So, since I've attended to the introduction to number theory course I couldn't stop thinking about this problem:  Prove that $\sqrt{2}$ is irrational using the Well Ordering Principle.  Ok, I started defining a set $X = \{ x = z \sqrt{2}; z \in \mathbb{Z} \}$ and supposing that there exists an $a \in \mathbb{Z}$ such that $a &lt; x$, $\forall x \in X$. Hence, there exists an $x' \in X$ such that $ x' \leq x$, $\forall x \in X$.  And that's when I can't proceed further. Is there a problem in the way I've defined the set or it is just a "leap" that I'm not seeing?  Thanks in advance.
123218
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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Is my proof beneath perfect and complete?  I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: $$\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.$$  I start by assuming that $m^{\frac 1n}$ is rational and non-integer. So there exist  co-prime integers $a,b$ so that $$\sqrt[n]{m}=\frac{a}{b}$$ $$\implies  m=\frac{a^n}{b^n}\in\mathbb{N}.$$ But since $a$ and $b$ have no common factor, $a^n$ and $b^n$ also have no common factor. So: $$\frac{a^n}{b^n}\notin\mathbb{N},$$ a contradiction.
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Prove that between any two different real numbers there is a rational number and an irrational number.  My attempt at a proof: (Throughout this proof it is assumed without any loss of generality that $x &lt; y$)  1.) $x$ is irrational and $y$ is irrational.  In this case the arithmetic mean $\frac{x+y}{2}$ is always an irrational number between $x$ and $y$. Also, if $y$ is irrational then it has a non-terminating decimal expansion and so we can always find a positive integer $n$ such that $x &lt; y - 10^{-n}y &lt; y$. The middle of the three numbers here will have a terminating decimal expansion, and hence it is rational. It also clearly lies between $x$ and $y$. These examples of rational and irrational numbers should also work in the cases:  2.) $x$ is irrational and $y$ is rational.  3.) $x$ is rational and $y$ is irrational.  There is one last case to consider.  4.) $x$ is rational and $y$ is rational.  In this case the arithmetic mean of $x$ and $y$ is rational. An irrational number for example could be the ratio $\frac{ex + \pi y}{e + \pi}$  I'm not sure if this proof is a good method or if it is even valid, so some feedback would be appreciated.
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We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:  Take $(\sqrt{2})^{\sqrt{2}}$  If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.  I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?   p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$.
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I'm being asked to prove that the set of irrational number is dense in the real numbers.  While I do understand the general idea of the proof:   Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$.  By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} &lt; p &lt; \frac{y}{\sqrt{2}}.$$  I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational.  What does this have to do with anything and how does it prove denseness of irrationals?  Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction.
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This question is from Apostol's Mathematical Analysis. I have a solution for it at   The question is:  Prove that $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational for every integer n ≥ 1.  And the proof I have is  Now, I can't understand how to get h=$3/2$ and k=$1/2$ ?  Please help.
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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Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational   My attempt:-     Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$ Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$ and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
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I've looked for help with this question but I have not found anything, I hope this is not a duplicate.  Define the set $A=\{\mid x+y\sqrt{2}\mid \ : x,y\in \mathbb{Z}\ \mbox{and} \mid x+y\sqrt{2}\mid\gt0 \}$, prove that the infimum of this set is zero.  This is what I've thought: Proving that $[y\sqrt{2}]$ (where $[\cdot ]$ is the floor function) can be close to zero as we want would prove the proposition, so I suppose that this doesn't hold and I want to arrive to the contradiction that for some $y_0$: $[y_0\sqrt{2}]=0$, this would contradict the fact that $\sqrt{2}$ is irrational.  Can someone give me a hand? Thanks! (This problem arised in the context of integer lattices)
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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 taking 'Introduction to Mathematical Thinking' on Coursera platform and following proof steps are given :      Proof of $\sqrt{2}$ is irrational.      Assume $\sqrt{2}$ is rational. $\sqrt{2}=p/q$      p and q have no common factors.   Why do p and q have no common factors? Is this a consequence of a property of the rational numbers? As p and q can be rational numbers we can set p = 6, q = 9 so p, q have common factors?
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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!
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I'm not sure how to solve this one. Thank you!  $2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational. (a) Given $n\in\mathbb N$ show that $\{\,k\alpha\bmod1\mid k\in\mathbb N\,\}\cap\left[0,\frac{1}{n}\right]\neq\emptyset$ (b) Prove that $\{\,n\alpha\bmod 1\mid n\in\mathbb N\,\}$ is dense in $[0,1]$.
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Can anyone please help me in this question:     $(X,\leq )$ is a well-ordered set. $\forall x\in X$, either $x$ is the largest element of $X$ or there exists $y\in X$ such that   (i) $x&lt;y$ and   (ii) if $x_0\in X$ such that  $x\leq x_0\leq y$, then either $x = x_0$ or $x_0 = y$.   Will this be sufficient to start the proof:  If $x$ is the largest element then there is nothing to prove. Suppose $x$ is not the largest element. $$A=\{ z\in X \mid x&lt;z\}\neq \emptyset$$ How do I proceed?
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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 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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From  I came to know that it is hard to conclude $\pi+e$ is an irrational? Can somebody discuss with reference "Why this is hard ?"   Is it still an open problem ? If yes it will be helpful to any student what kind ideas already used but ultimately failed to conclude this.