My teacher solved this using inclusion-exclusion formulas to count the number of surjections from a set of 8 elemets (containing items) to a set of 5 elements (containing boxes). However, I thought of a different solution. But I have a hunch it's wrong.
What I thought is to first make sure every...
I am interested in the following theorem:
Every field of zero characteristics has a prime subfield isomorphic to ℚ.
I am following the usual proof, where we identify every p∈ℚ as a/b , a∈ℤ,beℕ, and define h:ℚ→P as h(a/b)=(a*1)(b*1)-1 (where a*1=1+1+1... a times) I have worked out the...
Let (A,+) be an Abelian group. Consider the ring E=End(A,A) of endomorphisms on the set A, with binary operations +, and *, where (f+g)(x)=f(x) + g(x), and (f*g)=f∘g.
I have tried to find zero divisors in this ring, but I just couldn't come up with an example.
I gave little information, and I am sorry.
To skip the settings, here's straight to the problem. Say we want to prove that limit of the function f(x)=sinx/x as x approaches 0 is 1. We can play around and get that cosx<sinx/x<1 for 0<x<π/4. Since the limit of cosx as x approaches 0 is 1, and...
When we define a limit of a function at point c, we talk about an open interval. The question is, can it occur that function has a limit on a certain interval, but it's extension does not? To me it seems obvious that an extension will have the same limit at c, since there is already infinitely...
So, I recently came across this example: let us "define" a function as ƒ(x)=-x3-2x -3. If given a matrix A, compute ƒ(A). The soution proceedes in finding -A3-2A-3I where I is the multiplicative identity matrix.
Now , I understand that you can't add a scalar and a matrix, so the way I see it is...
So is my 2nd post incorrect? I know that the given projection is not injective, but it is surjective. Can we then restrict it to an injective one, and get a new function from a subset of the domain to the codomain? From here it follows that the second set is either finite or countable. Anyways...
This book is in Croatian, and you can say it is not really a book, more like a compilation of notes made by one of our professors.
Also, sorry for posting twice. I didn't know about the convention.
So, I thought about this, and this is what I have concluded. Since there is a surjection from ℤxℤ* to ℚ, then there is injection from S⊂ℤxℤ* to ℚ, which means that there is a bijection from S to ℚ. Since ℤ⊂ℚ, ℚ is infinite, but then S is infinite too. Since S is an infinite subset of a countable...
I know there are many proofs of this I can google, but I am interested in a particular one my book proposed. Also, by countable, I mean that there is a bijection from A to ℕ (*), since this is the definition my book decided to stick to. The reasoning is as follows:
ℤ is countable, and so iz ℤxℤ...
This may be a silly way to approach it, but I thought of this.
(∀n∈M) s(n)∈M is by definition equivalent to (∀n)(n∈M →s(n)∈M), which is obviously not equivalent to (∀n∈ℕ)(n∈M →s(n)∈M). Another way to think about it is that these can become equivalent if we consider a few things, which is not...
So , what I was wondering about was a slight difference in notation, for which I am not certain if correct (mine, in particular.).
The induction axiom says: If M is a subset of ℕ, and if holds that:
a)1∈M
b)(∨n∈ℕ)(n∈M→s(n)∈M)
then M=ℕ.
Now my question is: why do we write (∨n∈ℕ)(n∈M→s(n)∈M)...