By - Swim_in_poo
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It could also be that x and y do not commute. If they were matricies for example this might be the case. But probably wolframalpha says false because it interprets it as a function
Nice answer. I completely agree, after all it's a knowledge engine and needs more information than this provided above.
But matrices are represented by capital letters. So they are not matrices tbh
In linear algebra textbooks, totally. In other contexts, not necessarily.. Still a novice to them but I’ve seen writing in functional analysis and group theory where authors don’t bother with capitalization because you’re just interested in a general set/class of functions, of which some members might be matrices. For example, you might be interested reasoning about a set of functions that share some of the same properties as matrices (eg non commutativity) but are not linear maps per se. In this case, capitalization doesn’t add much information in the notation so it seems common to ignore it.
This is not generally the case, and besides I used matricies because it is the most straightforward explanation of a non-commutative algebra. An other example would be quaternions, where if we look at the basis elements \[i,j\] = 2k, whereas for a commutative algebra this would vanish. As pointed out by others group elements are an other example where x and y would be expected, but xy does not always equal yx. One simple (finite) example is S3 the symmetry group of order 3. This group contains all the permutations of three numbers, i.e. 123, 312, 231, 132, 321, 213. The composition rule here is that we reorder the numbers depending on their position. For example (312)(132) = (213). Notice how if we swap the two terms (132)(312) = (321), which is distinctly different. But again the elements of this group are generally labeled something like e = (123), r = (312), r\^2 = (231), m = (132), rm = (213), r\^2m = (321), with no regard for capitalisation.
Minor edit: Technically S3 does not contain the permutations of 1,2,3 but that is the easiest representation of the group to explain in a comment. The group elements just follow the algebra described and usually are just written either as the elements I already mentioned (e,r,r^2 etc) or simply a,b,c,d,e,f.
Wolfram alpha interprets A(B) as a function A dependant on B by default. If you want to get what you are thinking then you should use the [ ] brackets.
Just curious, would this be viable alternatives? (xy)(xy), or (xy)\^2
edit; honestly I could've just went to wolfram alpha and try it myself.. lol
If with x(y) you mean x as a function of y then thats just x so x²=x²*y² => y²=1.
If the brackets are just to change the order and you meant x*y*x*y = x²*y² that is in fact always true.
If you meant something else try to explain it better.
I disagree with the 2nd para, if x and y are non commutable then the equation doesn't hold true
Sure but unless otherwise specified it's reasonable in this context to assume that we're talking about standard multiplication which is commutative.
Which wolfram also assumes in my limited experience.
Try writing it [differently](https://www.wolframalpha.com/input?i=is+x*y*x*y+%3D+x%5E2+*+y%5E2%3F).
You're confusing the calculator
I only came to say "don't sell yourself short."
Just because you don't know something doesn't mean you're dumb. The fact that you seek understanding means you **aren't** dumb.
(My wife is a school teacher and I suffer from Imposter Syndrome, so I see this a lot.)
I thought because of commutativity and associativity you could just do x(y)x(y) = xx(y)(y) = x^2 * y^2
Like u/jakobocepek said. If you are interpreting them as two x's and two y's being multiplied together then this is true. The issue is that X(Y) could mean 'X as a function of Y'. The parenthesis are making the problem more confusing in this case.
Oh ok. So the parenthesis makes it look like X could be a function in which case it would be (x(y))^2, right? I see. Thank you
why do you put a single variable in parenthesis? i am not sure which order you are trying to impose, but you can write `(xy)(xy)` or `xyxy` or `x(yxy)` or, ... Your use of parenthesis likely confused the software, it also confused the humans here. Some say "We don't know if you are talking about (real)numbers or some other non-commutative algebra, so if x and y were matrices this wouldn't always be true" but i think you are assuming x and y are real numbers, or natural numbers, or something like that.
Oh ok. I understand it now. For whatever reason I became used to writing multiplication with a parenthesis like I write 2(4)x instead of 2 * 4 * x while solving a problem so sometimes it ends up like in the picture midway through. But yeah, it is probably because I am mostly doing simple math so only real numbers involved.
Instead of a(b)c I would suggest you write (a)(b)(c) instead. It looks cleaner and, to me, it's easier to understand.
it takes longer to write that way, but it does look better
Yea just do x * y * x * y
And as long as you're using real numbers and standard multiplication it will always be true
In addition to being functions, they could also just be noncommutative. For example if they are matrices, in general a*b does not equal b*a, or if they are quaternions ij =-ji
It’s not always true, since it depends what x(y) means.
X(y) is a function, and can mean anything.
It could mean x(y) = x*y in this case then it is true.
But it could also mean that x(y) = x+y
In which case it isn’t true.
It depends on the definition of x(y)
Enter just "x(y)" and you'll see why. Wolfram thinks this is a function.
"is (x*y) (x*y) =x^(2)*y^(2)"
Is asking what I think you want to ask it.
In words does that mean that it is because y could be a polynomial instead of just a single number and a negative number in the polynomial would mess things up?
if x is a function writing x² without a variable doesnt make sense
only time that wouldn’t be true is if x and y don’t commute
The way you've formatted it, it thinks x is a function of y
If [x,y]\neq 0 so xy-yx = A. x(y)x(y) so switching x and y in the middle we find xxyy the difference between xxyy and xyxy = x[x,y]y \neq 0
Don't forget that sometimes x\^2 y\^2 = (-x)(-x) \* (-y)(-y)