[math-fun] Re: Even Jeb Bush is smarter than the NYT
Mostly to Eugene Salamin:
I don't fault Jeb for not knowing the answer. Governors are not usually elected for their mathematical skills. ... You seem to have missed the essential point of my message. It was not an attack on Gov. Bush, or on an 18-year old student, or even public education. It was an attack on the New York Times.
Newspaper reporters aren't selected for their math skills, either. So why do you consider it OK for a governor not to know, but you consider it bad that a reporter didn't know? As usual, your own biases determined the conclusion that you drew from the story. Dean Hickerson dean@math.ucdavis.edu
--- Dean Hickerson <dean@math.ucdavis.edu> wrote:
Mostly to Eugene Salamin:
I don't fault Jeb for not knowing the answer. Governors are not usually elected for their mathematical skills. ... You seem to have missed the essential point of my message. It was not an attack on Gov. Bush, or on an 18-year old student, or even public education. It was an attack on the New York Times.
Newspaper reporters aren't selected for their math skills, either. So why do you consider it OK for a governor not to know, but you consider it bad that a reporter didn't know?
The news brief was presented as if it were factually correct, and with the intent of taking a jab at Jeb. A quality newspaper would have scientifically literate staff to validate submissions from dubious sources.
As usual, your own biases determined the conclusion that you drew from the story.
Dean Hickerson dean@math.ucdavis.edu
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Gene Salamin includes The news brief was presented as if it were factually correct, and with the intent of taking a jab at Jeb. A quality newspaper would have scientifically literate staff to validate submissions from dubious sources. Alas, there is no quality newspaper in the world.
Alas, there is no quality newspaper in the world.
Well, the St Petersburg Times got it right. Yesterday afternoon I got a call from one of their reporters and was surprised to find myself quoted in the morning paper with all my "Um..well..ah..gee". Fortunately I had Maple open in front of me when she called so could give an answer. I offered to give her 10 or more decimal places for each of the non-90 degree angles, but she was happy with just 2. Their story is accompanied by a nice drawing of the 3-4-5 and 30-60-90 triangles. http://www.sptimes.com/2004/07/07/State/Query_stumps_governor.shtml One of my colleagues said he would have just answered, "Of course, the answer is pi/2, arctan(3/4), pi/2-arctan(3/4)." Alas, I couldn't think that fast. --Edwin
Is there an easy way to tell when a + b*sqrt(k) is a perfect square in the extension field Q(sqrt(k))? For example, 3 + 2*sqrt(2) is the square of 1 + sqrt(2). I'm particularly interested in when a, b, k are integers, but a and b rationals would be OK, too. Thanks, Kristin Cordwell
On Wed, 7 Jul 2004, Bill & Rosemary Cordwell wrote:
Is there an easy way to tell when a + b*sqrt(k) is a perfect square in the extension field Q(sqrt(k))? For example, 3 + 2*sqrt(2) is the square of 1 + sqrt(2). I'm particularly interested in when a, b, k are integers, but a and b rationals would be OK, too.
Here's a quick way. If the square root is c + dx (x = sqrt(k), you evaluate the square roots of a+bx and a-bx as complex numbers, and up to sign they'll be c + dx and c-dx, so their semisum and semidifference will be 2c and 2dx in some order. You look at these to see which looks like an ordinary integer and which like an integer multiple of x. You can still do this if you're asking about a rational answer, provided you expect the denominators to be fairly small, and it's probably the quickest way, in practice. If you do need to do it exactly and theoretically, though, that's also fairly easy. You just solve the equations (c +- dx)^2 = a +- bx algebrically, which leads to a quartic equation for c, with rational coefficients. Make these integers, say A,B,C,D,E - then the numerator of c divides E and its denominator divides A, so it can be found. John Conway
participants (6)
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Bill & Rosemary Cordwell -
Dean Hickerson -
Edwin Clark -
Eugene Salamin -
John Conway -
John McCarthy