Oh, so it's not implying multiplying every element of the set by a constant. Well then that's easy enough to show lol.
I guess I just got confused by what the notation was referring to ~~
I had thought that, but then it seemed as though multiplication weren't quite distribution with that since
Z = ..., -2, -1, 0, 1, 2, ...
1/2 + Z = ..., -1 1/2, -1/2, 1/2, 1 1/2, 2 1/2, ...
2(1/2 + Z) = ..., -3, -1, 1, 3, 5, .... != Z which makes it seem as though distributivity does not apply...
Homework Statement
View Z as a subgroup of the additive group of rational numbers Q. Show that given an element \bar{x} \in Q/Z there exists an integer n \geq 1 such that n \bar{x} = 0.
Homework Equations
The Attempt at a Solution
As we are working in an additive group, it is...
Interesting... (P \lor Q) \rightarrow R
What is the truth table for this? That is one way of showing it. Namely, what is the value of R for various values of P and Q.
Similarly, what are those values in the case of (P \rightarrow R) \wedge (Q \rightarrow R)
Hint: Wolfram Alpha is...
Right, and as further elucidation, something like a+b = b+a is actually not always true... so you've got to be sure. That's why you have theorems that you might think are trivial. Turns out they might not be so trivial.
Well, one can do this algebraically or graphically. Since I think you want a visualization of it, I am going to describe it graphically.
Imagine a basic continuous curve of some sort defined by f(x)=y. For example any quadritic/cubic/etc will work.
Now let's pick two random x-values, say...
Wait, in that case it seems your question is not what the question states - because the question states, what theorem does this statement refer to. Whereas, your question seems to be... "what is the meaning of this theorem"? Am I understanding this correctly now or no?
How to explain this, you're not really messing up, cause like, there's 1 step to the problem.
This is because the problem is: "Here is the statement of a theorem, now name the theorem".
Basically it would have been written verbatim somewhere in your book, and if you were unsure, you could...
http://en.wikipedia.org/wiki/Intermediate_value_theorem
I'm pretty sure that's what you're looking for. I mean... it is trivial enough to not be able to really help without just telling you haha.
Technically he just showed Id_{x} is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that Id_{x} is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And...
Yea, reducing it to a^{5}b-ab^{5} is actually unhelpful as far as I can see, given how I proved it, as it would make things more complicated.. You want to do as Mark said and consider whether or not it is divisible by 2, 3, and 5. If it is divisible by any one of them, you don't have to worry...