My take this is a flawed study also interpreted by someone who doesn't understand science.

The conclusion is suspect, IMHO. How the heck can you only study people without cardiovascular disease and legitimately conclude, anything about cardiovascular disease? Their actual conclusion is "In this large middle-aged cohort without cardiovascular disease, moderate to heavy coffee consumption was not associated with significant changes in arterial stiffness measured by AoD and ASI compared with individuals who drink 1 cups of coffee/day." They only mention arterial stiffness and say nothing about the many other intermediate or possible symptoms of heart or circulatory issues.

The original says nothing about 25 cups a day being safe. One group drank between 3 cups a day and 25 cups a day. They did specifically mention a desire to study the 3 to 25 cups a day group more thoroughly.

The scientific version is here.
https://heart.bmj.com/content/...

Errhhh, both ability to teach and subject matter are required. Merely taking a few years past the level they are teaching is completely insufficient. Someone who has significant experience beyond the level they are teaching has multiple advantages:
A. fewer mistakes. Keep in mind a single mistake can confuse 30 students very quickly.
B. More thorough understanding of the subject matter is excellent for the student who asks a more advanced question or about a specific topic. Plus keeping the top 10% of the class engaged is excellent for the class, because sometimes students learn better from each other. Ex. As a teacher I use the precise, correct terminology, etc., but students know other students might not communicate it exactly.
C. understanding what will be essential further down the road.
Ex. 1. Although calculators may do addition and subtraction of number fractions, being able to do them by hand is crucial to learning addition and subtraction of fractions with variables.
Ex. 2 Long division of decimals is hard to learn and hard to teach, but essential for students to learn division of polynomials. Yes advanced calculators and WolframAlpha will do all of this, but that starts to get off topic. Briefly- when a computer can do all the math a person knows, what will their career be?
D.I learned calculus far more thoroughly while tutoring it, than I did in the course. And my current students learn much easier from me because I know Calculus far better than when I started.

Hmmmm, yahoo has had this capability for maybe a decade. For $20 a year, I get up to 500 email addresses. http://antispam.yahoo.com/addressguardtour

Ahhhh, an excellent, creative and paradoxical problem. Thank you. First I dissect and disprove U17, then present the reconciliation of the paradox. TO U17: The trouble with math is someone can write an excellent reasoning with advanced techniques and notation. However, if there's a single critical mistake, the whole thing is spoiled. Who would want to drink filtered purified, cold refreshing water, after someone put in a drop of used motor oil? It's perfectly acceptable to multiply by a non zero variable or expression. This spoils U17's otherwise beautiful reasoning. Multiplying by an expression which IS zero, (x = 0) can create extra solutions. Check x =-1 on the step P3, (-1) ^3 ?=? (-1)^2 - (-1), -1 ?=? 1 +1 doesn't work even after we multiplied by x. The extra solution must have come from somewhere else. SPOILER ALERT- THE RESOLUTION OF THE PARADOX IS BELOW. Consider working on the question a while, per Lockhart, then after you've worked on it for a while. Try to rediscover the challenge & creativity of math, then come back here and read. The truth is x^3 = -1 contains the solutions. We need to delve into complex numbers though. This particular equation has 3 solutions in the set of complex numbers. http://www.wolframalpha.com/input/?i=x^3%20%3D%20-1&t=ff3tb01 x = -1 is real, and reals are a subset of complex numbers. http://www.wolframalpha.com/input/?i={x%20%3D%3D%20%28-1%29^%281%2F3%29}%2C%20{x%20%3D%3D%20-%28-1%29^%282%2F3%29}&t=ff3tb01 x = (1 +- i * root 3) / 2 are two solutions. (should See DeMoivre's theorem (convert to polar coordinates or see the graph), multiply it out by hand or check with WolframAlpha.com to confirm x^3 =-1 And the two complex solutions are our answer. http://www.wolframalpha.com/input/?i=x%3D%3D+1%2F2+(1-i+sqrt(3))%2C+x^2+-+x++%2B+1%3D The original reasoning has two oversights. A. x = -1 is an extraneous solution akin to y = 5, y^2 = 25, y = +- 5 and assuming y can be -5 or +5. Apparently when taking a cube root, we need to check for all solutions if they are extraneous (extra) solutions which rules out x = -1. Every Day Math B. Failure to consider complex solutions when taking a cube root. (Not commonly taught, but hidden right next to the complex properties which are commonly taught)

Ahhhh, an excellent, creative and paradoxical problem. Thank you. First I dissect and disprove U17, then present the reconciliation of the paradox. TO U17: The trouble with math is someone can write an excellent reasoning with advanced techniques and notation. However, if there's a single critical mistake, the whole thing is spoiled. Who would want to drink filtered purified, cold refreshing water, after someone put in a drop of used motor oil? It's perfectly acceptable to multiply by a non zero variable or expression. This spoils U17's otherwise beautiful reasoning. Multiplying by an expression which IS zero, (x = 0) can create extra solutions. Check x =-1 on the step P3, (-1) ^3 ?=? (-1)^2 - (-1), -1 ?=? 1 +1 doesn't work even after we multiplied by x. The extra solution must have come from somewhere else. SPOILER ALERT- THE RESOLUTION OF THE PARADOX IS BELOW. Consider working on the question a while, per Lockhart, then after you've worked on it for a while. Try to rediscover the challenge & creativity of math, then come back here to read. Hint: The truth is x^3 = -1 contains the solutions. We need to delve into complex numbers though. This particular equation has 3 solutions in the set of complex numbers. http://www.wolframalpha.com/input/?i=x^3%20%3D%20-1&t=ff3tb01 x = -1 is real, and reals are a subset of complex numbers. http://www.wolframalpha.com/input/?i={x%20%3D%3D%20%28-1%29^%281%2F3%29}%2C%20{x%20%3D%3D%20-%28-1%29^%282%2F3%29}&t=ff3tb01 x = (1 +- i * root 3) / 2 are two solutions. (should See DeMoivre's theorem (convert to polar coordinates or see the graph), multiply it out by hand or check with WolframAlpha.com to confirm x^3 =-1 And the two complex solutions are our answer. http://www.wolframalpha.com/input/?i=x%3D%3D+1%2F2+(1-i+sqrt(3))%2C+x^2+-+x++%2B+1%3D The original reasoning has two oversights. A. x = -1 is an extraneous solution akin to y = 5, y^2 = 25, y = +- 5 and assuming y can be -5 or +5. Apparently when taking a cube root, we need to check for all solutions if they are extraneous (extra) solutions which rules out x = -1. B. Failure to consider complex solutions when taking a cube root. (Not commonly taught, but hidden right next to the complex properties which are commonly taught) Aughh, this properly formatted text doesn't display properly in the preview screen.