# Recent content by UNChaneul

1. ### Left Cosets of Z in Q

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 ~~
2. ### Left Cosets of Z in Q

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...
3. ### Left Cosets of Z in Q

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...
4. ### Bridge to abstract math: what is wrong with following proof

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...
5. ### Question about correct theorem, for a bunch of jibberish

Do you really want to start generating the positive integers via sets?
6. ### Question about correct theorem, for a bunch of jibberish

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.
7. ### Question about correct theorem, for a bunch of jibberish

yes, with the extra caveat that the upper and lower bounds of y are set by f(y) at the two points you choose to create the interval.
8. ### Proof 30 divides ab(a^2 -b^2)(a^2 +b^2)

Ah I see, good 'ol testing things (but hopefully not "proving" them via American Induction, as my math prof used to joke about).
9. ### Question about correct theorem, for a bunch of jibberish

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...
10. ### Question about correct theorem, for a bunch of jibberish

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?
11. ### Isomorphism and Binary operation

English needs a pronoun for when one does not know gender xD. Using "they" sounds so awkward when referring to one person!
12. ### Question about correct theorem, for a bunch of jibberish

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...
13. ### Question about correct theorem, for a bunch of jibberish

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.
14. ### Isomorphism and Binary operation

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...
15. ### Proof 30 divides ab(a^2 -b^2)(a^2 +b^2)

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...