Recent content by ruip

That's problem 25 of chapter 2. I'll try to do that one too. :)

The set P is the set of positive numbers but , from the given properties, the only information we have about P is the trichotomy law, the close of addition over P, and the close of multiplication over P. I understand now that you assumed that 1 is in P but in reality that must be proved first...

Sorry, but I don't understand how is this "shorter". It looks exactly the same, since you depend on the fact that 1 is in P to conclude that (x+1)-x = 1 is in P. Since 1 in P is the same as 1-0 in P, that is 1 > 0, then we need to prove first that 1 > 0. The way you wrote it, looks like...

One of the properties I gave was "if a and b in P, then a*b is also in P", and so if a*a in P for any a ≠ 0 (as shown in previous post), then a*a - 0 is in P, and by definition a*a > 0. Since 1*1 = 1 and 1*1 > 0 then 1 > 0. To prove this I depend on the fact that 1 > 0 as proved previously...

Well, for any number a, if a is positive then a*a > 0 (by definition). Also if a < 0, then (-a) > 0 and (-a)*(-a) > 0. Since (-a)(-a) = a*a[1], then a*a > 0. So, for any number a ≠ 0, a2 > 0, and in particular 12 = 1 > 0. Given that 1 > 0, then -1 < 0. Also 1 + 1 > 0 + 1, that is 2 > 1. Also 1...

Let me restate the argument from the beginning. If an integer n is even then there is an integer k so that n = 2k. Also, if n is odd then there is an integer m so that n = 2m+1. Now, if that is true then 2k = 2m+1 and so 2(k-m) = 1. When (k-m)<= 0, I can easily get a contradiction. When...

I guess you are right. But I must add that the even definition talks about integer numbers also, namely, a natural number n is even if there exists _an integer k_ so that n = 2k. So I think I can use the trichotomy law for the integer 'k' in this definition. Maybe this isn't enough anyway.

No, in fact the problem is about natural numbers. Prove that a natural number is either even or odd. Sorry about the confusion. I have no definition for natural number besides the properties they respect. Quoting Spivak "The simplest numbers are the "counting numbers" 1,2,3,... The fundamental...

Sorry, I misunderstood you when you said "whole numbers". You were referring to the integers. Ignore my comment "My "whole numbers" are the real numbers and so," and consider the rest. I remove that from the previous post.

Hello, I have the following three basic properties regarding inequalities: 1. Trichotomy law. For every number a, one and only one of the following holds: a = 0 a is in collection of positive numbers P -a is in collection of positive numbers P 2. If a and b are in P, then a + b is in P 3...

Sorry, but I don't know how to prove that if k > 0 and k is an integer, then k >= 1. I don't see how can I show that there aren't integers between 0 and 1.

I can rewrite it as 2(k-m) = 1 and make a similar argument as in the post above. If k-m = 0, then 2(k-m) = 0 if k-m < 0, then 2(k-m) < 0 The problem is with k-m > 0. 2(k-m) > 0 and 1 > 0 so everything looks fine and I don't get a contradiction. I'm missing something. :\

Yes, I can use inequality properties: a number n is positive, n = 0, or -n is positive; if n and m are positive then n+m and n*m are also positive. The example you give was proved from the previous properties and so I can use that and similar properties too. Following your example with the...

Hello, I'm re-studying calculus using Spivak's Calculus 4ed and I'm stuck in one of the problems. Any help is appreciated. Homework Statement The theorem to prove is "every natural number is either even or odd". The definition of even given by Spivak is the following: A natural number n is...