And here’s one of the key people basing their investment advice on Bart Simpson’s haircut. This is reportedly one of the more trusted, regulated, and credible companies that retail investors can turn to when looking into crypto. Take a moment to consider that Luno says it has more than 10 million customers in 40 countries. The Bart is a chart pattern where the crypto market rallies strongly before experiencing a brief period of sideways movement until it suddenly recaptures all of its previous gains,” was Kopelman’s imparted wisdom. “ marks a reappearance of the earlier bear market in a so-called Bart formulation. While the value of the most prominent digital coin fell below $30,000, Kopelman was quick to reassure punters that Bitcoin would soon be trading at $34,500 due to the Bart Formula (sic) reappearance. Sam Kopelman, UK manager at crypto brokerage Luno, rushed to deliver an expert commentary on Bitcoin’s fresh declines to clients on Wednesday. Correct answers win a sticker depicting a vomiting camel formation.
Readers are invited to guess which is which. One was intended as a serious research paper distributed to journalists the other, a broad parody. (At least) two potentially groundbreaking papers on digital assets were published this week. But with a set like $(0,10)$, which has no smallest or largest element, you have to look a bit more closely to see what’s happening.Truth is often stranger than fiction, but nowhere more so than in the cryptocurrency world, where the lines between the two are blurring at an alarming rate. Similarly, if it has a largest element, that will be the least upper bound. If a set has a smallest element, that element is always the greatest lower bound. Thus, no number smaller than $10$ is an upper bound for $(0,10)$, and $10$ is an upper bound, so it must be the least upper bound. ($\frac12(x+10)$ is just the average of $x$ and $10$, which is halfway between $x$ and $10$.) In each case $x$ cannot be an upper bound for $(0,10)$, because something in $(0,10)$ is bigger than $x$. On the other hand, if $x<10$, then either $x<9$, in which case $9$ is a member of $(0,10)$ bigger than $x$, or $$x<\frac12(x+10)<10\ ,$$ and $\frac12(x+10)$ is a member of the interval bigger than $x$. Every real number $u\ge 10$ is clearly an upper bound. The reasoning showing that $10$ is the least (smallest) upper bound of $(0,10)$ is similar. On the other hand, $0 This shows that $(0,10)$ has no smallest element. Let $s=\frac2$ is also in the interval and is smaller than $x$. If $b'\le 0$, then clearly $b'$ is not an upper bound for $S$. We show that $b'$ is not an upper bound of $S$. (i) It is obvious from the definition of $S$ that any $s\in S$ is $\le 10$. That means that for any $b'\lt b$, there is an $s$ in $S$ such that $s\gt b'$. We need to show that (i) $s\le b$ for every $s$ in $S$ and (ii) there is no "cheaper" upper bound than $b$. But the problem naturally breaks up into two parts, which in general should be handled separately. There is no general way to prove that $b$ is the least upper bound of a set $S$. In particular, $9.93$ is in $(0,10)$, so $9$ is not an upper bound for our set. The interval $(0,10)$ consists of all real numbers $x$ such that $0\lt x\lt 10$.
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