I am working out of Calculus by Spivak (4th ed.) and have come across a question which I would like some further insight on.   If $a$ and $b$ are both irrational, is $a+b$ necessarily irrational?  I have already proved that if $a$ is rational and $b$ is irrational then $a+b$ is irrational.  Thanks in advance!
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If $x$ and $y$ are irrational, is $x + y$ irrational? Is $x - y$ irrational?
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I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
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I found the following proof arguing for the irrationality of $\sqrt{2}$.     Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q &gt; 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 &lt; q' &lt; q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.   Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:  Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.  Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.  Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2&gt;\sqrt{2} &gt; 1$ , we get $$1&gt;\sqrt{2}-1&gt;0$$ proving that $q'$ is a non-negative integer.  As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.  Am I missing something or something is wrong with how I proceeded?  Here is a link to the textbook where I found this proof: , (p 32, problem 1.14).  P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented.
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Can someone point me to a proof that the set of irrational numbers is uncountable? I know how to show that the set $\mathbb{Q}$ of rational numbers is countable, but how would you show that the irrationals are uncountable?
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Possible Duplicate:          I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?
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Prove that $\sqrt[31]{12} +\sqrt[12]{31}$ is irrational.   I would assume that $\sqrt[31]{12} +\sqrt[12]{31}$ is rational and try to find a contradiction.  However, I don't know where to start. Can someone give me a tip on how to approach this problem?
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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?
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Prove : $$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$$  I think the way to go here is to falsely assume that $x$ is rational, thus having $x=\frac mn$ with $m,n$ integers and go with that, we also know that $x$ with the celling belongs to the integers and that $\frac 1x$ with the celling gotta be a $0$. Still can't prove it. Thank in advance !
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Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?
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If a.b.c,d are rationals and x is irrational number, then prove that $\frac{ax+b}{cx+d}$ is usually a irrational number. When do exceptions occur?  This is a homework Question. i did this by assuming that given fraction is rational say $\frac{p}{q}$. Now i write $$\frac{ax+b}{cx+d} = \frac{p}{q}$$. After cross multiplying i get   $$\frac{qb-pd}{pc-qa}=x$$  So this contradiction as right hand side is rational and left hand side is irrational. Now the question asks where does exceptions occur. I see that exceptions can occur when $a=c=0$ and $a=b=c=d$. But what are other exceptions and is my working method correct?  Thanks
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My friend and I were discussing this and we couldn't figure out how to prove it one way or another.  The only rational values I can figure out for $\sin(x)$ (or $\cos(x)$, etc...) come about when $x$ is some product of a fraction of $\pi$.  Is $\sin(x) $ (or other trigonometric function) necessarily irrational if $ x $ is rational?  Edit: Excluding the trivial solution of 0.
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Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.  Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.
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I know and I have proved more than once that the set of irrational numbers ($\mathbb{I}$) is uncountable, but now I'm given to solve this problem:   Show that $|\mathbb{I}|=|\mathbb{R}|$,  How can I do that?  Do I need to assume the Continuum Hypothesis in order to that statement to be true?  Thanks
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Just a curiosity. I just saw a video where pi was being calculated by the series  $$1-(1/3)+(1/5)-(1/7)+...$$  My question is how can $\pi$ be irrational if we can calculate it using rational numbers?
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I understand that the set of irrational numbers with multiplication does not form a group (clearly, $\sqrt{2}\sqrt{2}=2$, so the set is not closed). But is there a proof or a counter-example that the irrationals with addition form (or do not form) a group?  Thank you!  Edit: In particular, I am wondering if the set is closed with respect to addition.
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how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational? I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields $\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
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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!