Linked Questions

0
votes
1answer
614 views

Intuition for a proof that the rationals are incomplete. [duplicate]

Let A be a set of positive rationals $p$ such that $p^2<2$. Now this set contains no upper bound. To prove this, for every rational $p$, a number $p- \frac{p^2-2}{p+2}$ is associated. This number (...
1
vote
3answers
288 views

Choice of $\xi$ [duplicate]

Possible Duplicate: Rational Numbers Suppose $\{x \in \mathbb{Q}|x>0,x^2<2\}$ has a supremum. Call this supremum $c$. In order to show that this cannot be the case, we learned that we need ...
8
votes
0answers
255 views

Constructing a larger rational whose square is also less than two [duplicate]

Possible Duplicate: Rational Numbers Baby Rudin has a very nice construction showing that, given a positive rational number whose square is less than (greater than) two, one can always find a ...
2
votes
0answers
59 views

In the following proof that square root of 2 is non-terminating, where does the expression for q come from? [duplicate]

I understand the following proof but how did the author come up with the following expression? $$q = p - \frac{p^2 -2}{p + 2}$$ $q$ being defined that way is crucial for the proof but what's so ...
1
vote
0answers
21 views

A Small Question about Rudin's Gap Construction [duplicate]

In Rudin's PMA, he shows the following. $A=$ { $ p \in R^+ | p^2<2$}, B= { $ p \in R^+ | p^2>2$} Let $q = p - \frac{p^2 -2}{p+2}= \frac{2p+2}{p+2}$, then $q^2 -2= \frac{2(p^2- 2)}{(p+2)^...
84
votes
7answers
44k views

What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6?

Problem 6 of the 1988 International Mathematical Olympiad notoriously asked: Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect ...
19
votes
5answers
5k views

Irrationality proofs not by contradiction

Per now, I have basically come upon proofs of the irrationality of $\sqrt{2}$ (and so on) and the proof of the irrationality of $e$. However, both proofs were by contradiction. When thinking about it,...
11
votes
4answers
703 views

Rudin's proof on the Analytic Incompleteness of Rationals [duplicate]

In Rudin's classical "Principles of Mathematical Analysis," he gave a proof like this: Claim: Let $A= \{p\in \mathbb{Q} | p>0, p^2 <2\}$. Then A contains no largest number. Proof: Given ...
1
vote
2answers
2k views

Prove that the set of positive rational values that are less than $ \sqrt{2}$ has no maximum value

Its been a while since I've written a proof and would appreciate some feedback on this one. Question: Given the set of rational positive values, $\{q | q \in \mathbb{Q} \wedge 0 \lt q \lt \sqrt{2}\}$...
2
votes
2answers
674 views

Let E = $\{x\in \mathbb{Q}: x>0$ and $x^2 <2\}$. Prove E does not have a largest element.

I'm pretty sure the way to go is with a contradiction, letting $x = \sup E$. Then I think I need to either show that $x < x^2\in E$ (contradicting $x = \sup E$ in that an element of E is greater ...
2
votes
4answers
220 views

Very basic intro to real analysis question

In page 2 of baby Rudin, Rudin wants to prove that $ \forall p \in \mathbb{Q} $ such that $ p>0 $ and $p^2 < 2$ there exists a $q \in \mathbb{Q}$ such that $q>p$ and $q^2<2$ In his ...
0
votes
1answer
91 views

Prove that $(G, \circ)$ is group, $G=\{x, -1<x<1 \}$ and $x\circ y =(x+y)/(1+x*y)$

If $1<x, y< 1$ then $-1<x*y<1$ and $0<1+x*y <2$, but then I can't prove that $-1<x\circ y =(x+y)/(1+x*y)<1$. Help please.
2
votes
1answer
134 views

Rudin's proof that rationals have no GLB (or LUB) [duplicate]

In Rudin's 3e of Principles of Mathematical analysis, he associates $q = (2p+1)/(p+1)$. How does he create this number? What is the method for discovering this answer?
0
votes
1answer
111 views

construct two rational numbers

Can one explain me a bit or more about how to construct the two rational numbers? From 《Principles of Mathematical Analysis》page 2 in proving $\sqrt{2}$ is not a rational number. $p>0$ $$q=p-\...