Cardinality of the basis for $\mathbb{R}$ over $\mathbb{Q}$? This question came up as a discussion I had with a friend. Clearly, the basis is not of finite cardinality since that would imply the set $\mathbb{R}$ has the cardinality $\aleph_{0}$ which is false. Therefore the basis clearly has to be infinite. However, the question that came up was that is the basis countably infinite or uncountably infinite. So I'd like to know the cardinality of the basis and also appreciate some pointers to the proof.  Thanks in advance. :)
Take $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$ I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.  Proof: Suppse that $|B|&lt;2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $ \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.  What do you think?  Thank you, Shir
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How do I make that space actually available to Fedora? Can I use my Schengen visa for a completely different purpose and entry point? Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved. How to mount EXT4 disk on 10.15? Since I have osxfuse installed it would be nice if I could use it. There are a  (heavily) outdated  here, saying that it is not . Is this still the case? Why was Dumbledore worried by the cut in Harry's arm? In the Goblet of Fire, Harry is cut in the arm by Pettigrew. When Dumbledore learns it, he's extremely worried and wants to check Harry's arm.  Why?     When Harry told of Wormtail piercing his arm with the dagger, however, Sirius let out a vehement exclamation and Dumbledore stood up so quickly that Harry   started. Dumbledore walked around the desk and told Harry to stretch out his arm.   Harry showed them both the place where his robes were torn and the cut beneath   them.      "He said my blood would make him stronger than if he'd used someone else's,"   Harry told Dumbledore. "He said the protection my - my mother left in me - he'd   have it too. And he was right - he could touch me without hurting himself, he   touched my face."      For a fleeting instant, Harry thought he saw a gleam of something like triumph in   Dumbledore's eyes. But next second. Harry was sure he had imagined it, for when   Dumbledore had returned to his seat behind the desk, he looked as old and weary   as Harry had ever seen him.      "Very well," he said, sitting down again. "Voldemort has overcome that particular   barrier. Harry, continue, please.     Edit:  Plz explain this sentence specifically:     "Dumbledore walked around the desk and told Harry to stretch out his   arm." Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation $$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$ has a (real) root, prove the inequality $$ a^2 + b^2 \geq 8. $$  I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out. No group of order 400 is simple There's a fairly standard two or three-semester curriculum for introductory quantum field theory, which covers topics such as:   classical field theory background canonical quantization, path integrals the Dirac field quantum electrodynamics computing $S$-matrix elements in perturbation theory, decay rates, cross sections renormalization at one loop Yang-Mills theory spontaneous symmetry breaking the Standard Model   What is a good, complete and comprehensive book that covers topics such as these? Colored borders around windows after update Order Bars in ggplot2 bar graph scanf not taking in data Rollback Committed transaction in SQL Server Is it possible to rollback a committed transaction?  I have deleted all the data in a table. The delete statement is inside Begin Transaction and I have committed the transaction with a commit transaction command. Now is it possible to rollback and get the data back? What can I do if my USB flash drive is write-protected or read-only? Write access without read access Is it possible for a user to have a write access to a file and not be able to read it? How is it possible?   I tried the following commands:  debianbox@debian:~/posix/io$ touch filetest debianbox@debian:~/posix/io$ ls -l filetest -rw-r--r-- 1 debianbox debianbox 0 14 oct.  03:10 filetest  debianbox@debian:~/posix/io$ echo "Hello World" &gt; filetest debianbox@debian:~/posix/io$ cat filetest Hello World  debianbox@debian:~/posix/io$ chmod u-r filetest debianbox@debian:~/posix/io$ cat filetest cat: filetest: Permission forbidden  debianbox@debian:~/posix/io$   As you can see here, I have write but not read access on this file. How can this be possible? Is this considered as a bug? If not, in what situation would this be useful? Compactness of $\operatorname{Spec}(A)$ In an exercise in Atiyah-Macdonald it asks to prove that the prime spectrum $\operatorname{Spec}(A)$ of a commutative ring $A$ as a topological space $X$ (with the Zariski Topology) is compact.   Now because the basic open sets $X_f = \{\mathfrak{p} \in \operatorname{Spec} (A) : \{f\} \not\subseteq \mathfrak{p} \}$ form a basis for the Zariski Topology it suffices to consider the case when   $$X = \bigcup_{i \in I} X_{f_i}$$   where $I$ is some index set. Then taking the complement on both sides we get that   $$\emptyset = \bigcap_{i \in I} X_{f_i}^c$$  so there is no prime ideal $\mathfrak{p}$ of $A$ such that all the $f_i$'s are in $\mathfrak{p}$. Now from here I am able to show that the ideal generated by the $f_i$'s is the whole ring as follows. Since there is no prime ideal $\mathfrak{p}$ such that all the $f_i \in \mathfrak{p}$, it is clear that there is no $\mathfrak{p}$ such that $(f_i) \subseteq \mathfrak{p}$ for all $i \in I.$ Taking a sum over all the $i$ then gives $$\sum_{i \in I} (f_i) = (1).$$  Now here's the problem:          How do I show from here that there is an equation of the form $1 = \sum_{i \in J} f_ig_i,$ where $g_i \in A$ and $J$ some finite subset of $I$?       This part has been giving me a headache. I am not sure if the finiteness bit has to do with algebra, topology or the fact that we are dealing with prime ideals.  This is not a homework problem but rather for self-study.  $\textbf{Edit:}$ I have posted my answer below after the discussion with Dylan and Pierre. How to get the last value of an ArrayList Notes:  Answers will occasionally be moved to the  when there are too many of them. When using them to report bugs, keep an eye on the post.    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Answers that are considered annoying or obnoxious, or that cause breakage for users, will be deleted. This is . Why do we use the object instead of the subject pronoun in constructions like "stupid me"? I'm trying to find out how come we say lucky me and stupid us rather than lucky I and stupid we. My understanding is that this is not a recent invention, but a relic from the distant past where it was a dative construction. However, I don't see the reason for using any case other than the nominative in this construction to begin with.        He went surfing. → Lucky him went surfing.   I got pulled over. → Stupid me got pulled over.   They bought a house. → Thoughtless them bought a house.      This is rather intriguing. After thinking about it for a while, I couldn't come up with any other language, Germanic or otherwise, where simply introducing an adjective would dictate a change of grammatical case. This appears to be an English peculiarity. How far can it be traced back? Can anyone comment on its background? TL2018 (MacTeX) and TikZ calligraphy library