Recent content by brntspawn

Yeah that is pretty much where I am at now, realizing that there is no one correct way to write a proof and finding my own style for mine when I write them. My school used Stewart for its calculus text and I came away with it feeling as though I didn't really understand much. I hated Calculus...

Thanks guys. I was actually wondering about the assume part because I know it was drilled into us in class that we should specifically state it. I assume at some point, perhaps when you move from undergrad classes to grad classes you do not specifically need to state it, but I am not sure. Are...

Homework Statement Prove: If a < b and c < d then a+c < b+d Homework Equations The Attempt at a Solution Proof Assume a < b and c < d then a+c < b+c and b+c < b+d so a+c < b+c < b+d therefore a+c < b+d Q.E.D. Pointers and suggestions are also welcome. I am looking for...

Ok, let's see if this is any better: Assume |S|=1 so -1\leqS\leq1 since 1\in\Re and \foralls\inS, 1\geqs, then by definition 1 is an upper bound since 1\inS then 1=max(S) Thus true for 1, now consider the case for m+1 Assume |S|=m+1 then -m-1\leqS\leqm+1 \forallm\inS, m+1>m and since...

Thanks Dick, I will work with that, I replied before seeing your post.

What I posted was the entire problem. Yes R was for the set of Reals. It had a hint with it: Use induction, which is why I was trying to use induction. Like I said I struggled in the intro class, so it is not surprising to me that my proof looks confusing. I assume starting with S={1} is where...

Homework Statement If S\subsetR is finite and non-empty, then S has a maximum. Can someone look over this? I struggled a bit in my first proof class, which is why I am asking for help, so I really am unsure if this is right at all. Let S={1} So 1\inR such that for all x\inS, 1\geqx So...

Great. Thanks for the help.

Homework Statement \exists M \in R such that \forall x\in S, x\leqM Write in symbolic for the negation of the statement. The Attempt at a Solution \forall M\in R, \exists x\in S such that x\geqM Is this correct?

Thanks for the help. :)

(\sqrt y - \sqrt x)^2>0 From this I can get: x<y \sqrt{x}<\sqrt{y} \sqrt{x}-\sqrt{x}<\sqrt{y}-\sqrt{x} 0<\sqrt{y}-\sqrt{x} 0<(\sqrt{y}-\sqrt{x})^{2} 0<y-2\sqrt{xy}+x 2\sqrt{xy}<y+x \sqrt{xy}<1/2(y+x) Assuming I am correct with this I have two more questions: 1. How did you know to...

For 0<x<y, show that x<\sqrt{xy}<1/2(x+y)<y I have no difficulty showing that x<\sqrt{xy} and 1/2(x+y)<y. I am having difficulty with \sqrt{xy}<1/2(x+y). x<y xx<xy x^{2}<xy x<\sqrt{xy} and x<y x+y<y+y x+y<2y 1/2(x+y)<y