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Harvard 1869 entrance exam

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41–50 of 141 posts

Re: Harvard 1869 entrance exam

#41

In a way, this reinforces my hypothesis that Latin in traditional Western higher education was never quite so much about Latin itself as it was about gaining a deeper understanding and greater praxis of your native tongue by reading its source code.

A better way to do this is to learn French or German.

Re: Harvard 1869 entrance exam

#43
post #34

> What is the reason that when different powers of the same quantity are multiplied together their exponents are added? As a math professor, I think this is a great question. Students learn that math is about manipulating formulas and equations, or about excessive formalities. But being able to explain simple arithmetic facts in clear and plain English is often neglected, and is of the utmost value.

Who here can come up with the most concise explanation?

(a^n) * (a^m)

  =

  (a*a*a*a*a) * (a*a*a*a*a*a*a)

      ^              ^

   n times        m times
I don't see what is simpler than this, It comes from the basic definition of what a power is.

Re: Harvard 1869 entrance exam

#44
post #38

Earlier quoted context omitted.

Me, me! Let me try! X^N * X^M = N copies of X, multiplied by M copies of X = N+M copies of X multiplied together = X^(N+M)

That's not very rigorous, especially with fractional (or God forbid, irrational) exponents.

Yeah, and I'd be even more interested in a similarly intuitive explanation for the case of complex exponents ;-)

Re: Harvard 1869 entrance exam

#45

In a way, this reinforces my hypothesis that Latin in traditional Western higher education was never quite so much about Latin itself as it was about gaining a deeper understanding and greater praxis of your native tongue by reading its source code.

I don't agree with you. I did 2 years of latin and one year of greek. I don't feel like it really helped me. I actually found it pretty useless except the syntax/grammar part which can be good to understand new languages easier.

It's exactly like learning scheme. Seriously who fucking cares about scheme?

I see latin and greek in a Harvard test as a part of distinguishing highly educated kids from the others.

Re: Harvard 1869 entrance exam

#46
post #18
post #10

Earlier quoted context omitted.

Only because it demands regurgitation of specific facts you don't happen to have memorized.

There were only one or two questions in the mathematical part of the test that I would consider regurgitation -- the rest is stuff modern students should be able to do.

Are you only referring to the mathematics section? Most of the things in other sections were either recollection, or knowing Latin and Greek. In the mathematics section, most of it was simply performing computation, which is still just knowing a simple algorithm that hasn't been very relevant since calculators became commonplace.

Still, I don't think this is necessarily a bad thing, since it's fair to make a test that selects for students that have been well-educated in that time period. I just don't think it's necessarily more difficult than a modern equivalent test would be.

Re: Harvard 1869 entrance exam

#47
post #37

> What is the reason that when different powers of the same quantity are multiplied together their exponents are added? As a math professor, I think this is a great question. Students learn that math is about manipulating formulas and equations, or about excessive formalities. But being able to explain simple arithmetic facts in clear and plain English is often neglected, and is of the utmost value.

That question stood out to me as a particularly bad question. What is the answer supposed to be? I completely understand how multiplication of exponents works, but I have no idea how to describe the "reason." You can give a simple algebraic proof quite easily (especially if we're just dealing with integer exponents), but unless "reason" had a more specific mathematical meaning in that time, it seems like a very vague…

That's the point -- you don't understand it well enough to explain it.

Re: Harvard 1869 entrance exam

#48

In a way, this reinforces my hypothesis that Latin in traditional Western higher education was never quite so much about Latin itself as it was about gaining a deeper understanding and greater praxis of your native tongue by reading its source code.

Uhm, what? You couldn't be more off. Traditionally, any liberal arts education would include extensive familiarity with the classics. That means Plato & Dante, at the minimum, whether a BA or BS. You were expected to know Latin and Greek because you were expected to read Latin and Greek. If you were pursuing a BS, then perhaps you would read Euclid's "Elements" instead of Thucydides' "The History of the Peloponnesian…

I think it was also a bit of holdover from a previous time, when Latin and Greek texts represented the pinnacle of human knowledge.

Re: Harvard 1869 entrance exam

#49
post #38

Earlier quoted context omitted.

That's not very rigorous, especially with fractional (or God forbid, irrational) exponents.

Yeah, and I'd be even more interested in a similarly intuitive explanation for the case of complex exponents ;-)

That would be difficult, since a formal construction of even the real numbers is a somewhat advanced (3rd or 4th year college mathematics) topic. I forget the details, but I believe a^n for real a and complex n is formally defined using the exponential function (e^x).

Re: Harvard 1869 entrance exam

#50
post #34

> What is the reason that when different powers of the same quantity are multiplied together their exponents are added? As a math professor, I think this is a great question. Students learn that math is about manipulating formulas and equations, or about excessive formalities. But being able to explain simple arithmetic facts in clear and plain English is often neglected, and is of the utmost value.

Who here can come up with the most concise explanation?

The process of explanation by example (though I agree with others that it is really intrinsic in the meaning of a power - id like to hear impendia's explanation):

a^(n+1)=a^n * a

a^(n+2)=a^n * a^2

a^(n+3)=a^n * a^3

a^(n+0)=a^n * a^0

therefore

a^(n+m)=a^n * a^m

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