Recent content by bins4wins

Thanks. I would use induction but I have too prove this for all real numbers k > c. Also, to clarify what I can use, I can use, like MarneMath said, the properties of fields, the least upper bound property of real numbers, the properties of metric spaces regarding the reals, and...

First off, there EXISTS a c. It is not any c. So you cannot pick any c so that the inequality should hold. I'm sorry, but I really think there's a continuous misunderstanding between you and I, I just want to see what others can offer. No disrespect, just clearly miscommunication. Also you...

All you did was reorder terms in your proof. I'm sorry, but I really don't see where you're going with it. How does m(m^c - 1) > 0 imply that m^k > km for all k > c.

But the property you suggested is a simple consequence of the assumption and doesn't prove the result I need. If you choose c = 2, then we already know 2m < m^2. If for example we chose c = 2, then we have to show that km < m^k for all k > 2. But our c must remain arbitrary.

Yeah I think there's a misunderstanding. I edited the statement. It was just one question.

I understand what you are saying, but please carefully read what I asked for. The assumption is something we take as true, and the following statement is what we want to prove given that the assumption is true.

Yes its true in some cases, but not in all. But I'm trying to prove that its true in all cases given the assumption.

No, because that statement is not true. The assumption is important.

No Logarithms, no exponential function, continuity of functions cannot be used, and calculus cannot be used. Mainly algebraic manipulations.

Prove that if ##m > 1## such that there exists a ##c > 1## that satisfies $$cm < m^c$$ then for any ##k > c## $$km < m^k$$ holds. Prove this without using logarithms or exponents or calculus. Basically using the properties of real numbers to prove this. One attempt I have tried, but didn't...

Thanks, I was wondering why it wasn't working...

Personally, I find the idea of the notation p(A) fairly loose in terms of rigor, since the definition of p as a function has a domain of the reals. The most amount of rigor you can put in plugging in A is just by defining what it exactly means to plug in A and that is to plug it into the...

We know p(x) is the characteristic polynomial of A. The meaning of p(A) is to take the input A into the characteristic polynomial, not into the formula d(A - xI). Also, note that the 0 in the expression p(A) = 0 isn't referring to a scalar 0, its referring to the zero matrix.

For any real ##x > 0##, prove that the sequence ##n^{n^{-x}}## is bounded (and if possible, monotonically decreasing after some point). The catch is that logarithms and the exponential constant cannot be used. We must arrive at the proof using fairly "primitive tools" If you look at the graph...