It is easy to see that $\sqrt{2}$ and $\sqrt{2}+\sqrt{3}$ are irrational. So $\sqrt{2}+\sqrt{3} + \sqrt{4}$ is irrational. But what about $\sqrt{2}+\sqrt{3} + \sqrt{4} + \sqrt{5}$? I suspect that $$\sum_{n=2}^{k} \sqrt{n}$$ is always irrational, is it true and is there a simple way to proof that?
100293
Related:    Except for the obvious cases $n=0,1$, are there any values of $n$ such that $\sum_{k=1}^n\sqrt k$ is an integer? How does one even approach such a problem? (This is not homework - just a problem I thought up.)
6503
I am studying the proof that $\sqrt 2$ is an irrational number. Now I understand most of the proof, but I lack an understanding of the main idea  which is:     We assume $\frac{m^2}{n^2} = 2$. Then both $m$ and $n$ can't be even.   I do not understand, why can't both $m$ and $n$ be even?
109334
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?
36779
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.
68178
Suppose we wish to prove that the square root of two is irrational. We begin by assuming that it is rational.  Namely, where both $a$ and $b$ are integers $$\frac{a}{b} = \sqrt 2  % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GH9aqpdaGcaaqaaiaaikdaaSqabaaaaa!31D3! $$  Why is it so important that both $a$ and $b$ be coprime and the fraction be irreducible?  That is, if $a$ and $b$ are not coprime and the fraction is indeed reducible, then$$\frac{a}{b} \ne \sqrt 2  % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GHGjsUdaGcaaqaaiaaikdaaSqabaaaaa!3294! $$
47345
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?
83192
My question looks quite obvious, but I'm looking for a strict proof for this:     Why can't the sum of two square roots of non-perfect squares be an integer?   For example: $\sqrt8+\sqrt{15}$ isn't an integer. Well, I know this looks obvious, but I can't prove it...
7195
The proof that shows the square root of 2 is irrational starts by assuming, for a contradiction, that it is rational. It starts with the assumption that it can be written as p/q where p and q have no factors in common.  Why do we start by assuming p and q have no factors in common? If we went on to find that they have factors in common, we could cancel them later on?
88828
Is my proof that the square root of a positive integer is either an integer or an irrational number correct?  The proof goes like this:  Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is  for some integer a.  Therefore, if n is a square number, then $\sqrt{n}$ is rational.  Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.  We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.   Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.    Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n &gt; 1$, then $a&gt;b&gt;1$.  By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes.  Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.  However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes.  Therefore, there exists at least one prime in the product of $nb^2$  that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.  Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.  Is this sufficient? Or is there any parts I did not explain well?
4939
If $p$ is a prime number, then $\sqrt{p}$ is irrational.  I know that this question has been asked but I just want to make sure that my method is clear. My method is as follows:  Let us assume that the square root of the prime number $p$ is rational. Hence we can write $\sqrt{p} = \frac{a}{b}$. (In their lowest form.) Then $p = \frac{a^2}{b^2}$, and so $p b^2 = a^2$. Hence $p$ divides $a^2$, so $p$ divides $a$. Substitute $a$ by $pk$. Find out that $p$ divides $b$. Hence this is a contradiction as they should be relatively prime, i.e., gcd$(a,b)=1$.
5075
Inspired by , I was wondering if there is a simple logical argument to   Show that $ a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q  \implies  \sqrt{a},\sqrt{b} \in\mathbb Q $  Note that the original link is using a computational method, where as I am looking for a simple logical argument.  I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example $a=1-\sqrt{2},b= 1+\sqrt{2} $)
7010
I have read a few proofs that $\sqrt{2}$ is irrational.  I have never, however, been able to really grasp what they were talking about.  Is there a simplified proof that $\sqrt{2}$ is irrational?
7410
For example how come $\zeta(2)=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$. It seems counter intuitive that you can add numbers in $\mathbb{Q}$ and get an irrational number.
90592
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.
87684
I am not sure how to show this. This is what I tried.   Proposition: If x is rational number and y is an irrational number then x+y is an irrational number.  Logically speaking this is  P=x is rational  Q= Y is irrational  R= x+y is irrational  $P\wedge Q \rightarrow R$  Explanation: What I did  $x=\frac{p}{q}$ P and Q integer  Y cannot be written as the a rational by definition  So therefore  $x+y=\frac{p}{q}+y$  Will not be a rational number as there cannot be a common denominator.
51542
Why $\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}$ is a rational number?
25556
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!!