Well, you’re on the right track with your answer to #2, but you need to elaborate just a little bit.

I’m intending to use this thread to post some interesting exercise questions on various math topics (and various math levels); some of the questions will be interrelated. Thus I’ve posted my first three exercise questions.

Correct.

Excellent job.

Now someone try #1? It’s not that difficult.

1. every number that isnt a prime can be factorized in its prime factors, for example 36 can be factorized in 2*2*3*3.
so the number c can be factorized in the prime factors P1*P2*P3.....Pn
a and b are relatively prime so they cant have one same prime factor, but since c is divisible by both a and b, they must be created by one or some of the prime factors in c. therefor a*b can never be more than c.

for example a can be P1*P2 and b can be P3*P4. if we divide c with a and b there will still be P5*P6*....*Pn left.
but if a is P1*P2, b cant be for example P2*P3 since then both will have P2 as a prime factor.

would this be a correct proof?? still practising

Last edited by JaneFairfax (2007-03-20 18:19:07)

0⁰ is defined as 1. Try it on a calculator.

And

Last edited by JaneFairfax (2007-03-20 18:22:26)

JaneFairfax wrote:

You’re more or less on the correct line of thought. However the problem is to show that c is divisible by ab. It’s true that ab is less than or equal to c, but this doesn’t show that c is divisible by ab.

I’ve already given a short solution using Bézout’s identity.

well since ab is created by the prime factors in c, c must be divisible by ab?

Letting 0⁰ be 1 is a standard convention, which helps eliminate special cases where some theorems break down. There are some combinatorial and set-theoretic justifications for this convention. However, in analysis 0⁰ is treated as an indeterminate. Moreover, consider x^x = e^{x ln x}. There is no real number corresponding to ln 0, so x^x is not defined at x = 0; in other words 0⁰ is undefined. The calculator says it is equal to 1 because of the common convention mentioned above; but a real decent calculator would notify you something like "Warning: 0^0 replaced by 1" .

Well, I’ve been trying to create some more exercises for this thread. Here is something I just this moment made up.

7. Let f(x) be a periodic odd function with period n. In other words, f(x) satisfies f(x) = −f(x) (odd function) and f(x+n) = f(x) (periodic with period n). If n is an odd integer greater than 1, prove that

Last edited by JaneFairfax (2007-04-17 08:27:46)

9. Prove that a monotone-increasing sequence of real numbers that is bounded above converges to its least upper bound.

10. Give an example of a monotone-increasing sequence of rational numbers that is bounded above which does not converge to a rational number.

11. Show that nevertheless a monotone-increasing sequence of rational numbers that is bounded above is a Cauchy sequence.