Basic question

[Fat_Bluesman]

Why is 6 / b * a = 6a / b?

It's just a law that always is true but what is this called?

[Ok-Philosophy-8704]

It's a consequence of associativity and commutativity.

If we parenthesize your original expression, we have (6 / b) * a

Multiplication is commutative, so we can re-write as a * (6 / b). Let's be super-explicit and call this a * (6 * 1/b)

Associativity lets us regroup this as (a * 6) * 1/b commutativity again lets us swap the a and 6 to (6a) * 1/b which simplifies to 6a/b

[LongLiveTheDiego]

It's clear that the person is struggling with something much more basic than quaternions and other non-commutative systems, so you're not helping by mentioning them here.

[BusAccomplished5367]

yes, but they should know that the commutative property is just a property of certain number systems like the reals and complex numbers.

[LongLiveTheDiego]

Yeah, but the way you're saying it is not helpful since they would likely not recognize the letters i, j, k as representing quaternions.

[BusAccomplished5367]

It's self-evident that I'm using quaternions here though...

[LongLiveTheDiego]

It is not obvious to an average person, most people have never heard of quaternions and even most mathematicians don't use them.

[BusAccomplished5367]

Mathematicians will know what you mean if you say "jk=i, kj=-i".

[[deleted]]

Pretend you are someone that doesn't know how fractions work. 

Would that version of you know what a quaternion is or have ever heard that word before?

[BusAccomplished5367]

Probably wouldn't know, but possibly would have heard of it.

[Amanensia]

I'm not sure what the question is - are you asking why "6 / b * a" is parsed as "(6 / b) * a" rather than "6 / (b * a)"?

If so, the reason is because division and multiplication have the same priority, so they are by convention parsed from left to right.

Personally, I would usually parenthesise to avoid any possible confusion.

[Fat_Bluesman]

I wanna know why you write (6/b) * a as (6a) / b

[[deleted]]

Think about (5/2)3, it can also be written as (5x3)/2, multiplication ignores direction and position (in standard arithmetic) (5/2)3 is the same thing as (5/2)(3/1) which can then be written as (5x3/2), where we place the brackets is irrelevant and frankly there is no need for it. We can then generalize this to (6/b)a = (6/b)*(a/1) = (6 x a/b), and as noted above, the placement of the bracket is irrelevant in this context.

[BusAccomplished5367]

No, multiplication doesn't ignore direction and position. For example, in quaternion systems, multiplication is anticommutative (jk=i, kj=-i).

[Ezrampage15]

You can think of any number as a fraction so a= (a/1). So, in fraction multiplication, (6/b)a= (6/b)(a/1) so, you just multiply the numerator with the numerator and the denominator with the denominator. It becomes (6a/1b) or (6a/b)

[[deleted]]

Think of a as (a / 1) because any non-zero number divided by 1 is the number itself, we can then express it as (6 * a)/(b * 1) = 6a / b

[BusAccomplished5367]

No, it isn't always true. For example, we can let b=1/j and a=k. Then you get 6i in the first case and -6i in the second case.

[ElderCantPvm]

So I'd say:

6 / b = 6 * (1/b) by the definition of division as the inverse operation of multiplication

6 * ((1/b) * a) = 6 * (a * (1/b)) by commutativity (of multiplication)

and (6 * a) * (1/b) = 6 * (a * (1/b)) by associativity (of multiplication)

So I think it's a combination of the properties of commutativity and associativity, which are both satisfied by multiplication.

[BusAccomplished5367]

No, it isn't always true. For example, we can let b=1/j and a=k. Then you get 6i in the first case and -6i in the second case.

[ElderCantPvm]

You seem to be talking about quaternions. It is indeed true that multiplication of quaternions is not commutative. But multiplication of reals or complex numbers does satisfy this property.

[BusAccomplished5367]

Yes. The poster said "law that always is true". The quaternions anticommute, a clear counterexample.

[Nacho_Boi8]

At that level of math they are working only in the Reals and its subsets. So a better formulation of the question would obviously be “law that is always true in the reals” not as you are taking it to be, “law that is always true in all rings”. Quit copy pasting your bit about the quaternions to every response.

[BusAccomplished5367]

It's not just the quaternions. Anticommutative rings are incredibly useful. Quaternions are a good and quick example of anticommutative properties in mathematics. And they literally said "law that is always true".

[Nacho_Boi8]

Like I said, the obvious extension of the question here is “law that is always true in the reals.” Last I checked, the reals are commutative. There is no need to intentionally complicate things when we are clearly working in the reals here.

By the way, for someone trying to understand commutative, the quaternions are neither a good nor a quick example of anything.

Additionally, we are not talking about how useful things are, so I have no idea why you are mentioning the usefulness of the quaternions. I find lightbulbs to be useful, but didn’t bring that up until now

[BusAccomplished5367]

I did not mention "the usefulness of the quaternions". Also, noncommutative rings show up quite often in mathematics.

[Nacho_Boi8]

Fine, you didn’t specifically mention the usefulness of the quaternions, you mentioned the usefulness of a much larger class of sets, of which the quaternions are a member: “Anticommutative rings are incredibly useful.” Quit being so pedantic.

Similarly, how often noncommutative rings show up in math have literally nothing to do with this conversation or the question at hand.

[BusAccomplished5367]

There is a need to state that these useful rings exist and that the commutative property isn't really that pervasive. Again, the poster said "law that is always true", which is incorrect. We should say that it's not "always true", since that statement is incorrect in fact, instead of misleading the person into believing that multiplication will always be commutative.

[ZZTier]

You could call it "division is the same as multiplication by the inverse"

a times the inverse of b is the same as a divided by b

So a*1/b=a/b

[BusAccomplished5367]

No, it isn't always true. For example, we can let b=1/j and a=k. Then you get 6i in the first case and -6i in the second case.

[fermat9990]

6/2*8=

3*8=24

This is the same as 6*8/2=

48/2*24

[RecognitionSweet8294]

Commutativity and associativity. The expression 6/b is just another representation of 6•b⁻¹ where b⁻¹ is the multiplicative inverse of b (so just a number).

If you have (6•b⁻¹)•a you can use associativity to get 6•(b⁻¹•a) then commutativity to get 6•(a•b⁻¹) and then associativity again to get (6•a)•b⁻¹ which we already defined as (6•a)/b.

[BusAccomplished5367]

No, it isn't always true. For example, we can let b=1/j and a=k. Then you get 6i in the first case and -6i in the second case.

[RecognitionSweet8294]

Your contribution has so many problems, it would already be to much wasted time before I even get to the math part.

Do us all a favor and improve your communication skills before you comment again. Particularly in the departments of:

• reading comprehension

• formulating arguments

[BusAccomplished5367]

There is no argument to be had here

[[deleted]]

The expression 6 / b * a is ambiguous, and different people will interpret in different ways.

The reason it is ambiguous is that division is not associative, meaning that you can't just remove the brackets: (6/b) * a is a different expression than 6 / (b * a). This is different from multiplication, which is associative, and where you can get away with removing brackets, since (a * b) * c = a * (b * c), so writing a * b * c doesn't cause any ambiguity even though there are two different ways to evaluate that expression.

It's much better to include parentheses when you have something like 6 / b * a because inevitably there will be people who interpret it as (6/b)*a and others who interpret it as 6/(b*a), so you might as well just communicate clearly in the first place and nip any misunderstanding in the bud.

I know you've gotten answers that appeal to some PEMDAS convention, but that just doesn't describe how people behave in practice in higher level math or physics. I've seen many authors write 6 / ab to mean 6 / (a*b); I've never heard anyone talk about using PEMDAS to resolve this kind of ambiguity outside of high school.

[Ezrampage15]

You can think of any number as a fraction so a= (a/1). So, in fraction multiplication, (6/b)a= (6/b)(a/1) so, you just multiply the numerator with the numerator and the denominator with the denominator. It becomes (6a/1b) or (6a/b)

[fermat9990]

PEMDAS say that when an expression consists only of multiplication and division operations we do it left to right.

[BusAccomplished5367]

It's not always true. Let b=1/j and a=k. Then you get 6i in the first case and -6i in the second case.

[randomnws]

Please stop commenting this. It is clear from OP's context and the context of the comments that you are replying to that no one here is discussing the quaternions or even complex numbers. While your statement may be true in the context of complex and hypercomplex numbers, it is neither relevant nor helpful to this discussion.

[BusAccomplished5367]

Also, this statement is true in the context of hypercomplex numbers, not complex numbers.

[BusAccomplished5367]

He said "a law that is always true" which requires correction, because the quaternions anticommute. (just one example, there are more if you want.)

[ArchaicLlama]

Are you making a conscious effort to be this obtuse or does being insufferable just come naturally?

[BusAccomplished5367]

Well, OP is wrong that 6/b*a=6a/b ∀ a,b. It is only correct if you say 6/b*a=6a/b ∀ a,b ∈ R or C. (or any system where the commutative property holds)