I am working on this exercise:  $$ \forall \varepsilon &gt; 0, \ \exists q \in Q  \text{ where } 0 &lt; |r - q| &lt; \varepsilon $$  To clarify, r is a real number, q is a rational number. This is what I have so far:  $$ 0 &lt; |r - q| &lt; \varepsilon $$ $$ -\varepsilon &lt; (r-q) &lt; \varepsilon $$ $$ r - \varepsilon &lt; q &lt; r + \varepsilon $$  Then making use of a given theorem that "Between any two distinct real numbers there is a rational number and an irrational number," I conclude the proof by claiming that the inequality must hold by the theorem and given that $r - \epsilon$ and $r + \epsilon$ are two distinct real numbers.  Is this sufficient for the proof? Otherwise, would appreciate some help on how to proof the above inequality more rigorously.   I recall the Archimedean Property that for each positive real number r, there exists a positive integer n such that $\frac{1}{n} &lt; r$. There is a hint to use this condition, but I am unsure how to apply it into the proof.   Thanks in advance for any help!
124084
I'm having difficulties making progress in proving: $$\forall \varepsilon &gt; 0, \ \exists q \in Q  \text{ where } 0 &lt; |r - q| &lt; \varepsilon $$  To clarify, $r$ is a real number and $q$ is a rational number.   Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ \exists n \in Z^{+}$ such that $na&gt;b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.  I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!
105755
I think I have the first part, as I have shown that $q &lt; q + (q'-q)/2 &lt; q'$, but I have trouble in proving that $|q-q'|&lt; \epsilon$.  Could someone tell me if there is a better way of showing the first part and how to show the second part?  For those asking x is a real number
105031
I just started real analysis.  I don't have a background in proofs or logic, simply calculus.  So I'm trying to learn more about proofs--so forgive the basic question, please.  How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$.  This number is given by $x=b+(-a)$.  First part of the proof I understand--it's simple.  We simply use some axioms and do the following:  $a+b-a=b$, which of course is true.  This is just true from plugging in the $b+(-a)$ for $x$.  But how about the uniqueness issue?  In my text, it says:  $$(a+x)+(-a)=b+(-a) x=b+(-a)$$  How does this prove uniqueness?  Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?  How would I have known to set this up as part of the proof if I was asked to prove something is unique?  Sorry for the basic question--have to start somewhere.
59929
I need to prove or disprove the following:  Suppose that $g$ is a positive real valued function of a real number. If $a &lt; b$ are real numbers, then there is a finite sequence $a = t_0 &lt; t_1 \dots &lt; t_n = b$ of real numbers such that in each interval $[t_k,t_{k+1}]$ there is a point where the value of the function $g$ is greater than the length of the interval.   With out the assumption of continuity, I'm not sure how to proceed. Any advice?
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I would like to verify my proof of the following:  Let $A=\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$.  I want to show that $-1$ and $1$ are the infimum and supremum respectively.  First I will show that $-1$ is the infimum.  To show this I will first demonstrate that it is indeed a lower bound.  Observe: $$\frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-1 \geq -1.$$  I now claim that $-1$ is the greatest lower bound.  So, by the archimedean property, there exists an $x \in \mathbb{R}$, $x&gt;0$ and $n' \in \mathbb{N}$ such that $1&lt;n'x$.  Thus, $\frac{1}{n'}&lt;x$ and $$\frac{1}{n'}-1&lt;x-1.$$  From the above it is clear that $\frac{1}{n'}-1 \in A$.  Also, it is clear that $-1&lt;-1+x$.   Let $y=\inf(A)$.  So we let $$-1&lt;y\leq-1+x.$$  By completeness there exists an $r\in\mathbb{Q}$ such that $$-1&lt;r&lt;y\leq-1+x.$$  Thus $y$ is not the infimum.  I was going to note that the supremum was 1 through the relationship between $$\inf(A)=-\sup(-A).$$
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I am self-studying real analysis for fun and came across the following question (from A First Course in Real Analysis by Berberian).  (i) Prove the following: If $r$ is a real number such that $0 \leq r \leq 1$, then there exists a unique real number $s$ such that $0 \leq s \leq 1$ and $s^2 = r$. (ii) Generalize (i) to the following: If $a \in \mathbb{R}$, $a \geq 0$, then there exists a unique $b \in \mathbb{R}$, $b \geq 0$ such that $b^2 = a$.  Put $y = 1-r$ and $x = 1-s$. Then we want to find a real number $x \in [0,1]$ such that $(1-x)^2 = 1-y$. I am trying to find an increasing sequence $(x_n)$ such that $0 \leq x_n \leq 1$ (defined recursively). For (ii) how could I use (i) to get the result?
87288
The question is: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x) = 0$ whenever $x$ is rational, then $f(x) = 0$ for all $x\in\mathbb{R}$.  My proof:  Let $ x\in\mathbb{R} $. If $ x\in\mathbb{Q}$, then $ f(x)=0 $.  Otherwise $ x\in\mathbb{R\backslash Q} $. Then there exists a sequence of rational numbers $ (q_n)_{n\in\mathbb{N}}$ (for example $q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}$) such that $q_n\to x$ as $n\to \infty$. By the properties of $f$, we have $f(q_n)=0$ for all $n\in\mathbb{N}$. So $f(q_n)\to 0$ as $n\to \infty$. Since $f$ is continuous, $q_n\to x$ implies that $f(q_n)\to f(x)=0$. So $f(x) = 0$ for all $x\in\mathbb{R}$.   Thank you for any help.
110715
Let $x_1, x_2, . .  , x_n$ real numbers from $[0, 1]$. Prove there is   $x \in [0, 1]$ so that $|x - x_1| + |x - x_2| + . . . + |x - x_n| =\frac n 2$     My attempt  Let $f:[0,1] \rightarrow R, f(x)=|x - x_1| + |x - x_2| + . . . + |x - x_n|$. Because $f$ continous, the image of $f$ is an interval. To prove $\exists x \in [0,1]$ so that $f(x)=\frac n 2$ it is enough to find two values $a,b \in [0,1]$ so that $f(a) \le \frac n 2 \le f(b)$. I failed to find these values.  Also, a solution without function continuity will be appreciated.
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Let $x$ be a positive rational number, strictly between $0$ and $1$. Prove that there is a finite strictly increasing list of positive integers $2 \leq q_1&lt;q_2&lt;\cdots&lt;q_N $ such that $$x={1\over q_1}+{1\over q_2}+ \cdots+{1\over q_N}.$$  I have tried many methods, such as mathematical induction. I know it is obvious that $\frac{1}{m}$ is right for assumption, but when I begin to prove $\frac{2}{m}$, I feel it is hard to find some patterns for $\frac{2}{m}$. Thus, maybe my thoughts are wrong.  Also, I still tried to change the fraction so that the numerator will be smaller and smaller. But I still cannot find a way to lower the numerator. Can someone help me solve the question? Or, can someone give me some hints. I will appreciate you very much!
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Let $a$ and $b$ be reals with $a&lt;b$. Show that there are infinitely many rationals $x$ such that $a&lt;x&lt;b$.  My plan of action was to assume that $x$ is the smallest such rational and find another rational in the interval $(a, x)$, but I am struggling to make it work. A hint will be much preferred to a full solution.
107245
Let us start by clarifying this a bit.  I am aware of  .  My issue comes by way of some properties of the reals. These issues can be summed up by the combination of the following questions:   Is it true that between any two rationals one may find at least one irrational? Is it true that between any two irrationals one may find at least one rational? Why are the reals uncountable?   I've been talking with a friend about why the answer of these three questions can be the case when they somewhat seem to contradict each other.  I seek clarification on the subject.  Herein lies a summary of the discussion:  Person A:  By  way of  it can be shown that the reals are uncountable.  Person B:  But is it not also the case that one may find at least one rational between any two irrationals and vice versa?  Person A:  That seems logical, I can't pose a counterexample... but why does that matter?  Person B:  Wouldn't that imply that for every irrational there is a corresponding rational?  And from this the Reals would be equivalent to 2 elements for every element of the rationals?  Person A:  That implies that the Reals are countable, but we have already shown that they weren't... where is the hole in our reasoning?  And so I pose it to you... where is the hole in our reasoning?
102600
Finding the closure of A=$\mathbb Q \cap [0,1]$.  My attempt: Between every two rational numbers, there is a rational number. That means that for  any arbitrarily small $\epsilon \gt0$, and some $q\in \mathbb Q \cap [0,1]$  there is $x \notin \mathbb Q$, $x\in [0,1]$ so that $x\in (q-\epsilon, q+\epsilon)$. That means - and I'm not sure if that's correct and how to show it - there is a sequence $\{a_n\}_{n=1}^\infty \subseteq \mathbb Q \cap [0,1]$ so that $\lim_{n \rightarrow \infty} a_n=x$.   Therefore, the closure of A is $[0,1]$.   Is this correct? And how do I show my missing step (if it's valid?).  Thanks for any assistance in advance!
114439
Let $\alpha$ be an irrational real number and let $a_j$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_j$ is a fraction expressed in lowest terms: $a_j = \alpha_j / \beta_j$. Prove that the $\beta_j$ tend to $\infty$  Attempt: AFSOC that $\beta_j$ does not tend to infinity, then it is bounded by some $M$. We can also find an interval $\alpha \in (k, k+1)$. Let $\epsilon &gt; 0$, $\exists N_0$ such that for all $N &gt; N_0$ we have $|a_N - \alpha|  &lt; \epsilon$. So $|\frac{p_N}{q_N} - \alpha| &lt; \epsilon$. We try to use the fact that $q$ is bounded but fail to derive a contradiction. I do not know how to use the fact that $\alpha$ is between two integers, although it might not be relevant at all.  Ideas?
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I am trying to show that for any real number a, there exist infinitely many rational numbers m/n with $ |a - m/n| &lt; 1 /n^{2} $. I've tried to attempt the question by assuming there are finite rational numbers and finding a contradiction.
11031
Possible Duplicate:          I am trying to prove that there exists an irrational number between any two real numbers a and b. I already know that a rational number between the two of them exists. My idea was to say represent a and b as $a = \sqrt{2} x, b = \sqrt{2} y$ for some $x, y \in \mathbb{R}.$ We know that there exists $\frac{m}{n}$ s.t. $x &lt; \frac{m}{n} &lt; y,$ so multiplying everything by the root of 2, we have $$\sqrt{2}x = a &lt; \frac{\sqrt{2}m}{n} &lt; \sqrt{2}y = b,$$ and so we have an irrational number in between two reals. Is this 'legit?' Are there 'better/'more elegant ways to go about this? I would really appreciate seeing alternatives proofs of this, as many as possible.
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I have the following problem from the book "How to Prove it" by Daniel J.Velleman.  Prove that $$ \exists z\in \mathbb{R} \forall x \in \mathbb{R}^+ \left[\exists y \in \mathbb{R} (y-x=y/x) \leftrightarrow x \neq z \right]. $$  I have no idea how to deal with the existential quantifier at the beginning. I tried to negate it, but the rest is too complicated. Do you have strategic hints how to proceed? Allowed is first-order logic.  EDIT  Based and the response, I would propose the following proof and kindly ask the community to check if it is correct.  Proof: Let $z=1$, and let x be some arbitrary real number.   $(\rightarrow)$ Suppose $\exists y \in \mathbb{R} (y-x=y/x)$ and choose $y_0$ to be some real number such that $(y_0-x=y_0/x)$. Suppose $x=z$. Since $x=z$ and $z=1$, $x=1$. However, plugging $x=1$ into $y_0-x=y_0/x$ yields a contradiction, and thus, $x\neq z$. Hence, $\exists y \in \mathbb{R} (y-x=y/x) \rightarrow x \neq z$. Since $x$ was arbitrary, it follows that $\forall x \in \mathbb{R}^+ \left[\exists y \in \mathbb{R} (y-x=y/x) \rightarrow x \neq z \right]$. ($\leftarrow$) Suppose $x \neq z$. Since $z=1$, $x \neq 1$. Let $y_0=\frac{x^2}{x-1}$, which is defined because $x \neq 1$. Plugging into the equation $y_0-x=y_0/x$, we can verify that the equation holds for this particular value $y_0$. Hence, $x \neq z \rightarrow \exists y \in \mathbb{R} (y-x=y/x)$. Since x was arbitrary, it follows that $\forall x \in \mathbb{R}^+ \left[\exists y \in \mathbb{R} (y-x=y/x) \leftarrow x \neq z \right]$.   Therefore, the theorem follows.
105641
This is a problem from Rudin, but I wanted to add my own intuition to it. It uses Rudin's definition of Archimedean property. I'd just like to know if my version holds  If $x \in \mathbb R$, $y\in \mathbb R$ and $x&lt;y$, then $\exists p \in \mathbb Q$ such that $x &lt; p &lt; y$  Since $x &lt; y$, then $y-x&gt;0$. Applying Archimedean property, we see $\exists n \in \mathbb Z^+$ such that $n(y-x)&gt;1$. Allow $A=\{i\in \mathbb Z\mid i &gt; nx\}$. Obviously $A$ is bounded by $nx$, so take $m = \inf A$. So we have $m &gt; nx$ and $m-1 \leq nx$ by set and $\inf$ definition. This gives us $m-1 \leq nx &lt; m$. Combining and rearranging with $n(y-x) &gt; 1$ gives us $nx &lt; m \leq nx +1 &lt; ny \implies nx &lt; m &lt; ny$. Thus $x &lt; \frac{m}{n} &lt; y$