I am trying to prove this and have looked at similar questions to gauge how to approach this. I have: Suppose that there exists a smallest rational number greater than $\sqrt{3}$. We shall call that number $n$, which, as it is rational, can be expressed as $\frac{p}{q}$ $\frac{\sqrt{3}+n}{\sqrt{3}}$ is a number greater than $\sqrt{3}$ but less than $n$, but this number would no longer be rational now, would it?
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I am working on a proof about Dedekind cuts on the positive rational numbers. I have been stuck for a while on the following point and would appreciate any help.  Given $x\in \mathbb{Q}^+$ such that $x^2&lt;2$, how to find $y\in \mathbb{Q}^+$ such $y&gt;x$ and $y^2&lt;2$?
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This might sounds stupid, but I really don't know can I show Irrational numbers in proves? And if so, how to show it?  For example, when I want to show Rational numbers, I do this:  $\frac{m}{n} $ , $m, n $ are integer $,$ $n\ne0$  Can I do something like that with Irrational numbers?
110389
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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A popular proof of the irrationality of $\sqrt2$ is to first assume that the number is rational. This means that $\sqrt2=a/b$ where $a$ and $b$ are integers. Another assumption is that $a$ and $b$ are coprime. It turns out that this leads to a contradiction. And the conclusion is that $\sqrt2$ is not rational (because the assumption of rationality led to a contradiction). But what about the second assumption? Logically, the conclusion could also be that $a$ and $b$ are not coprime..  How can this be resolved?  Edit:  Thank you all for your comments. It cleared up my mind.  Three aspects of this kind of proof are: mathematical, logical, and didactical.  Mathematically, it is the case that any rational number can be written as a fraction with coprime numerator and denominator. But this point is often described in a way that is either literally a logical assumption, or in a way that is natural to interpret as a hidden logical assumption. In this case, the proof will have a flawed logic, because any of the two assumptions could be the cause of the contradiction. Didactically, this can cause confusion in the mind of the student, who might not be so accepting of the proofs conclusion. I used to be that student.  So, how to resolve it? Many of you provided explanations on the mathematical aspect, which are all correct. But my main point is not of not understanding the proof, but of the use of flawed logic which leads to not understanding the proof. I am sure that a lot of teachers think or say something along the lines of WLOG, but for many students, this makes the proof weaker, because it is not so convincing. When we present this proof to our students, we need to have this in mind.  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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I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| &gt; x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
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I think this is true because of prime factorisations, i.e.  If $3$ a factor of the prime factorisation of $q^3$, then $3$ is a factor of the prime factorisation of $q$.  Therefore If $3$ divides $q^3$, then $3$ divides $q$.  Or for a general case, if $n$ divides $q^3$; then $n$ divides $q$. Is this correct?
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The sum of a finite number of rational numbers is of course a rational number, but the sum of an infinite number of rational numbers might be an irrational number.  Can someone give me some intuition why this sum might be irrational?  I just "don't feel it."
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Let $x=\frac{A}{B}$ be a positive rational number in lowers terms (i.e., $A, B\in\mathbb{N}$ and $hcf(A,B)=1$). Prove that $\sqrt{x}$ is rational if and only if $A$ and $B$ are both perfect squares. (Remember that your proof should work for cases like $A=40$, where $A$ is not a perfect square but has a factor that is a perfect square.) I know prime factorization is involved but that's basically it. Proving that $\sqrt{y}$ (for some number y) is rational if and only if $y$ is a perfect square I can do. Applying that that to the fraction mentioned above? Not so much. Any and all help would be appreciated. Thank you.
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I have heard  called the most irrational number. Numbers are either irrational or not though, one cannot be more "irrational" in the sense of a number that can not be represented as a ratio of integers. What is meant by most irrational? Define what we mean by saying one number is more irrational than another, and then prove that there is no $x$ such that $x$ is more irrational than $\varphi$.  Note: I have heard about defining irrationality by how well the number can be approximated by rational ones, but that would need to formalized.
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I am trying to figure out how to prove that all rational numbers are either terminating decimal or repeating decimal numerals, but I am having a great difficulty in doing so. Any help will be greatly appreciated.
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The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is rational, $$\begin{align}        \sqrt{4} &amp;=p/q        \qquad\text{where pand q are coprimes} \\         4 &amp;=p^2/q^2\\       4q^2&amp;=p^2       \tag{1} \\          4&amp;\mid p^2\\          4&amp;\mid p\\      \text {let }p&amp;=4m  \qquad\text{for some natural no. m}  \\        p^2&amp;=16m^2\\       4q^2&amp;=16m^2   \qquad\text{(from (1) )}\\        q^2&amp;=4m^2\\          4&amp; \mid q^2\\          4&amp;\mid q \end{align} $$ but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence  $\sqrt{4}$ is not rational. But we know that it is a rational. Why?
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It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
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Possible Duplicate:          I have this homework problem that I can't seem to be able to figure out:Prove: If $n\in\mathbb{N}$ is not the square of some other $m\in\mathbb{N}$, then $\sqrt{n}$ must be irrational. I know that a number being irrational means that it cannot be written in the form $\displaystyle\frac{a}{b}: a, b\in\mathbb{N}$ $b\neq0$ (in this case, ordinarily it'd be $a\in\mathbb{Z}$, $b\in\mathbb{Z}\setminus\{0\}$) but how would I go about proving this? Would a proof by contradiction work here?  Thanks!!
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Show that if the smallest prime factor $p$ of the positive integer $n$ exceeds $\sqrt[3]{n}$, then $\frac{n}{p}$ must be prime or 1.  I'm stuck trying to prove this. I tried this using contradiction, but I haven't been able to prove it.  My approach:  Let $p$ be the smallest prime factor of the positive integer $n$.  Let $p &gt; \sqrt[3]{n}$  Then we must show that $\frac{n}{p}$ is either a prime number or 1.  We know that $n = pe$ for some $e \in \mathbb{Z}$.  Thus, $\frac{n}{p} = e$  We also know that $p$ > $\sqrt[3]{n}$, so $p^3 &gt; n$, and $p^2 &gt; \frac{n}{p}$  So, assume $\frac{n}{p} = e$ is not prime nor 1, meaning it's a composite number. Let's try to get a contradiction.  If $\frac{n}{p}$ is composite, then $\exists a,b \neq 1: \frac{n}{p} = ab $  Now, here's where I get stuck. What can I do with this info? $\frac{n}{p} = ab $ and $p^2 &gt; \frac{n}{p}$.
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Show that a number has a terminating decimal expansion if and only if, it is rational and when in lowest terms, its denominator is coprime to all primes other than $2$ and $5$.   This is an unsolved question in my lecture notes. I can only seem to prove the converse direction for this. Would appreciate a solution for the other direction.     For converse direction:   Let the number, in lowest terms, be given by $\frac{l}{m}$ where $m = 2^ \alpha 5^\beta$, for some positive integers, $\alpha$ and $\beta$.    If $\alpha &gt; \beta$, let $k = 5^{(\alpha-\beta)}$. If $\alpha &lt; \beta$, let $k = 2^{(\beta-\alpha)}$.   Then $\frac{l}{m} = \frac{kl}{k2^\alpha5^\beta}=\frac{kl}{10^q}$ where $q = \max(\alpha,\beta)$.   Hence $\frac{l}{m}$ is a terminating decimal.
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We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$?