Is function composition commutative?

musicbachv7 2022-08-05 Answered
Is function composition commutative?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Jake Landry
Answered 2022-08-06 Author has 22 answers
The functions g and f are said to commute with each other if g???f = f???g. In general, composition of functions will not be commutative.
Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, |x| + 3 = |x + 3| only when x ? 0.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-05-29
suppose A is a C -algebra and x is a normal element in A. C -algebra generated by x denote by A [ x ]. then
1) A [ x ] is commutative.
2) A [ x ] is the clouser of polynomials of two variable x and x .
asked 2022-06-21
Is there an algebraic way to construct the coproduct, as for example some form of topological tensor product or similar?
Let A be a commutative unital C*-Algebra and let X = S p e c ( A ) be the corresponding compact Hausdorff space of characters. By Gelfand-Naimark duality we know that
X × X = S p e c ( A A )
or in other words
A A = C ( X × X ) .
asked 2022-06-20
Let A be some F-Algebra, for some field F, with the property that A is finite dimensional over F. Is A always commutative?
asked 2022-04-10
An associative magma is said to be a semigroup. A semigroup with identity is said to be a monoid... Is there a name for an associative and commutative algebra?
asked 2022-07-14
Let A B be rings, B integral over A; let q , q be prime ideals of B such that q q and q c = q c = p say. Then q = q .
Question 1. Why n c = n c = m?
My attempt: Since p q, we have m = S 1 p S 1 q = n B p . But m is maximal in A p , which is not necessarily maximal in B p . I can't get m = n by this.
Question 2. When we use that notation A p , which means the localization S 1 A of A at the prime ideal p of A. But in this corollary, p doesn't necessarily be a prime ideal of B. Why can he write B p ? Should we write S 1 B rigorously?
asked 2022-05-24
Show that R can be written as a direct product of two or more (nonzero) rings iff R contains a non-trivial idempotent. Show that if e is an idempotent, then R = R e × R ( 1 e ) and that R e may be realized as a localization, R e = R [ e 1 ].
asked 2022-06-24
Is it true that for simple C -algebras, meaning that they don't have non-trivial two-sided ideals, it holds that they are necessarily non-commutative? And why?

New questions

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question