }{y z } *z }\]

Answer 21

What...?
What happened to your x? :(

Answer 22

And why is the fraction on the left side?

Answer 23

Sorry, I put in the wrong key. the z is suppose to be the x

Answer 24

I'm not sure why you put the x's on the right side...
that's rather confusing.
And also, do not use * for multiplication, that's extra confusing :D lol

Answer 25

\[\large\rm x*(y*z)=x*\frac{y+z}{yz}=\frac{x+\frac{y+z}{yz}}{x\cdot\frac{y+z}{yz}}\]ok good.\[\large\rm (x*y)*z=\frac{x+y}{xy}*z=\frac{\frac{x+y}{xy}+z}{\left(\frac{x+y}{xy}\right)z}\]We need to determine whether or not these big fractions that we ended up with are equivalent. So I guess we need to simplify them further :[

Answer 26

Err maybe we don't need to.
I think it's pretty clear from this point that they are not going to be equal.
Umm

Answer 27

Thank you so much so staying on this problem so long to help me!

So we should keep it in variables to find if the operation is commutative or associative rather than substituting numbers in?

Answer 28

Yes. If you plug in specific numbers,
you're able to determine that you have associativity *sometimes*,
but not necessarily all the time.

You may have just gotten lucky and found a few numbers that are associative.
Leave them unknown so they can represent any natural from the set.

So what did we end up with here?
I think we determined that * is `associative` but not `commutative`?
I hope we did everything alright. :P

Answer 29

Is this like.. Abstract Algebra or something?
Maybe ask someone or teacher about the (3*2)*2 thing.
This isn't my strongest subject so it'd be good to get another opinion :D

Answer 30

It's part of the section addition and multiplication properties of real numbers.

Did you mean to say that it is commutative but it's not associative?

Answer 31

Ahh yes, good call.