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Jan
1
comment V.I. Arnold says Russian students can't solve this problem, but American students can — why?
@djechlin, I was wearing a monocle but it fell off when I read your post, which is what put me in such a bad mood.
Dec
31
comment V.I. Arnold says Russian students can't solve this problem, but American students can — why?
(contd) book of word games that we are clued in to further examine the superfluous information about the triangle contained in the (reported) question. As to "applying the nearest formula available", you mischaracterise an entirely reasonable application of a basic fact about geometry (again, charitably granting that the student may assume competence and good faith on the part of the examiner). Your insult to the American education system - no matter what its faults might be - is unwarranted and makes you appear childish. It is cynicism without intelligence or insight.
Dec
31
comment V.I. Arnold says Russian students can't solve this problem, but American students can — why?
@djechlin - While it is vacuously true that "the area would also be -18 or any number", the answer of 30 square inches is in some sense distinguished from others in that it's the answer that you would arrive at if you ignored the information stated in the question that the triangle is right-angled. Ignoring this information is not unreasonable; if we start off with the assumption that the question is being posed in good faith by a competent examiner, then we might as well discard that information without examining it any further! It is only because the (meta-)question here is from a (contd)
Dec
20
awarded  Notable Question
Jul
18
comment Tangent to the curve
Saying "the above answer" or things like it is error-prone on stack exchange sites, because although the answer you're referring to is above this answer at the present time, answers can in principle change order. For example if someone else posted an answer and that answer got a higher score than yours, but the accepted answer didn't change, the new answer would be displayed above yours.
May
18
comment If you draw two cards, what is the probability that the second card is a queen?
"Your probability" here means the probability the asker was asking about, namely the probability that when two cards are drawn from the deck in sequence, the second is a queen. The point of this answer is to get the reader to understand that in the absence of any information about the first card drawn, the second card drawn has an equal chance of being any card in the deck, just as in a shuffled deck the second card in the deck has an equal chance of being any card.
Mar
23
accepted Is it true that “there is no such thing as the square root of minus one”?
Jan
29
awarded  Yearling
Aug
28
comment Why are all non-prime numbers divisible by a prime number?
@krowe - What do you mean by "NULL"? Perhaps you would find this Wikipedia article: en.wikipedia.org/wiki/Empty_product useful in helping you to understand why mathematicians find it natural to make the convention that the product of no numbers at all is 1.
Jul
2
awarded  Curious
Jun
25
asked Is it true that “there is no such thing as the square root of minus one”?
Apr
16
comment “Here's a cool problem”: a collection of short questions with clever solutions
@Steven, that is one way to do it. You might also consider this diagram: junk.orderofthehammer.com/j2014/floor-tiling-puzzle-hint.png
Mar
28
comment A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$
Yes, I understood. Are you assuming that $b$ is an integer in your first step (comparison with $g$)? I suppose (writing $b = c/d$) it is possible to find $\lfloor c\sqrt{2}\rfloor$ and then divide by $d$ and correct for any error introduced.
Mar
28
accepted A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$
Mar
27
answered Finding a coefficient of $x^6$ in the expansion $(x-1)^5 (x+1)^5$
Mar
26
comment A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$
This is some great advice, thank you. Having the method related to the Pell numbers is interesting. I will leave off accepting for a little longer but will probably accept this answer.
Mar
26
answered Is there a solution to this Seating Plan problem?
Mar
26
answered “Here's a cool problem”: a collection of short questions with clever solutions
Mar
26
comment A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$
Bill, I had not. I see that one might potentially approach the problem by applying one of the square root algorithms on that page (maybe the "Babylonian method") to calculate a series of approximate square roots of $2b^2$, which when it has converged on the square root to within an integer gives a candidate for $\lfloor b\sqrt{2}\rfloor$ which should not be far off the correct answer.
Mar
26
asked A general method to efficiently calculate the floor of an element of $\mathbb{Q}[\sqrt{2}]$