Recent content by snes_nerd

Okay so I am given a 3D figure with 5 points. Keep in mind this model has the hyperbolic parallel property and satisfies the incidence axioms. The question is to construct the dual geometry and then to prove or disprove that it is an incidence geometry. My question is how do I go about...

Proof If a < b, then a + b < 2b (axiom or reasoning I can assume this?) Then with simple algebra, (a+b)/2 < b (definition of multiplicative inverse) If a < b, then 2a < a + b (axiom or reasoning I can assume this?) Then a < (a+b)/2 ( definition of multiplicative inverse Thus, a <...

So x3=(x2+b)/2 implies that a < x1 < x2 < x3 < b?.

I am sure it is very straightforward once I actually know what it means. In fact, its probably really easy. I feel kind of dumb not seeing it, and I am sure once I see it, I will feel even more dumb. Maybe its because I don't understand why x1 = (a+b)/2 and (xi+b)/2. (a+b)/2 tells me a...

Given any real numbers a and b such that a < b, prove that for any natural number n, there are real numbers x1, x2, x3, ... , xn such that a < x1 < x2 < x3 < ... < xn < b. The hint I was given says : Define xi recursively by x1 = (a+b)/2 and x(i+1) = (xi +b)/2. Prove that xi < xi + 1 < b...

Thanks everyone for the help. I didnt realize it was that simple and straightforward. So for proving their is an integer that is not the least element I go about it like this: Suppose there is an integer that has a least element n. Then n-1 is an integer to. But n-1 < n which is a contraction...

I can say that n+1 is an integer to and n+1 > n. So this would contradict the proposition that n is the greatest element. Is this the right idea?

Prove that the set of integers has neither a greatest nor a least element. I was given a hint: There are 2 different non existence results to prove, so prove them as separate propositions or claims. Divide into cases using the definition of the set of integers. So I was kind of confused...

Im not quite sure how to prove that. I mean I can look at this problem and figure it out, but to logically prove it is kind of different. Never had to do something like this before. x+y= x = x+y' implies that y=y' which is what I am trying to prove since that would be the conclusion. However...

A unique element for yεℝ such that for all xεℝ, x+y=x

Homework Statement Prove there is a unique element y such that for all x, x+y=x Homework Equations I also have to prove their is a unique element y such that for all x, xy=x The Attempt at a Solution x+y=x, x+y-x=x-x, y=0. xy=x, xy(1/x)=x(1/x). y=1. The problem...