mtey wrote:Today I managed to get them both wrong however after the test I solved them and they are actually not so tough.
Since f'' has only one root - and it is increasing, f' is decreasing in and increasing in . Thus from f'(0) < 0 follows that f' has only one root(it is continuous since f'' exists) and it is unbounded from the mean-value theorem. The same argument is valid for f. So f is unbounded and has a unique solution, but it is NOT always increasing (just like f' too).
I. You can easily verify that this is a subring by checking that 1, 0, addition and multiplication of elements are in it.
II. This is what I got wrong. Here is a simple counter example p1(x) = -x^2 + x, p2(x) = x^2; p1(x) + p2(x) = x is not in this subset of R[x]
III. Now it is a well known fact that Q[x] is ring of the polynomials with rational coefficients, but if you don't know that you can check it easily.
We cannot go out of the ring by using multiplication and addition and it contains 1 and 0 too.
feroz_apple wrote:I dont think you need to show the existence of 1 for it to be a subring
origin415 wrote:If 1 wasn't in the set, it wouldnt be a ring, and therefore not a subring.
And it is possible to have 0, addition, and multiplication to be defined on a subset of a ring but not include 1, such is the case for any proper ideal.
origin415 wrote:In my abstract algebra class, rings always had unity, I've never heard it any other way.
It appears the Princeton Review book doesn't require unity, so I suppose the GRE doesn't either, my mistake.
Users browsing this forum: No registered users and 6 guests