Recent content by gottfried

Sorry I missed that but thanks for the clarification.

Cool, that is a very good thing to know.

I see. I was under the impression total meant one or the other rather than atleast one or the other.

Why is that total?

According to wikipedia a total order ≤ on a set X is one such that If a ≤ b and b ≤ a then a = b (antisymmetry); If a ≤ b and b ≤ c then a ≤ c (transitivity); a ≤ b or b ≤ a (totality). I'm wondering why antisymmetry is a condition since, as far as I can see, totality discounts...

Clearly using n+2 was a mistake but I originally used it to see if this would result in an easier inequality for me to manipulate. I see your proof and it makes sense. I didnt see it originally but my original proof works for n+1 aswell. 2n+1 =2n2 ≥2(1+n)2=2n2+4n+2≥n2+4n+4=(n+2)2 But...

I've actually just realized this can be done with n+1 and I'm just being silly. Sorry

Homework Statement Find all natural numbers such that 2n ≥ (1+n)2, and prove your answer. 2. The attempt at a solution I can see this is true for n=0 and n>5. I try to prove this using induction as follows 20 =1≥ 1=(1+0)2 base case: 26 =64≥ 49=(1+6)2 so it is true for n=6 and suppose 2n...

The proof is at a level that I don't really understand but I'm pretty sure the proof wasn't constructive.

According to a result of Paul Cohen in a mathematical model without the axiom of choice there exists an infinite set of real numbers without a countable subset. The proof that every infinite set has a countable subset (http://www.proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset) is...

That makes total sense. It has clicked into place

I can see that my questions are unclear but I'm not sure how to phrase them better, probably because my understanding of what I'm trying to say is pretty flakey but thanks for bearing with me. So if the 5 sets aren't measurable why does it make the proof false. What are the implications of a set...

It seemed too trivial, to be true. Thanks.

I'm trying to understand which aspects of the proof were true before we had a measure which no longer are if we put a measure on the set.

Let me see if I have this right. The rigid motions are measure preserving in the case of the paradox because no measure is defined on the sets being moved but if we define a measure on the sets then the rigid motions, as constructed in the proof, no longer exists because they aren't measure...